Mechanics Notes

Kinematics ..................................................................................................................................................... 2 Introduction to Motion in One Dimension ............................................................................................... 2 Motion under Gravity ............................................................................................................................. 11 Motion in Two Dimensions ..................................................................................................................... 22 Circular Motion ....................................................................................................................................... 35 Relative Motion....................................................................................................................................... 41 Solved Examples on Mechanics:- ............................................................................................................ 49 Newton’s Laws of Motion:- ......................................................................................................................... 63 Classification of Forces:- ......................................................................................................................... 64 Newton’s First Law of Motion:- .............................................................................................................. 67 Newton’s Second Law of Motion:-.......................................................................................................... 69 Newton’s Third Law of Motion:- ............................................................................................................. 77 Friction:- .................................................................................................................................................. 90 Free body diagram .................................................................................................................................. 97 Centripetal Force .................................................................................................................................. 108 Conservation of Linear Momentum:- ....................................................................................................... 130 Work, Energy and Power:- ........................................................................................................................ 138 Work:- ................................................................................................................................................... 139 Energy (E):- ............................................................................................................................................ 147 Power:- .................................................................................................................................................. 174 Gravitation ................................................................................................................................................ 191 Gravitational Field and Intensity ........................................................................................................... 194 Satellites and Planetary Motion............................................................................................................ 206 Fluid Mechanics ........................................................................................................................................ 213 Thermal Physics:- ...................................................................................................................................... 218

Kinematics Introduction to Motion in One Dimension A body:A certain amount of matter limited in all directions and consequently having a finite size, shape and occupying some definite space is called a body. Particle:A particle is defined as a portion of matter infinitesimally small in size so that for the purpose of investigation, the distance between its different parts may be neglected. Thus, a particle has only a definite position, but no dimension. In the problems we are going to discuss, we will consider a body to be a particle for the sake of simplicity. Motion:The position of object can change on a straight line (like on x-axis with respect to origin) or on a plane with respect to some fixed point on frame. So we can define motion as follows:An object or a body is said to be in motion if its position continuously changes with time with reference to a fixed point (or fixed frame of reference). But note that, the moving object is either a particle, a point object (such as an electron) or an object that moves like a particle. A body is said to be moving like if every portion of it moves in the same direction and at the same rate. Motion in one dimension:When the position of object changes on a straight line i.e. motion of object along straight line is called motion in one dimension. To understand the essential concepts of one dimensional motion we have to go through some basic definitions. Frame of reference:One can see the platform from a running train, and it seems that all the objects placed on platform are continuously changing their position. But one, who is on platform, concludes that the objects on the platform are at rest. It means if we will take the trains are reference frame the objects are not stationary and taking reference frame as platform the objects are stationary. So the study of motion is a combined property of the object under study and the observer. Hence there is a need to define a frame of reference under which we have to study the motion of an object.

Definition A frame of reference is a set of coordinate axes which is fixed with respect to a space point (a body or an object can also be treated as a point mass therefore it can become a site for fixing a reference frame), which we have arbitrarily chosen as per our observer's requirement. The essential requirement for a frame of reference is that, it should be rigid. Position of an object The position of an object is defined with respect to some frame of reference. As a convention, we define position of a point (essentially we treat body as a point mass) with the help of three co-ordinates X, Y and Z. Hence X, Y, Z is a set of coordinate axes

representing a 3-dimensional space and each point in this space can be uniquely defined with the help of a set of X, Y and Z coordinate, all three axes being mutually perpendicular to each other. The line drawn from origin to the point represents the position vector of that point. Position vector It describes the instantaneous position of a particle with respect to the chosen frame of reference. It is a vector joining the origin to the particle. If at any time, (x, y, z) be the Cartesian coordinates of the particle then its position vector is given by vector

= xi + yj

+ zk. In one-dimensional motion: vector

= xi,

y = z = 0 (along x-axis)

In two-dimensional motion: vector

= xi + yj

(in x-y plane z = 0)

In the figure above, the position of a point P is specified and vector OP is called the position vector.

Displacement Consider a case in which the position of an object changes with time. Suppose at certain instant 't' the position of an object is x1 along the x axis and some other instant 'T' the position is x1 then the displacement Δx is defined as, Δx = x2 - x1

It can be seen in the figure above where x1 and x2 are instantaneous position of the object at that time. Now consider the motion of a point A with respect to a reference point O. The motion of point A makes its radius vector vary in the general case both in magnitude and in direction as shown in figure above. Suppose the point A travels from point 1 to point 2 in the time interval Δt. It is seen from the figure that the displacement vector Δ vector Δ

=

in time Δt: 2

-

t

of the point A represents the increment of

Difference between distance and displacement To understand the difference between distance and displacement, we study the motion of vertical throw of a ball

with

respect to point O, as shown in the left figure, to height h. After some time it will come again to the same point O. The displacement of ball is zero but there is some distance traversed by the ball. It's because distance is a scalar quantity but displacement is a vector quantity. Uniform and Non Uniform Motion

Speed is the rate of change of distance without regard to directions. Velocity is the rate at which the position vector of a particle changes with time. Velocity is a vector quantity whereas speed is scalar quantity but both are measured in the same unit m/sec. The motion of an object may be uniform or non-uniform depending upon its speed. In case of uniform motion the speed is constant, whereas in the non-uniform motion, the speed is variable. In uniform motion in one dimension the velocity (v) is mathematically defined as v = (x2 - x1)/(T-t)

...... (1)

Where x1 and x2 are instantaneous displacement as shown in figure above at time 't' and 'T' respectively. Graphical representation of the uniform motion Form the equation (1) we have the following equation

x2 = x1 + v(T - t) where v is constant. Take t = 0, the equation becomes x2 = x1 + vT, from this equation it follows that the graph of position of object 'x2' against 'T' is a straight line, cutting off x1 on the position axis where x1 is the distance of the particle from the origin at time t = 0. v = slope of the graph which is constant



If the graph is not a straight line, it will represent non-uniform motion.



Motion of a body cannot be correctly identified unless one knows the position of body as specified by a fixed frame of reference.

Velocity Vector in Non Uniform Motion In any non-uniform motion, we can define an average velocity over a time interval. Average velocity

is the ratio of the displacement Δx (that occurs during a particle time interval Δt)

to that interval of time i.e.

Now refer to the example, related to figure 2.3, the ratio of Δ direction of the vector <

> coincides with that of Δ

/Δt is called the average velocity <

> during the time interval Δt. The

. Average velocity is also a vector quantity.

Note: The ratio of total distance traveled and time taken during the motion is called average speed. Average speed is a scalar quantity. If

at

any

time

t1 position

vector

of

the

particle

is

1 and

at

time

2 position

vector

is

2 then

for

this

interval Instantaneous velocity Instantaneous velocity is defined as the rate of change of displacement.

Velocity The velocity at any instant is obtained from the average velocity shrinking the time interval closer to zero. As Δt tends to zero, the average velocity approaches a limiting value, which is the velocity at that instant, called instantaneous velocity, which is a vector quantity, mathematically we can define it as

The magnitude v of the instantaneous velocity is called the speed and is simple the absolute value of In the example related with figure given below, the instantaneous velocity is

Hence instantaneous velocity is the rate at which a particle's position is changing with respect to time at a given instant. The velocity of a particle at any instant is the slope (tangent) of its position curve at the point representing that instant of time, as shown in figure above. Speed Speed is defined as rate of change of distance with time. In any interval of time, average speed is defined as = (total distance)/(total time taken) = Δs/Δt. As Δs > |Δ Think

:

(i)Can

a

body

have

a

constant

|, hence > speed

(ii)Can a body have a constant velocity and still have a varying speed?

and

still

have

a

varying

velocity?



The dispalcement remains unaffected due to shifting of origin from one point to the other.



The displacement can have positive, negative or zero value.



The dispalcement is never greater than the actual distance travelled.



The displacement has unit of length.



Velocity can be considered to be a combination of speed and direction.



A change in either speed or direction of motion results in a change in velocity.



It is not possible for a particle to possess zero speed with a non-zero velocity.



A particle which completes one revolution, along a circular path, with uniform speed is said to possesss zero velocity and nonzero speed.



In case a body moves with uniform velocity, along a straight line, its average speed is equal to its instantaneous speed.

Acceleration Acceleration is the rate of change of velocity with time. The concept of acceleration is understood in non-uniform motion. It is a vector quantity. Average acceleration is the change in velocity per unit time over an interval of time.

Instantaneous acceleration is defined as

Acceleration vector in non uniform motion

Suppose that at the instant t1 a particle as in figure above, has velocity acceleration

during the motion is defined as

and at t2, velocity is

. The average

Variable Acceleration The acceleration at any instant is obtained from the average acceleration by shrinking the time interval closer zero. As Δt tends to zero average acceleration approaching a limiting value, which is the acceleration at that instant called instantaneous acceleration which is vector quantity.

i.e. the instantaneous acceleration is the derivative of velocity. Hence instantaneous acceleration of a particle at any instant is the rate at which its velocity is changing at that instant. Instantaneous acceleration at any point is the slope of the curve v (t) at that point as shown in figure above. Equations of motion The relationship among different parameter like displacement velocity, acceleration can be derived using the concept of average acceleration and concept of average acceleration and instantaneous acceleration. When acceleration is constant, a distinction between average acceleration and instantaneous acceleration loses its meaning, so we can write

where is the velocity at t = 0 and

is the velocity at some time t.

So,

.................. (2) This is the first useful equation of motion. Similarly for displacement .................. (3) in which of particle is

is the position of the particle at t0 and

is the average velocity between t0 and later time t. If at t0 and t the velocity

This is the second important equation of motion. Now from equation (2), square both side of this equation we get

This is another important equation of motion. Note:



The equation of motion derived above are possible only in uniformly accelerated motion i.e. the motion in which the acceleration is constant.

Graphical Representation and Equations of Motion While studying motion of bodies we have to keep two things in mind. We have to study various characteristics associated with moving bodies while we are to explore and study in detail, about the cause of producing motion. Kinematics is the branch of physics which deals only with the description of motion of bodies. Mainly there are two type of graphs are belonging to the kinematics. One is position-time graph and the other is velocity-time graph. These graphs will have validity only if motion under study is along a straight line. Then, displacement, velocity and acceleration vectors are collinear and can be treated as algebraic quantities. Let x-axis be the path of motion. Then, x-coordinate represents the magnitude of position vector. Position - Time Graph:-

If we plot time t along the x-axis and the corresponding position (say x) from the origin O on the y-axis, we get a graph which is called the position-time graph. This graph is very convenient to analyse different aspects of motion of a particle. Let us consider the following cases.

(i) In this case, position (x) remains constant but time changes. This indicates that the particle is stationary in the given reference frame. Hence, the straight line nature of position-time graph parallel to the time axis represents the state of rest. Note that its slope (tan θ) is zero. .

(ii) When the x-t graph is a straight line inclined at some angle (θ≠) with the time axis, the particle traverses equal displacement Δx in equal intervals of time Δt. The motion of the particle is said to be uniform rectilinear motion. The slope of the line measured by Δx/Δt = tθ represents the uniform velocity of the particle.

(iii)

When the x-t graph is a curve, motion is not uniform. It either speeds up or slows down depending upon whether the slope (tan

θ successively increases or decreases with time. As shown in the figure the motion speeds up from t = 0 to t=t1 (since the slope tan θ increases). From t=t1 to t=t2, AB represents a straight line indicating uniform motion. From t=t2 to t=t3, the motion slows down and for t>t3 the particle remains at rest in the reference frame. The Velocity - Time Graph:The velocity-time graph gives three types of information. (i) The instantaneous velocity. (ii) The slope of the tangent to the curve at any point gives instantaneous acceleration. a = dv/dt = tan θ (iii)

The area under the curve gives total displacement of the particle.

Now, let us consider the uniform acceleration. The velocity-time graph will be a straight line.

The acceleration of the object is the slope of the line CD. a = tan θ = BC/BD = (v-u)/t v = u +at

…... (1)

The total displacement of the object is area OABCD s = Area OABCD = Area of rectangle OABD + Area of triangle BCD s = ut + (1/2)at2

…...

(2)

Again, s = Area OABCD = 1/2(AC + OD) x OA = 1/2(v + u)xt The acceleration-time graph:Acceleration time curves give information about the variation of acceleration with time. Area under the acceleration time curve gives the change in velocity of the particle in the given time interval.

Refer this simulation that shows a car traveling along a graph:The purpose is to show that, the car can traverse the graph, then it's first derivative is a continuous function.



The displacement can have positive, negative or zero value.



The displacement is never greater than the actual distance travelled.



A change in either speed or direction of motion results in a change in velocity.



It is not possible for a particle to possess zero speed with a non-zero velocity.



The velocity-time graph for a uniformly accelerated body is a straight line.

Motion under Gravity Whenever a body is released from a height, it travels vertically downward towards the surface of earth. This is due to the force of gravitational attraction exerted on body by the earth. The acceleration produced by this force is called acceleration due to gravity and is denoted by „g‟. Value of „g‟ on the surface of earth is taken to the 9.8 m/s2 and it is same for all the bodies. It means all bodies (whether an iron ball or a piece of paper), when dropped (u=0) from same height should fall with same rapidity and should take same time to reach the earth. Our daily observation is contrary to this concept. We find that iron ball falls more rapidly than piece of paper. This is due to the presence of air which offers different resistance to them. In the absence of air both would have taken same time to reach the surface of earth. When a body is dropped from some height (earth's radius = 6400 km), it falls freely under gravity with constant acceleration g (= 9.8 m/s2) provided the air resistance is negligible small. The same set of three equations of kinematics (where the acceleration

remains constant) are used in solving such motion. Here, we replace

by

conveniently. When the y-axis is chosen positive along vertically downward direction, we take

and choose the direction of y-axis as positive and use the equation

as

v = u + gt, v2 = u2 + 2gh, and h = ut + 1/2gt2 where h is the displacement of the body and u is initial velocity of projection in the vertically downward direction. However, if an object is projected vertically upward with initial velocity u, we can take y-axis positive in the vertically upward direction and the set of equations reduces to v = u - gt, v2 = u2 - 2gh, and h = ut - 1/2gt2 In order to avoid confusion in selecting

as positive or negative, it is advisable to take the y-axis as positive along vertically upward

direction and point of projection as the origin. We can now write the set of three equations in the vector form:

,

and

where h is the displacement of the body. Illustration: The motion of a particle is described by the equation u = at. The distance travelled by the particle in the first 4 second. Solution: Because for the motion u = at. So acceleration is uniform which is equal to a. Therefore, Distance traveled = 1/2[(a)(4)2] = 8a Illustration: A body moving with a constant retardation in straight line travels 5.7 m and 3.9 m in the 6th and 9th second, respectively. When will the body come momentarily to rest? Solution: A body moving with initial velocity u and acceleration a, traverses distance Sn in nthsecond of its motion. Sn = u + (1/2)(2n - 1)a

=> 5.7 = u + (1/2)(2 x 6 - 1) a

or 5.7 = u + (11/2) a and 3.9 = u + (1/2)(2 x 9 - 1)a

or,

3.9 = u + (17/2) a

Solving eqns. (1) and (2) we get, u = 9 m/s and a = -0.6 m/s2. If the body stops moving after t seconds, then from the relation v=u+at Thus, 0 = 9 + (-0.6)t

or, t = (9/0.6)s = 15s

Illustration (JEE Advanced): A stone, thrown up is caught by the thrower after 6s. How high did it go and where was it 4 s after start? g = 9.8 m/s 2 Solution: Time to go up and come back = 6s Thus, time to reach the highest point = (6/2) s = 3s From point of projection to the highest point we have u = ?, v = 0, a = -9.8 m/s2, t = 3s Using the relation, v = u + at 0= u – 9.8 3 Thus, u = 29.4 m/s2 Maximum height, H = u2/2g = [(29.4)2/2(9.8)] m =44.1 m Let h = height of stone at 4s.

Using the relation, S = ut + ½ at2 So, h = [(29.4)(4)-1/2 (9.8) (4)2]m = [117.6-78.4] m = 39.2 m From the above observation we conclude that, the height would be 39.2 m.



Acceleration of all these bodies is constant.



Acceleration is always directed downward.



All bodies, when dropped (u=0) from same height should fall with same rapidity and should take same time to reach the earth.

Motion in a Straight Line with Acceleration

Velocity of a body is defined as the time rate of displacement, where as acceleration is defined as the time rate of change of velocity. Acceleration is a vector quantity. The motion may be uniformly accelerated motion or it may be non-uniformly accelerated, depending on how the velocity changes with time. Uniform acceleration The acceleration of a body is said to be uniform if its velocity changes by equal amounts in equal intervals. Non-uniform acceleration The acceleration of a body is said to be non-uniform if its velocity changes by unequal amounts in equal intervals of time. Average velocity

Average acceleration

Illustration: A particle moves with a velocity v(t) = (1/2)kt2 along a straight line. Find the average speed of the particle in time T. Solution:

Illustration: A particle having initial velocity is moving with a constant acceleration 'a' for a time t. (a)Find the displacement of the particle in the last 1 second.

(b)Evaluate it for u = 2 m/s, a = 1 m/s2 and t = 5 sec. Solution: (a) The displacement of a particle at time t is given s = ut + 1/2at2 At time (t - 1), the displacement of a particle is given by S' = u (t-1) + 1/2a(t-1)2 So, Displacement in the last 1 second is, St = S - S' = ut + 1/2 at2 – [u(t-1)+1/2 a(t-1)2 ] = ut + 1/2at2 - ut + u - 1/2a(t - 1)2 = 1/2at2 + u - 1/2 a (t+1-2t) = 1/2at2 + u - 1/2at2 - a/2 + at S = u + a/2(2t - 1) (b) Putting the values of u = 2 m/s, a = 1 m/s2 and t = 5 sec, we get S = 2 + 1/2(2 x 5 - 1) = 2 + 1/2 x 9 = 2 + 4.5 = 6.5 m Illustration: Position of a particle moving along x-axis is given by x = 3t - 4t2 + t3, where x is in meters and t in seconds. (a)Find the position of the particle at t = 2 s. (b)Find the displacement of the particle in the time interval from t = 0 to t = 4 s. (c)Find the average velocity of the particle in the time interval from t = 2s to t=4s. (d)Find the velocity of the particle at t = 2 s. Solution: (a) x(t) = 3t - 4t2 + t3 => x(2) = 3 x 2 - 4 x (2)2 + (2)3 = 6 - 4 x 4 + 8 = -2m. (b) x(o) = 0 X(4) = 3 x 4 - 4 x (4)2 + (4)3 = 12 m. Displacement = x(4) - x(0) = 12 m. (c) < v > = X(4)X(2)/(4-2) = (12-(-2))/2 m/s = 7 m/s (d) dx/dt = 3 - 8t + 3t2 v(2) (dx/dt)2 = 3 - 8 x 2 + 3 x (2)2 = -1m/s Illustration: Two trains take 3 sec to pass one another when going in the opposite direction but only 2.5 sec if the speed of the one is increased by 50%. The time one would take to pass the other when going in the same direction at their original speed is (a) 10 sec

(b) 12 sec

(c) 15 sec

(d) 18 sec

Solution: Using the equation,

t = d/vr We have, 3 = d/v1+v2 2.5 = d/1.5v1+v2 Solving we get, v1 = 2d/15 and v2 = d/5 When they are going in same direction, vr = v2 – v1 = d/15 Thus, t = d/vr = d/(d/15) = 15 s From the above observation we conclude that, option (c) is correct. Analysis of Uniformly Accelerated Motion

Case-I: For uniformly accelerated motion with initial velocity u and initial position x0. Velocity time graph

In every case tanθ = a0 Position time graph

Initial position x of the body in every case is x0 (> 0) Case II:

For uniformly retarded motion with initial velocity u and initial position x0. Velocity time graph

In every case tanθ = -a0 Position time graph

Initial position x of the body in every case is x0 (> 0) Illustration:

A particle is moving rectilinearly with a time varying acceleration a = 4 - 2t, where a is in m/s2 and t is in sec. If the particle is starting its motion with a velocity of -3 m/s from x = 0. Draw a-t, v-t and x-t curve for the particle. Solution: a = 4-2t

v = 4t-t2-3

x = 2t2 – t3/3 – 3t Acceleration Acceleration is the rate of change of velocity with time. The concept of acceleration is understood in non-uniform motion. It is a vector quantity.

Average acceleration is the change in velocity per unit time over an interval of time.

Instantaneous acceleration is defined as

Acceleration vector in non uniform motion

Suppose that at the instant t1 a particle as in figure above, has velocity acceleration

during the motion is defined as

and at t2, velocity is

. The average

Variable Acceleration

The acceleration at any instant is obtained from the average acceleration by shrinking the time interval closer zero. As Δt tends to zero average acceleration approaching a limiting value, which is the acceleration at that instant called instantaneous acceleration which is vector quantity.

i.e. the instantaneous acceleration is the derivative of velocity. Hence instantaneous acceleration of a particle at any instant is the rate at which its velocity is changing at that instant. Instantaneous acceleration at any point is the slope of the curve v (t) at that point as shown in figure above. Equations of motion The relationship among different parameter like displacement velocity, acceleration can be derived using the concept of average acceleration and concept of average acceleration and instantaneous acceleration. When acceleration is constant, a distinction between average acceleration and instantaneous acceleration loses its meaning, so we can write

where

is the velocity at t = 0 and

is the velocity at some time t

Now,

Hence, …... (2) This is the first useful equation of motion. Similarly for displacement …... (3) in which

is the position of the particle at t0 and

velocity of particle is

is the average velocity between t0 and later time t. If at t0 and t the

…... (4) From equation (3) and (4), we get,

…... (5) This is the second important equation of motion. Now from equation (2), square both side of this equation we get

[Using equation (4)] Using equation (3), we get, …... (6) This is another important equation of motion. Caution:The equation of motion derived above are possible only in uniformly accelerated motion i.e. the motion in which the acceleration is constant. Refer this simulation for motion in a straight line:Illustration: The nucleus of helium atom (alpha-particle) travels inside a straight hollow tube of length 2.0 meters long which forms part of a particle accelerator. (a) If one assumes uniform acceleration, how long is the particle in the tube if it enters at a speed of 1000 meter/sec and leaves at 9000 meter/sec? (b) What is its acceleration during this interval? Solution: (a) We choose x-axis parallel to the tube, its positive direction being that in which the particle is moving and its origin at the tube entrance. We are given x and vx and we seek t. The acceleration ax is not involved. Hence we use equation 3, x = x0 + t. We get x = v0 + ½ (vx0) + vx) t, with x0 = 0 or t = 2x/(vx0+vx), t = ((2)(2.0 meters))/((1000+9000)meters/sec) = 4.0/10-4 sec

Ans.

(b) The acceleration follows from equation 2, vx = vx0 + axt => ax = (v0-vx0)/t = ((9000-1000)meters/sec)/(4.0×10(-4) sec) = 2.0 × 107 meter/sec2 Ans. Pause: The above equations of motion are, however, universal and can be derived by using differential calculus as given below:

Or, Let at t = 0, then, Or, Further we know that,

or Integrating,

Or,

At, t = 0, x = x0 then c' = x0 Hence,

Thus, we have derived the same equation of motion using calculus. To understand the use of calculus in solving the kinematics problems we can look into the following illustrations. Illustration: The displacement x of a particle moving in one dimension, under the action of a constant force is related to the time t by the equation t = √x + 3 where x is in meter and t is in seconds. Find the displacement of the particle when its velocity is zero. Solution: Here t = √x + 3 => √x = t - 3 Squaring both sides, we get x = t2 - 6t + 9, As we know velocity, v = dx/dt Hence we get v = dx/dt = 2t - 6 Put v = 0, we get, 2t - 6 = 0 So, t = 3s

When t = 3s, x = t2 - 6t + 9 = 9 - 6(3) + 9 = 0 Hence the displacement of the particle is zero when its velocity is zero. Illustration: A particle starts from a point whose initial velocity is v1 and it reaches with final velocity v2, at point B which is at a distance 'd' from point A. The path is straight line. If acceleration is proportional to velocity, find the time taken by particle from A to B. Solution: Here acceleration a is proportional to velocity v. Hence a α v => a = kv, where k is constant => dv/dt = kv ............... (1) => (dv/ds)(ds/dt) = kv => (dv/ds) v = kv

From equation (1), dv/v = kdt

or, Or, ln (v2/v1) = kt Or, t = ln (v2/v1) /k = [d ln (v2/v1)/(v2-v1)]



The dispalcement remains unaffected due to shifting of origin from one point to the other.



The displacement can have positive, negative or zero value.



The dispalcement is never greater than the actual distance travelled.



The displacement has unit of length.



Velocity can be considered to be a combination of speed and direction.



A change in either speed or direction of motion results in a change in velocity.



It is not possible for a particle to possess zero speed with a non-zero velocity.



A particle which completes one revolution, along a circular path, with uniform speed is said to possesss zero velocity and non-zero speed.



In case a body moves with uniform velocity, along a straight line, its average speed is equal to its instantaneous speed.

Motion in Two Dimensions In this part, we discuss motions in two dimensions like the motion of a particle moving on a circular path or on a parabolic path. Consider a particle moving on x-y plane along a curved pat at time t, as shown in figure given below. Its displacement from origin is measured by vector

its velocity is indicated by vector

(tangent to the path of the particle) and acceleration

as shown in

the figure.

The vectors

,

and

are inter related and can be expressed in terms of their components, using unit vector notation as,

From the above relations we can say that two-dimensional motion is nothing but superimpose of two one dimension motions. This is for the reason that in any two-dimensional motion, all the parameters can be resolved in two mutually perpendicular directions. Frame of reference:-

A frame of reference is a set of coordinate axes which is fixed with respect to a space point (a body or n object can also be treated as a point mass and therefore it can become a site for fixing a reference frame), which we have arbitrarily chosen as per our observer requirement. It is constituted of two components: (i) Body on which observer is apparently sitting to observe the motion, i.e. the origin. (ii) A coordinate system fixed on this body so that whatever is being observed can be measured or mathematically determined with the help of the coordinates of space points, which are defined by a unique set of (x, y, z). To fix a frame of

reference one has to fix the origin to a chosen space point and then fix the co-ordinate system on it. Once the origin and coordinate system are fixed, then we have done the exercise of fixing the frame of reference and we can proceed to solve problems and study various parameters i.e. position, velocity, accelerations as functions o time. Here, an important thing to notice is that earth becomes a natural choice for the frame of reference as in most of the cases it is very easy to visualize the motion with respect to it. As it is a routine in our daily life we innocently (even in early childhood) refer and perceive all motion phenomenon with respect to earth only. While analyzing any motion we will take the help of the coordinate system. What we shall do is that we shall keep ourselves free to take it there. This is called a frame of reference because we shall refer to all positions in this coordinate system fixed on that body. Here,

remembers

that

we

fix

all

three

axis

of

coordinate

system

i.e.

X-axis. Y-axis and Z-axis as well as the origin on the body and the coordinates of origin are (0, 0, 0). It is the origin where we mostly assume that the observer is sitting to observe the motion in the concerned reference frame. Choice of a frame of reference:Let us come back to the concept of motion. Do you believe that all what you see moving is in motion and what you see not moving is at rest! Like vehicles on road! You should be warned that motion or observation of motion is really not that simple! Let us take an example: The moon in the night sky! Some time at night you see that moon is travelling across the clouds towards east or west. And in some other nights it doesn't move at all! We shall all agree that it cannot be. It cannot sometimes move faster and sometime slower and sometimes become stationary. Then, what is the phenomenon there? Motion in Two Dimensions:We start with the first point that whenever we speak of any motion it should be with respect to some fixed frame of reference. So, when we see the moon moving. It is not with certainly one can point out that in the night when it looked stationary there were no clouds. So it is an effect of the clouds. The fact is that if we have clouds in some nights moving towards west or east then we see moon travelling towards east or west in the sky respectively. How to explain it with the help of reference frames? It is like this. When we see two objects in the sky one moon and other a cloud, then due to relatively bigger size of clouds (as seen by us) one gets his eyes fixed on clouds and therefore frame of reference gets fixed on the clouds (Figure shown above). It is equivalent to say that we are sitting on the clouds itself, which looks stationary to us and the feeling comes, as if moon is travelling across this stationary cloud in the opposite direction to the clods motion (see figure given above). Similarly, we see things going backwards when we travel on a road. Now let us deliberately fix our eyes on the moon and see. What we are doing now is that we are fixing frame of reference on the moon. It is equivalent to say that we are sitting on the moon to observe the cloud's motion and now we can see clouds drifting away (which is the fact). See Figure shown above.

An important observation to remember is that once we choose an object as a frame of reference, we can't see it moving with respect to itself. Naturally, in most of the cases of observations the Earth becomes a natural choice for the reference frame. Also, based on the choice of reference frames one can have many apparent motion of an object under consideration. Here one important point to be noted is that in all previous discussion we have assumed that the reference frames are not accelerating, which means it's acceleration is zero. These are called inertial or un accelerated frames of reference. Now obviously, we are left with one more choice of class of reference frames that is the accelerated ones. These are called non-inertial or accelerated frame of reference. But in all the forthcoming exercises in this chapter we shall deal with and look only for inertial frames of reference.

Relative motion:Consider a observer S is fixed to the earth. The other observer S' is moving on the earth, say a passenger sitting on a bus. Hence other frame of reference S' is fixed with bus. Each observer is studying the motion of a cyclist in his or her own frame of reference. Each observer will record a displacement, velocity and acceleration for the cyclist measured relative to his reference frame. Now the question is "How will we compare their results"?. To do this we need the concept of relative motion. Let us do it. In figure shown above at the top, the reference frame S, represented by the x and y-axes, can be thought of as fixed to the earth. The other reference frame S', represented by x'- and y'-axes, which moves along the x-axis with a constant velocity u, measured in the s-system, fixed with the bus. Initially, a particle is at a point called A in the S-frame and called A' in the S'-frame. At a time t later, the reference frame S' has moved a distance ut to the right and particle has moved to B. The displacement of the particle from its position in the S-frame is the vector

from A to B. The displacement of the particle from its initial position in the S'-frame is the vector r' from A' to B. These are

different vectors because the reference point A' of the moving frame has been displaced a distance ut along the x-axis during the motion. From the second figure shown above we see that =

is the vector sum of

+

Differentiating equation leads to

But,

is the instantaneous velocity of the particle measured in the s-frame. and

is the instantaneous velocity of the same particle measured in the S' frame, so that =

'+

.................. [A]

and

:

Hence the velocity of the particle relative to the S-frame, the velocity

is the vector sum of the velocity of the particle to the S' frame,

, and

of the S' frame relative to the S-frame.

Once again differentiating equation [A] We get,

Hence the acceleration of particle is the same in all reference frame moving relative to one another with constant velocity



The relative velocity between S and S' must be a constant.



At any instant, the velocity the velocity of a particle as measured by S is equal to the velocity of the particle as measured by S'.



Law of transformation of velocity permits us to transform a measurement of velocity made by an observer in one frame reference to another frame of reference.



The reference frames which are not accelerating are called inertial frame of reference. Problem 1:A particle starts with initial velocity (3i + 4j) m/s and with constant acceleration (8i + 6j) m/s2. Find the final velocity and final displacement of the particle after time t = 4 sec. Solution:As the velocity and acceleration vectors are not parallel, the velocity and acceleration cannot be treated as a scalar. Now divide the whole phenomenon in two one-dimension motions. In x-direction, Initial velocity = vx0 = 3m/s, ax = 8 m/s2

Hence velocity after 4 second, vx = 3 + 8 × 4 = 35 m/s Displacement after 4 second sx = x = 3 × 4 + ½8 (4)2 = 76 m In y-direction, Initial velocity vy0 = 4m/s, ay = 6 m/s2 Velocity after 4 second vy = 4 + 6 × 4 = 28 m/s Displacement after 4 second sy = 4 × 4 + ½ 6 (4)2 = 64 m

It means final velocity v = vx i + vy j = (35 i + 28 j) m/s and displacement = s sx i + sy j = (76 i + 64 j) m

Problem 2:-

There are two boats B1 and B2. They are continuously moving towards a bank. Water flow velocity is 10m/sec away from the bank. When at one instant of time, observation was taken we found that B1 and b2 are at a position (figure is gen below) which is 100 m from the bank and after 1 sec, b1 is 50 away and B2 is 25m away from the bank. Find out all possible (motion) velocities observed from different reference frames choices. Solution:Here we have four (at least) immediate choices of reference frames with respect to which we can observe velocities. (i)

Bank (earth)

(ii)

B1

(iii)

B2

(iv) Water

(i) Bank: (See figure shown above) First we shall fix the frame of reference on the bank. Since B1 travels 50 m towards the bank in 1 sec., its velocity towards the Bank (that is in +y direction) is 50m/sec. Similarly B2 travels 75 m/s towards the bank Thus, its velocity with respect to bank or as observed by an observer sitting at rest on the bank is 75m/sec (+y direction) Since water moves away from bank at speed of 10m/sec. Therefore, water's velocity w.r.t. the bank is 10m/sec (-y direction). (ii) Boat B1: Now, let us fix frame of reference on Boat B1. See the right side figure:

(a) Since B1 cannot move w.r.t. itself (only in case of deformation, a part of this boat can be seen moving w.r.t. other parts) Thus, VB1 = 0 (relative velocity w.r.t. to itself) (b) at t = 0, B1 and B2 were in the same line and had no at t = 1 B2 leads B1 by 25 m that is B2 has moved 25m in 1 sec w.r.t. B1 Thus, VB2 = 25 m/sec (+ direction) (c) Bank, at t = 0 was away by 100 m. and at t = 1 it has come closer by 50 m towards B1 Therefore, bank is approaching B1 by a speed of Vbank = 50m/sec (-y direction) (d) To visualize the motion of water w.r.t. B1 we can imagine a wooden piece of negligible mass, floating on the water surface. Its velocity will be equal to that of water. At t = 0 the piece was in the same line joining B1 and B2 at a distance of 100 m from the bank. At t = 1 sec it will move away by 10m from the bank that is its distance from b1 will be 60m (see figure shown above) Following a similar approach find out various relative velocities as seen from frames of reference, which are fixed, on boat B2 (figure shown below) and when the reference frame is fixed on water. (Figure shown next after this given below)

Problem 3:A windblown rain is filling, 20 degrees from the vertical, at 5 m/s. Passengers in a car see the rain falling vertically. What is the magnitude and direction of car's velocity? Solution:-

Here we have two frames S and S'. Frame S is attached with ground while S' attached with car. To determine the magnitude and directions of car's velocity, we use the concept of relative motion discussed earlier. Let vRG = velocity of rain with respect to ground vRC= velocity of rain with respect to car vc= velocity of car with respect to ground From the equation for relative velocity in two frames, we get

From the vector diagram we see that car should be travelling along the inclined direction of rain with velocity vC = 5 sin 20o m/s

Motion of Projectile

Now we discuss some example of curved motion or two dimensional motion of constant acceleration such as the motion of constant acceleration such as the motion of a particle projected at certain angle with the horizontal in vertical x-y plane (this type of motion is called projectile motion). Air resistance to the motion of the body is to be assumed absent in this type of motion. A body projected into the space and is no longer being propelled by fuel is called a projectile.

To analyze the projectile motion we use the following concept "Resolution of two dimensional motion into two one dimension motion" as discussed earlier. Hence it is easier to analyze the motion of projectile as composed of two simultaneous rectilinear motions which are independent of each other: (a) Along the vertical y-axis with a uniform downward acceleration 'g' and (b) Along the horizontal x-axis with a uniform velocity forward. Consider a particle projected with an initial velocity u at an angle θ with the horizontal x-axis as shown in figure shown below. Velocity and accelerations can be resolved into two components:

Velocity along x-axis = ux = u cos θ Acceleration along x-axis ax = 0 Velocity along y-axis = uy = u sin θ Acceleration along y-axis ay = -g Here we use different equation of motions of one dimension derived earlier to get the different parameters. …... (a)

…... (b) 2

v =

v02

– 2g (y-y0)

…... (c)

Total Time of flight:When body returns to the same horizontal level, the resultant displacement in vertical y-direction is zero. Use equation b. Therefore, 0 = (u sin θ) t - (½)gt2, As t cannot equal to zero, then, total time of flight,

Horizontal Range:-

Horizontal Range (OA=X) = Horizontal velocity × Time of flight = u cos θ × 2 u sin θ/g So horizontal range,

Maximum Height:At the highest point of the trajectory, vertical component of velocity is zero. Therefore, 0 = (u sin θ)2 - 2g Hmax So, maximum height would be,

Equation of Trajectory:Assuming the point of projection as the origin of co-ordinates and horizontal direction as the x-axis and vertical direction as the yaxis. Let P (x, y) be the position of the particle at instant after t second. Then x = u cos θ.t and y = u sin θ.t - 1/2 gt2 Eliminating 't' form the above equations, we get, y = x tan θ - (gx2/2u2cos2θ) This is the equation of trajectory which is a parabola (y = ax + bx2). Problem 1 (JEE Advanced):The speed of a projectile when it is at its greatest height is √2/5 times its speed at half the maximum height. Find out the angle of projection. Solution :Let θ be the angle of projection and u its initial speed. Then maximum height will be,

H = u2sin2θ/2g So, gH = u2sin2θ/2 Now, vH = u cosθ Or, vH2 = u2 cos2θ vH/22

2

…... (1) 2

= u -2g(H/2) = u -gH

vH/22=

u2-(u2sin2θ/2)

…...(2)

Now it is given that, vH = [√2/5] vH/2 Or, vH2 = (2/5) vH/22 Substituting the values from equations (1) and (2), we get, (u2 cos2θ) = 2/5 [u2-(u2sin2θ/2)] Or, 5cos2θ = 2[1-(sin2θ/2)] Or, 5(1-sin2θ) = 2-sin2θ Or, sin2θ = ¾ Or, sinθ = (√3)/2, Or, θ = 60o Thus from the above observation, we conclude that, the angle of projection would be 60o. Problem 2:A gun moving at a speed of 30m/sec fires at an angle 30o with a velocity 150m/s relative to the gun. Find the distance between the gun and the projectile when projectile hits the ground. (g = 10 m/sec) Solution :Vertical component of velocity = 150 sin 30o = 75 m/sec Horizontal component of velocity relative to gun = 150 cos 30o = 75√3 m/sec Horizontal component of velocity relative to ground = 75√3 + 30 ≈ 160m/sec Time of flight = (2 75)/g = 15 s Range of projectile = 160 × 15 = 2400 m Distance moved by the gun and projectile = 2400 - 450 = 1950 m. Horizontal projection:-

Consider a particle projected horizontally with a velocity

from a point O as shown in side figure.

Assuming the point of projection O as the origin of coordinates and horizontal direction as the X-axis and vertical direction as Y-axis. Let P (x, y) be the position of the particle after t seconds. So, x = horizontal distance covered in time t = ut.

............... (1)

y = vertical distance covered in time t = ½gt2

............... (2)

Eliminate t from equations (1) and (2) then we get, y = (1/2)(g/u2) (x2) This is the equation of parabola passing through the origin, with its vertex at the origin O. Hence the trajectory is a parabola. Problem 3:-

A stone is thrown at a speed of 19.6 m/sec at an angle 30o above the horizontal from a tower of height 490 meter. Find the time during which the stone will be in air. Also find the distance from the foot of the tower to the point where stone hits the ground? Solution :Let us consider the motion of stone in the horizontal and vertical directions separately. (i) Vertical motion (downward direction negative) : Initial vertical velocity y = 19.6 sin 30o Acceleration a = g = -9.8 m/s2 Vertical distance covered = h = 490 m Using, h = ut + 1/2gt2 We have, 490 = - 9.8t + (1/2) 9.8t2 100 = - 2t + t2

or

t2 - 2t - 100 = 0

t = 11.05 sec (ii) Horizontal motion: Initial horizontal velocity y = 19.6 sin 30o = 9.8 m/s Hence distance from the foot of tower to the point where stone hits the ground = Horizontal component × time of flight = 19.6 cos 30o × 11.05 = 188 m Projectile Motion on an inclined plane:Let the particle strike the plane at A so that OA is the range of the projectile on inclined plane. This initial velocity into two components:

can be resolved

(i) u cos (α - β) along the plane (ii) u sin (α - β) perpendicular to the plane. The acceleration due to gravity g can be resolved into two components: (i) g sin β parallel to the plane (ii) g cos β perpendicular to the plane. Time of Flight:Let t be the time taken by the particle to go from A to B. In this time the displacement of the projectile to the plane is zero. Hence, 0 = u sin (α-β) t - ½g β t2 => t = 2u sin(α-β)/gcosβ Range:During time of flight, the horizontal velocity u cos α remains constant. Hence, horizontal distance, OB = (ucosα) t = 2u2sin(α-β)cosα/gcosβ Now, OA = OB/cosβ = 2u2sin(α-β)cosα/gcosβ The greatest distance of the projectile from the inclined plane is u2sin2 (α-β)/2gcosβ . Problem 4:-

A Particle is projected with a velocity 39.2 m/sec at an angle of 30o to an inclined plane (inclined at an angle of 45o to the horizontal). Find the range on the incline (a) when it is projected upward (b) when it is projected downward. Solution :Time of flight will be same in both cases because the acceleration perpendicular to the plane is same. Therefore, 0 = 39.2 sin 30o t - (½) g cos 45o t2 Or, t = (2×39.2 sin 30)/(g cos 45) = 4√2 sec

(a) Range upward = 39.2 cos 30o t - (½) g sin 30o t2 = 39.2 × √3/2 × 4√2- (1/2) × 9.8 × (1/2) × (4√2)2 = 113.7m (b) Range downward = 39.2 cos 30o × t + (½) g sin 30o t2 = 39.2 × √3/2 × 4√2 + (1/2) × 9.8 × (1/2) × (4√2)2 = 270.5m Ans. Motion down the plane:-

Let the particle be thrown at a velocity v0 at angle „α‟ with the horizontal as shown in figure. v0 sin (α+β)T- 1/2 gcosβT2=0 [for y'=0] => T = (2v0 sin(α+β))/gcosβ R = v0 cos(α+ β)T+ 1/2 g sin βT2= (v02)/g [(sin(2α+β)+sinβ)/(1-sin2β)] Since α is the variable and maximum value of sin function is 1, therefore for R to be maximum, sin (2α+β)=1 and Rmax (v02)/g [(1+sinβ)/(1-sin2β)]= (v02)/(g(1-sinβ)) down the plane.



Every projectile experiences one single force and that is due to gravity only.



Horizontal velocity of a projectile remains the same throughout its flight (it may be zero also).



No projectile ever experiences any acceleration in the horizontal direction.



Vertical acceleration of every projectile is „-g‟.



The path of projectile is parabolic except for those projected along vertical direction. In that case it is a straight line.



The horizontal and vertical motion of a projectile are independent of each other.

Circular Motion Now we shall discuss another example of two-dimensional motion that is motion of a particle on a circular path. This type of motion is called circular motion. The motion of a body is said to be circular if it moves in such a way that its distance from a certain fixed point always remains the same. Direction of motion of body at any instant:If the string breaks suddenly, the stone shall fly tangentially to the path of motion. So, instantaneous direction of motion of the body is always along the tangent to the curve at that point. Consider a particle P is moving on circle of radius r on X-Y plane with origin O as centre. The position of the particle at a given instant may be described by angle θ, called angular position of the particle, measured in radian. As the particle moves on the path, its angular position θ changes. The rate of change of angular position is called angular velocity, ω, measured in radian per second.

= dθ/dt = ds/rdt = v/r The rate of change of angular velocity is called angular acceleration, measured in rad/s2. Thus, the angular acceleration is α = dω/dt = d2θ/dt2 Relation between These Parameters:-

It is easy to derive the equations of rotational kinematics for the case of constant angular acceleration with fixed axis of rotation. These equations are of the same form as those for on-dimensional transitional motion. ω = ω0 + αt ............ (a) ϕ = ϕ0 + ω0t + αt2/2

............ (b)

ω2 = ω02 + 2α (ϕ - ϕ0)

............ (c)

ϕ = ϕ0 + (ω0 + ω)/(2t)

............ (d)

Here, ϕ0 is the initial angle and ω0 is the initial angular speed. Problem 1:

(a)

What is the angular velocity of the minute and hour hands of a clock?

(b)

Suppose the clock starts malfunctioning at 7 AM which decelerates the minute hand at the rate of 4Π radians/day. How much

time would the clock loose by 7 AM next day? Solution:(a) Angular speed of, minute hand : ωmh = 2π rad/hr = 48π rad/day = (Π/1800) rad/sec hour hand : ωhh = (π/6) rad/hr = 4π rad/day = (Π/21600) rad/sec (b) Assume at t = 0, ϕ0 = 0, when the clock begins to malfunction. Use equation (ii) to get the angle covered by the minute hand in one day. So, ϕ = ω0(1 day) 1/2α(1 day)2 = 46π rad Hence the minute hand complete 23 revolutions, so the clock losses 1 hour. Problem 2: A particle is rotating in a circular path having initial angular velocity 5 rad/sec and the angular acceleration α = 0.5 ω, where ω is angular velocity at that instant. Find the angular velocity, after it moved an angle π? Solution:Here angular acceleration is α = 0.5 ω => dω/dt = 0.5ω => (dω/dθ) (dθ/dt) = 0.5ω => ω dω/dθ = 0.5ω Or,

=> ω - 5 = 0.5 × π => ω = 5 + 0.5 ×π = 6.57 rad/sec. Hence, when acceleration is not constant, use the method of calculus as shown in above illustration.

Motion of a particle in a circular path:-

It is a special kind of two-dimensional motion in which the particle's position vector always lies on the circumference of a circle. In order to calculate the acceleration parameter it is helpful to first consider circular motion with constant speed, called uniform circular motion. Let there be a particle moving along a circle of radius r with a velocity shown in figure given below, such that

, as

= v = constant. For this particle, it is our aim to calculate the magnitude and direction of

its acceleration. We know that,

Now, we have to find an expression for points A and B. Displacing =

in terms of known quantities. For this, consider the particle velocity vector at two

, parallel to itself and placing it back to back with

–

Consider ΔAOB, angle between OA and OB is same as angle between and

, as shown in figure given below. We have

is also perpendicular to vector OB.

OB = OA = r and

So, ΔAOB is similar to the triangle formed by

,

and

and

because

is perpendicular to vector OA

Thus,From geometry we have, Δv/v = AB/r Now AB is approximately equal to vΔt.

In the limit Δt → 0 the above relation becomes exact, we have

This is the magnitude of the acceleration. The direction because of this

is instantaneously along a radius inward towards the centre of the circle,

is called radial or centripetal acceleration. The acceleration vector of a particle in uniform circular motion

averaged over one cycle is a null vector. Problem 3: The moon revolved about the earth making a complete revolution in 2.36 mega second. Assume that the orbit is circular and has a radius of 385 mega meter. What is the magnitude of the acceleration of the moon towards the earth? Solution:Here first of all we calculate speed v of the moon which is given by v = 2πR/T where R = 385 mega meter = 385 × 106 m and T = 2.36 mega second = 2.36 × 106 sec. Hence v = 1020 m/sec. The magnitude of centripetal acceleration is a = v2/R = 0.00273 m/sec2. From the above observation we conclude that, the magnitude of the acceleration of the moon towards the earth would be 0.00273 m/sec2. In the previous enquiry we have discussed the uniform circular motion in which the particle has constant speed. If the particle's speed varies with time then the motion will be no more uniform but a non-uniform circular motion. Let us discuss about this motion using the concept of vectors.

Simulation for Car and Curves:-

This animation is used to explain why a passenger slides to the "outside" of a curve while riding inside a car is NOT an example of centrifugal forces. Instead is is a combination of centripetal force and inertia. It emphasize that when an object moves to the outside of a circle it is because of a lack of enough centripetal force and inertia keeps it moving in a straight line. Non uniform circular motion:-

Let us use the vector method to discuss non-uniform circular motion. In the side figure,

and

are unit vectors along radius and tangent vector respectively. In terms of er and eθ the motion of a

particle moving counter clockwise in a circle about the origin in figure 2.30 can be described be the vector equation.

In this case, not only

=(d

)/dt =

but v also varies with time. We can obtain instantaneous acceleration as,

θ dθ/dt + v(d

θ)/dt

Again,

Here, aT = dv/dt and aR = v2/r The first term,

is the vector component of

that is tangential to the path of the particle and arises from a change in the

magnitude of the velocity in circular motion, called tangential acceleration whereas aR centripetal acceleration. The magnitude of

Problem 4:

is

Point A travels along an arc of a circle of radius r as shown in figure given below. Its velocity depends on the arc coordinates l as v = A √l where A is a constant. Let us calculate the angle α between the vectors of the total acceleration and of the velocity of the point as a function of the coordinate l. Solution:It is seen from figure shown above that the angle α can be found by means of the formula tan α = aR/aT. Let us find aR and aT. aR = v2/r=(A2 l)/r; aT= dv/dt = dv/dl = A/(2√l)×A√l = A2/2 Hence tan α = 2l/r. Problem 5 (JEE Main): Position vector of a particle moving in x-y palne at time t is . What will be the path of the particle? Solution:x = a (1-cos ωt) y = a sinωt From first equation, we get acos ωt = a-X

…... (1)

From second equations, we get, a sin ωt = y Squaring and adding Eqs. (1) and (2), we get, (X-a)2 + y2 = a2 This is an equation of a circle of radius a and ceter at (a,0).



Centripetal and centrifugal forces are equal in magnitude and opposite in direction.



Centripetal and centrifugal forces cannot be termed as action and reaction since action and reaction never act on same body.



Basic requirement for a body to complete motion in a vertical circle, under limiting conditions, is that the tension in the string must not vanish before it reaches the highest point. If it vanishes earlier, it will be devoid of the necessary centripetal force required to keep the body moving in a circle.



Cetrifugal force is the fictitious force which acts on a body, rotating with uniform velocity in a circle, along the radius away from the centre.

Relative Motion Relative Velocity

In spite of our best efforts we could not find a fixed point with respect to which we should study absolute rest and absolute motion. So, the study of relative rest and relative motion assumes importance. Consider two cars A and B running parallel to each other on same road with same velocities. No doubt the speedo meter of each indicates motion but the two drivers will always find themselves facing each other. Relative velocity of a body A with respect to another body B, when both are in motion, is the velocity with which A appears to move to B. The position, velocity and acceleration of a particle depend on the reference frame chosen. A particle P is moving and is observed from two frames S and S'. The frame S is stationary and the frame S' is in motion. Let at any time position vector of the particle P with respect to S is,

Position vector of the origin of S' with respect to S is,

From vector triangle OO'P, we get,

Physical Significance of Relative Velocity

Let two cars move unidirectional. Two persons A and B are sitting in the vehicles as shown in figure. Assume, VA = 10 m/s & VB = 4 m/s. The person A notices person B to be moving towards him with a speed of (10-4) m/s = 6 m/sec. That is the velocity of B with respect to (or relative to) A. That means

is directed from B to A.

Similarly A seems to move towards B with a speed 6 m/sec. That means the velocity of A relative to B ( m/sec & directed from A to B as shown in the figure.

Therefore,

In general, So,

vAB = √ vA2 + vB2 - 2vAvB cosθ and θ = tan-1{(vBsinθ)/(vA – vBcosθ)}

) has the magnitude 6

Relative Motion between Rain and Man

We know that, vr = vrg = velocity of rain w.r.t. ground, vm ≡ vmg. Velocity of man w.r.t. ground and

velocity of rain w.r.t. man. So, That means the vector addition of the velocity of rain with respect of man ( actual velocity of rain 2

. The magnitude and direction of

) and the velocity of man (vehicle) (

) yield the

can be given as,

2

vr = √((vrm) +(vm) + 2vrm vm cosθ) ϕ = tan-1((vrm sinθ)/(vrm cosθ+ vm )) with horizontal

Illustration (JEE Main): Four particles A, B, C and D are in motion. The velocities of one with respect to other are given as north,

is 20 m/s towards east and

(a) 20 m/s towards north (b) 20 m/s towards south (c) 20 m/s towards east (d) 20 m/s towards west Solution: From the question, we know that,

is 20 m/s towards south. Then

is

is 20 m/s towards

…... (1) …... (2) …... (3) Equation (1) – (2) + (3) gives:

or That is,

is 20 m/s towards west.

Thus from the above observation we conclude that, option (d) is correct. Illustration: A stationary person observes that rain is falling vertically down at 30 km/hr. A cyclist is moving on the level road, at 10 km/hr. In which direction should the cyclist hold his umbrella to project himself from rain? Solution: Relative to stationary frame, velocity of rain is 30 km/hr downward. Take horizontal axis as x-axis and vertical axis as y-axis and i,j are the unit vectors along x and y axes respectively.

= 0-30ĵ,

= 10î

= -30j - 10i = -10i - 30j If angle between horizontal and

is θ, then,

tan θ = -30/-10 = 3 =>

θ

=

tan-1 3

=>

θ=72°

Therefore, to protect himself from rain the cyclist should hold the umbrella at an angle of 72° from horizontal. Illustration: A man walking eastward at 5 m/s observes that wind is blowing from the north. On doubling his speed eastward, he observes that wind is blowing from north-east. Find the velocity of the wind. Solution: Let velocity of the wind is, vw=(v1i+v2j)m/s And velocity of the man is,

vm=5i So, vwm = vw- vm=(v1-5)i + v2j In first case, v1- 5 = 0 => v1= 5 m/s. In the second case, tan 45o = v2/(v1- 10) => v2= v1 - 10 = -5 m/s. => vw= (5i - 5j) m/s. Illustration: From a lift moving upward with a uniform acceleration 'a', a man throws a ball vertically upwards with a velocity v relative to the lift. The ball comes back to the man after a time t. Show that a + g = 2 v/t Solution: Let us consider all the motion from lift frame. Then the acceleration, displacement and velocity everything will be considered from the lift frame itself. As the ball comes back to the man, therefore displacement from the lift frame is zero. Again, the velocity with respect to the lift frame is v. g - (-a) = a + g (↓) Now, s = ut + 1/2at

downwards 2

=>

0 = vt - 1/2 (a+g)t2

or a + g = 2(v/t) . Relative Motion of a Swimmer in Flowing Water Take

= velocity of man = velocity of flow of river,

= velocity of swimmer w.r.t. river can be found by the velocity addition of

and

.

Crossing of the River with Minimum Drift Case 1: vmw > w A man intends to reach the opposite bank at the point directly opposite to the stationary point. He has to swim at angle θ with a given speed of water

w.r.t. water, such that his actual velocity ).

will direct along AB, that is perpendicular to the bank (or velocity

=> For minimum drift,

⊥

You can realize the situation by a simple example. If you want to reach the directly opposite point or cross the river perpendicularly, a man, that is to say, Hari, must report you that, you are moving perpendicular to the shore. What does this report signify? Since Hari observes your actual velocity (

) to be perpendicular to the bank

is perpendicular to

.

Observing the vector-triangle vw = vmw sinθ & vm = vmw cosθ => θ = sin-1 (vw/vmw ) & vm = √((vmw )2- (vw)2) => The time of crossing, t = d/vm => t = d/√(vmw )2 - (vw)2) Case 2 : vw > vmw Let the man swim at an angle θ‟ with normal to the bank for minimum drift. Suppose the drift is equal to zero. For zero drift, the velocity of the man along the bank must be zero. => vm= vw- vmw sin θ' = 0 This gives, sinθ' = vw / vmw, since vw > vmw, sinθ' > 1 which is impossible. Therefore, the drift cannot be zero. Now, let the man swim at an angle θ with the normal to the bank to experience minimum drift. Suppose that the drifting of the man during

time

t

when

the

BC = x x = (vm)x (t) where t = AB/((vm )y cosθ) = d/(vmw cosθ) and (vm)x = vw – vmw sin θ Using (1), (2) & (3), we obtain x = (vw- vmw sinθ d/(vmw cosθ))

… (1) … (2) … (3)

reaches

the

opposite

bank

is,

= (vw/vmw sec θ-tanθ)d x = (vw/vmw sec θ-tanθ)d

… (4)

For x to be minimum, dx/dθ = (vw/vmw secθ - tanθ - sec2θ)d = 0 vw/vmw tanθ = (sec θ) => sinθ = vmw/vw θ = sin-1(vmw/vw) Substituting the value of θ in (4), we obtain, x = [√(vw2 – vmw2)/vmw] d Crossing of the River in Minimum Time Case 1: To reach the opposite bank for a given vmw Let the man swim at an angle θ with AB. We know that the component of the velocity of man along shore is not responsible for its crossing the river. Only the component of velocity of man (vm) along AB is responsible for its crossing along AB. The time of crossing = t = AB/(vmw cosθ) Time is minimum when cos θ is maximum.

The maximum value of cos θ is 1 for θ = 0. That means the man should swim perpendicular to the shore. =>

mw ⊥

w

=> Then tmin = d/(vmw cosθ)|(θ=0) = d/vmw => tmin= d/vmw Case 2: To reach directly opposite point on the other bank for a given vmw & velocity v of walking along the shore. To attain the direct opposite point B in the minimum time, let the man swim at an angle θ with the direction AB. The total time of journey t = the time taken from A to C and the time taken from C to B. => t = tAC + tCB

where tAC = AB/vmvcosθ & tCB = BC/v where v = walking speed of the man from C to B. => t = AB/vmvcosθ + BC/v Again BC = (vm)xt => BC = (vw - vmwsinθ) (AB/vmwcosθ) Using (1) & (2) we obtain,

t = AB/vmwcosθ + ((vw - vmwsinθ)/v(vmvcosθ)) => t = AB[(1+vw/v)secθ/vmv - tanθ/v] => t = d/vmv[(1+vw/v)secθ/vmv - tanθ/v] Putting dt/dθ = 0, For minimum t we get, dt/dθ = d/dθ[d/vmv (1+vw/v) secθ/vmv - tanθ/v] = [secθ/vmv - tanθ/v (1+vw/v) (sec2θ)/v] = 0 =>tanθ/vmv (1+vw/v) secθ/v => sin θ = (vmw/v+vw) => θ = sin-1(vmw/v+vw) This expression is obviously true when vmw < v + vw. Velocity of Separation/Approach, Relative Angular Velocity Let thane be two particles A and B with velocity

and

at any instant as visualized from ground frame.

If we visualize the motion of B from frame of A the velocity of particle B would be

.

If α, β be the angle made with line AB, then, VB cos β - VA cos α is relatively velocity of B w.r.t. A along line AB.



If VB cos β - VA cos α > 0; it is called as velocity of separation.



If VB cos β - VA cos α < 0; it is called as velocity of approach.

VB sin β - VA sin α is relative velocity of B w.r.t. A along direction perpendicular to AB. If length of AB is, then, angular velocity B w.r.t. A is (VB sinβ - VA sinα)/l

and relative angular velocity = (VB sinβ - VA sinα)/l.



The analytical method for determination of relative velocity of one body with respect to another which is discussed here is to general nature. This treatment is valid for motion in two dimensions and in three dimensions also.



Relative velocity of a body A with respect to another body B, when both are in motion, is the velocity with which A appears to move to B.



Relative velocity of a body „B‟ w.r.t. a body „A‟, when both are in motion, is the velocity with which „B‟ appears to move to „A‟.

Solved Examples on Mechanics:Example 1:A 5.1-kg block is pulled along a frictionless floor by a cord that exerts a force P = 12 N at angle θ=25º above the horizontal as shown in below figure. (a) What is the acceleration of the block? (b) The force P is slowly increased. What is the value of P just before the block is lifted off the floor? (c) What is the acceleration of the block just before it is lifted off the floor?

Concept:The figure below shows the forces acting on the block with mass m: Assume that the weight of the Block is given by W such that W = mg, the horizontal component of force the vertical component of force is Py = Psinθ (θ is the angle at which the force

is Px = Pcosθ whereas

acts).

(a) When the block does not lift off from the ground there is no acceleration in the vertical direction. However, the acceleration does exist on the horizontal direction and is given as: ax = Px/m Substitute Px = Pcosθ,

ax = Pcosθ/m Substitute the values of P, m and θ to get the desired result. (b) When the force

is increased, the magnitude of vertical component of force Py = Psinθ will also increase. The moment at

which the block is about to lift, the magnitude of the vertical component of force will be equal to the weight W, that is, Py = W Substitute Py = Psinθ, Psinθ = W P = W/sinθ Substitute the values of W and θ to get the desired result. (c) Just before the block is lifted there in no acceleration in the vertical direction however the acceleration exists in the horizontal direction, and is given as: ax = Px/m Substitute Px = Pcosθ, ax = Pcosθ/m Substitute the values of P, m and θ to get the desired result. Solution:(a) Substitute 12 N for P, 5.1 kg for m and 25º for θ in equation ax = Pcosθ/m, ax = Pcosθ/m = 12 N cos (25º) /5.1 kg = (10.87 N) [(1 kg.m/s2) / 1 N] / 5.1 kg = 2.2 m/s2 Therefore, the horizontal acceleration of the block is 2.2 m/s2. (b) The weight of the block is: W = mg Substitute 5.1 kg for m and 9.8 m/s2 for g, W = (5.1 kg) (9.8 m/s2) = (49.9 kg.m/s2) (1 N/1 kg.m/s2) = 49.9 N Substitute 49.9 N for W and 25º for θ in equation P = W/sinθ, P = 49.9 N/sin ( 25º) = 118.26 N Round off to three significant figures, P = 119 N Therefore the magnitude of net force on the block when it is about to lift is 119 N. (c) Substitute 119 N for P, 5.1 kg for m and 25º for θ in equation ax = Pcosθ/m, ax= 119 N (cos25º) / 5 kg = (107.8 N) [(1 kg.m/s2)/1 N] / 5.1 kg = 21.1 m/s2 Round off to two significant figures, ax = 21 m/s2 Therefore, the acceleration of the block when it is about to lift is 21 m/s2. _____________________________________________________________________________________________________

Example 2:A worker drags a crate across a factory floor by pulling on a rope tied to the crate. The rope, which is inclined at 38.0º above the horizontal, exerts a force of 450 N on the crate. The floor exerts a horizontal resistive force of 125 N, as shown in the below figure. Calculate the acceleration of the crate (a) if its mass is 96.0 kg, and (b) if its weight is 96.0 N.

Concept:The figure below shows the forces acting on the crate of mass M, when it is pulled by a worker. The force experienced by the crate is components as

, and can be resolved into horizontal and vertical

and

respectively.

Also assume that the resistive force from the ground is represented by the horizontal direction is given by

(refer figure above), and the acceleration of the crate in

and the acceleration in the vertical direction is

.

(a) Using Newton‟s second law of motion for the forces acting in the horizontal direction, one can write F cos (38.0º) – f = Max ax = [F cos (38.0º) – f]/ M

…… (1)

Therefore, the acceleration of the crate in the horizontal direction is ax = [F cos (38.0º) – f] / M. Similarly, apply Newton‟s second law in the vertical direction, F sin (38.0º) – Mg = May ay = F sin (38.0º) – Mg / M It can be seen from the equation above that if the term F sin (38.0º) is smaller than the weight Mg, the crate will rest on the ground. (b) If the weight of the crate is W, then its mass is W/g, and the acceleration given by equation (1) can be written as: a = [F cos (38.0º) – f]/ (W/g) Therefore, the acceleration of the crate of weight W can be calculated using relation a = [F cos (38.0º) – f]/ (W/g). Similarly, apply Newton‟s second law in the vertical direction, F sin (38.0º) – W = (W/g) ay ay = [F sin (38.0º) – W] / [W/g] Therefore the acceleration of the crate in the vertical direction is ay = [F sin (38.0º) – W] / [W/g]. Solution:(a) Substitute 450 N for F, 125 N for f and 96.0 kg for M in equation a = [F cos (38.0º) – f]/ M,

a = [F cos (38.0º) – f]/ M = [(450 N) cos (38.0º) – 125 N]/[96.0 kg] = (229.6 N) [(1 kg.m/s2)/1 N] / 96.0 kg = 2.40 m/s2 Therefore, the horizontal acceleration of the crate of mass 96.0 kg is 2.40 m/s2. Substitute 450 N for F in expression F sin (38.0º), F sin (38.0º) = (450 N) sin (38.0º) = 277 N Substitute 9.8 m/s2 for g and 96.0 kg for M in expression Mg, Mg = (96.0 kg) (9.8 m/s2) = (940.8 kg.m/s2) [1 N/(1 kg.m/s2)] = 940.8 N The weight of the crate is larger than the upward force F sin (38.0º), therefore the crate will rest on the ground. (b) Substitute 450 N for F, 125 N for f, 9.8 m/s2 for g and 96.0 N for W in equation a = [F cos (38.0º) – f]/ (W/g), a = [(450 N) cos (38.0º) – 125 N]/(96.0 N/9.8 m/s2) = (229.6 N) [(1 kg.m/s2)/1 N] / 9.79 kg = 23.4 m/s2 Therefore, the horizontal acceleration of the crate of weight 96.0 N is 23.4 m/s2. Substitute 450 N for F, 125 N for f, 9.8 m/s2 for g and 96.0 N for W in equation ay = [F sin (38.0º) – W] / [W/g], ay = [F sin (38.0º) – W] / [W/g] = (450 N) sin (38.0º) – (96.0 N)/(96.0 N/9.8 m/s2) = (181 N) [(1 kg.m/s2)/1 N] / 9.79 kg = 18.4 m/s2 Therefore, the crate has an upward acceleration of 18.4 m/s2. ____________________________________________________________________________________________________ Example 3:A 110-kg crate is pushed at constant speed up the frictionless 34º ramp as shown in the below figure. What horizontal force F is required? (Hint: Resolve forces into components parallel to the ramp.)

Concept:The figure below shows the component of forces

on the block, moving with a constant speed on a frictionless 34º ramp.

Assume that the x axis is along the length of the ramp, and y axis perpendicular to the ramp such that the component of force along the x axis is Fx = F cos (34º) whereas the component of force along the y axis is Fy = – Fsin (34º). The block moves up the ramp at constant speed, and therefore has no acceleration in the x direction. The Newton‟s third law of motion for the forces along the x axis gives, Mg sin (34º) – F cos (34º) = 0 Here M is the mass of the block and g is the acceleration due to gravity. Mg sin (34º) = F cos (34º) F = Mg sin (34º) / cos (34º) = Mg tan (34º) Therefore the force required to move the block with constant speed on a frictionless 34º ramp is given by relation F = Mg tan (34º). Solution:Substitute 110 kg for M and 9.8 m/s2 for g in equation F = Mg tan (34º), F = Mg tan (34º) = (110 kg) (9.8 m/s2) tan (34º) = (727 kg.m/s2) [1 N/(1 kg.m/s2)] = 727 N Therefore, the magnitude of force required to move the block with constant speed on a frictionless 34º ramp is 727 N. _________________________________________________________________________________________________ Example 4:A jet fighter takes off an angle of 27.0º with the horizontal, accelerating at 2.62 m/s2. The weight of the plane is 79,300 N. Find (a) the thrust T of the engine on the plane and (b) the lift force L exerted by the air perpendicular to the wings; see below figure. Ignore air resistance.

Concept:The figure below shows the forces acting on the jet fighter: Assume that the x direction is in the direction of motion of plane during its take off and the y direction perpendicular to it such that the weight W can be resolved into components along the x and y direction as Wx = W sinθ and Wy = W cosθ (θ is the angle between the jet and the ground). (a) If ax is the acceleration of the jet in x direction, then the net force acting on the jet is F = (W/g) ax (g is the free fall acceleration). Apply Newton‟s second Law of motion along the x direction, T – Wx = F Substitute Wx = W sinθ and F = (W/g) ax, T – Wsinθ = (W/g) ax T = (W/g) ax + Wsinθ Substitute the value of ax, W and θ to calculate the desired result. (b) The jet does not have any acceleration in the y direction, therefore the sum of the forces acting on it in that direction must be equal to zero, that is L – Wy = 0 L = Wy Substitute Wy = W cosθ, L = W cosθ Solution:(a)Substitute 79,300 N for W, 2.62 m/s2 for ax, 9.8 m/s2 for g and 27.0º for θ in equation T = (W/g) ax + Wsinθ, T = [(79,300 N)/(9.8 m/s2)] (2.62 m/s2) + (79,300 N) sin (27.0º) = 21,200.6 N + 36001.4 N = 57,202 N Round off to three significant figures, T = 57,000 N Therefore, the magnitude of thrust T is 57,000 N. (b) Substitute 79,300 N for W and 27.0º for θ in equation L = W cosθ, L = W cosθ

= (79,300 N) cos (27.0º) = 70656.8 N Round off to three significant figures, L = 70,600 N Therefore the magnitude of the force L is 70,600 N. ___________________________________________________________________________________________________ Example 5:A certain airplane has a speed of 180 mi/h and is diving at an angle of 27º below the horizontal when a radar decoy is released. The horizontal distance between the release point and the point where the decoy strikes the ground is 2300 ft. (a) How long was the decoy in the air? (b) How high was the plane when the decoy was released? See below figure.

Concept:The figure below shows the plane, decoy and the initial velocity v0 of the decoy when it was released. It can be seen from the figure that the horizontal component of velocity of the decoy when it was released is v0 cos (-θ) while the vertical component of the decoy is v0 sin (-θ). It is important to note that the negative sign in θ is due to the fact that the angle is measured in clockwise direction with respect to horizontal. (a) The time of flight of decoy can be calculated using relation, t = x/v0cosθ Here x is the range of the decoy, v0 is the initial velocity and θ is the angle between the decoy with respect to horizontal (refer figure above). (b) The vertical displacement of the decoy during its time of flight can be calculated using relation, y = (v0 sinθ) t – ½ gt2

Here g is the acceleration due to gravity and t is the time of flight calculated in part (a). Solution:(a) Substitute -27º for θ, 180 mi/h for v0 and 2300 ft for x in equation t = x/v0cosθ, t = [2300 ft] / [(180 mi/h) (5280 ft/ 1 mi) (1 h/3600 s) cos (27º)] = (2300 ft)/(235.2 ft/s) = 9.77 s Therefore, it takes 9.77 s for decoy to reach the target. (b) Substitute 9.77 s for t, 32 ft/s2 for g, -27º for θ and 180 mi/h for v0 in equation y = (v0 sinθ) t – ½ gt2 , y = ((180 mi/h) sin (27º)) (9.77 s) – ½ (32 ft/s2) (9.77 s)2 = [((180 mi/h) (5280 ft/1 mi) (1 h/3600 s) sin (-27º)) (9.77 s)] – ½ (32 ft/s)2 (9.77 s)2 = -1170.9 ft – 15272 ft = -2698 ft The negative sign accounts for the fact that the decoy moves closer to ground. Therefore, the vertical displacement of decoy is 2698 ft. ___________________________________________________________________________________________________ Example 6:In earlier days, horses pulled barges down canals in the manner shown in below figure. Suppose that the horse pulls a rope that exerts a horizontal force of 7900 N at an angle of 18º to the direction of motion of the barge, which is headed straight along the canal. The mass of the barge, which is headed straight along the canal. The mass of the barge is 9500 kg and its acceleration is 0.12 m/s2. Calculate the horizontal force exerted by the water on the barge.

Concept:The force say

acting on the barges of mass m is show in the figure below:

The horizontal component of force on the barge is Fx = F cosθ whereas the vertical component of force is Fy = F sinθ (θ is the angle made by force F relative to horizontal axis). Assume that the horizontal force exerted by the water on the barge is Px, then from Newton‟s second law one can write,

Fx – Px = max Px = Fx - max Here, ax is the acceleration of the barges in the horizontal direction. Substitute Fx = F cosθ, Px = F cosθ - max Substitute the value of F, m and a to get the desired result. Solution:Substitute 0.12 m/s2 for a, 9500 kg for m, 7900 N for F and 18º for θ in equation Px = F cosθ - max, Px = F cosθ - max = (7900 N) cos(18º) – (9500 kg) (0.12 m/s2) = 7513.3 N – (1140 kg.m/s2) [1 N/(1 kg.m/s2)] = 6373. 3 N Round off to four significant figures, Px = 6373 N Therefore the magnitude of horizontal force exerted by the water is 6373 N and it acts in a direction opposite to the horizontal motion of barges. __________________________________________________________________________________________ Example 7:A dart is thrown horizontally toward the bull‟s eye, point P on the dart board, with an initial speed of 10 m/s. It hits at point Q on the rim, vertically below P, 0.19 s later; as shown in below figure. (a) What is the distance PQ? (b) How far away from the dart board did the player stand?

Concept:(a) The distance PQ is the vertical distance travelled by the dart in time t under the action of gravitational force, and can be given as: y = - ½ gt2 The negative sign is due to the convention that the vectors pointing downwards are taken as negative. Here g is the acceleration due to gravity and t is the elapsed time. (b) The horizontal distance between the dart board and the player can be calculated using relation, x = v0xt Here v0x is the initial speed of the dart and t is the elapsed time. Solution:(a)Substitute 9.81 m/s2 for g and 0.19 s for t in equation y = - ½ gt2, y = - ½ (9.81 m/s2) (0.19 s)2 = - 0.17 m The negative sign indicates the fact that the displacement is in the negative y direction. Therefore the distance PQ is -0.17 m. (b) Substitute 0.19 s for t and 10 m/s for v0x in equation x = v0xt, x = (10 m/s) (0.19 s) = 1.9 m Therefore, the horizontal separation between the dart board and the player is 1.9 m. _________________________________________________________________________________________

Example 8:In a baseball game, a batter hits the ball at a height of 4.60 ft above the ground so that its angle of projection is 52.0º to the horizontal. The ball lands in the grandstand, 39.0 ft up from the bottom; see in below figure. The grandstand seats slope upward at 28.0º with the bottom seats 358 ft from home plate. Calculate the speed with which the ball left the bat. (Ignore air resistance.)

Concept:The figure below shows the trajectory of the ball: It can be seen from the figure above that the total horizontal displacement say x of the ball is x = 358 ft + 39 ft cos (28º) whereas the vertical displacement say y of the ball is y =39 ft sin (28º) – 4.6 ft . Assume that the ball left the ball with initial velocity v0 at an angle θ such that the horizontal component of initial velocity is v0x = v0 cosθ whereas the vertical component of initial velocity is v0y = v0 sinθ. The ball does not accelerate in the horizontal direction, therefore the horizontal distance travelled by the ball is: x = v0xt Substitute x = 358 ft + 39 ft cos (28º) and vox = v0cosθ, 358 ft + 39 ft cos (28º) = (v0cosθ) t t = [358 ft + 39 ft cos (28º)] / (v0cosθ) Similarly, the vertical displacement of the ball can be written as: y = v0yt – ½ gt2 Substitute v0y = v0 sinθ and y =39 ft sin (28º) – 4.6 ft, [39 ft sin (28º) – 4.6 ft] = (v0 sinθ)t – ½ gt2 Substitute t = [358 ft + 39 ft cos (28º)] / (v0cosθ) , [39 ft sin (28º) – 4.6 ft] = (v0 sinθ)[[358 ft + 39 ft cos (28º)] / (v0cosθ)]– ½ g{[358 ft + 39 ft cos (28º)] / (v0cosθ)}2 13.7 ft = (392.4 ft) tanθ – ½ g (392.4 ft/v0cosθ)2 Solve the equation for v0 to get the desired result. Solution:Substitute 52º for θ and 32.2 ft/s2 for g in equation 13.7 ft = (392.4 ft) tanθ – ½ g (392.4 ft/v0cosθ)2, 13.7 ft = (392.4 ft) tan (52º) – ½ g (392.4 ft/v0cos(52º))2, ½ (32.2 ft/s2) [392.4 ft/v0 cos (52º)]2 = 488.5 ft Round of to three significant figures, v0 = 115 ft/s

Therefore, the speed with which the ball leaves the bat is 115 ft/s. _____________________________________________________________________________________________ Example 9:You throw a ball with a speed of 25.3 m/s at an angle of 42.0º above the horizontal directly toward a a wall as shown in below figure. The wall is 21.8 m from the release point of the ball. (a) How long is the ball in the air before it hits the wall? (b) How far above the release point does the ball hit the wall? (c) What are the horizontal and vertical components of its velocity as it hits the wall? (d) Has it passed the heighest point on its trajectory when it hits? Concept:The figure below shows the velocity of the ball and its components along the horizontal and vertical directions respectively:

The horizontal component of initial velocity of ball is: vx = v0 cosθ Here, v0 is the magnitude of initial velocity of the ball, and θ is the angle at which the ball is projected relative to horizontal. The vertical component of initial velocity of the ball is: vy = v0 sinθ Here, v0 is the magnitude of initial velocity of the ball, and θ is the angle at which the ball is projected relative to horizontal. (a) The time of flight of the ball is given as: t = x/vx Here, x is the range of the projectile. Substitute vx = v0 cosθ, t = x/v0 cosθ Substitute the value of x, v0 and θ to get the desired result. (b) The vertical displacement of the ball corresponding to time t is: y = vyt – ½ gt2 Substitute vy = v0sinθ, y = (v0sinθ) t – ½ gt2 Substitute the calculated value of t, and the given values of v0, θ to obtain the desired result.

(c) The horizontal component of velocity remains the same as the ball hits the wall because the ball has no acceleration in the horizontal direction. However, the ball acceleration in the vertical direction and therefore the magnitude of its final vertical velocity changes to, vy‟ = vy – gt Substitute vy = v0 sin θ, vy‟ = v0 sin θ – gt Substitute the calculated value of t, and the given values of v0, θ and g to obtain the desired result. (d) If the magnitude of vy‟ has a positive sign, the ball is still heading up and has not reached its maximum height yet. However, if the sign is negative, the velocity vector of the ball has reversed its direction and the ball has started falling downwards depicting that the maximum height is being crossed. Solution:(a) Substitute 25.3 m/s for v0, 42.0º for θ and 21.8 m for x in equation t = x/v0 cosθ, t = [21.8 m]/[(25.3 m/s) cos (42.0º)] = 1.15 s Therefore, the time of flight of the ball is 1.15 s. (b) Substitute 25.3 m/s for v0, 42.0º for θ, 9.8 m/s2 for g, and 1.15 s for t in equation y = (v0sinθ) t – ½ gt2, y = (v0sinθ) t – ½ gt2 = ((25.3 m/s) sin (42.0º)) (1.15 s) – ½ (9.8 m/s)2 (1.15 s)2 = 12.9 m Round off to two significant figures, y = 13 m Therefore, the vertical distance travelled by the ball in 1.15 s is 13 m. (c) Substitute 25.3 m/s for v0, 42.0º for θ, 9.8 m/s2 for g, and 1.15 s for t in equation vy‟ = v0 sin θ – gt, vy‟ = v0 sin θ – gt = (25.3 m/s) sin (42.0º) – (9.8 m/s2) (1.15 s) = 5.65 m/s Therefore, the vertical component of final velocity of the ball before hitting the wall is 5.65 m/s. The horizontal component of velocity remains unchanged during its flight, vx‟ = vx = v0 cosθ Substitute 25.3 m/s for v0 and 42.0º for θ, vx‟ = (25.3 m/s) cos (42.0º) = 18.8 m/s Therefore, the horizontal component of final velocity of ball, before hitting the wall is 18.8 m/s. (d) From part (c) it is clear that vertical component of final velocity of ball is 5.65 m/s, and is positive. Therefore the ball is yet to reach its maximum height but collided with the wall before doing the same. _____________________________________________________________________________________________ Example 10:(a) In Galileo‟s Two New Sciences, the author states that “for elevations [ angles of projection] which exceed or fall short of 45º by equal amounts, the ranges are equal.” Prove this statement. See below figure. (b) For an initial speed of 30.0 m/s and a range of 20.0 m, find the two possible elevation

Concept:Assume the angle by which one of the projectiles falls shorter of 45º is θ whereas the other one exceeds 45º by the same account. The range of the projectile with angle of projection 45º – θ is given as: R1 = v02 sin ( 45º – θ) / g Here, v0 is the initial speed of the projectile, and g is the acceleration due to gravity. Similarly, the range of the projectile with angle of projection 45º + θ is: R2 = v02 sin 2 ( 45º + θ) / g Here, v0 is the initial speed of the projectile, and g is the acceleration due to gravity. The term v02/ g remains the same in R1 and R2, therefore if the values sin 2 (45º - θ) is equal to sin 2 (45º - θ), then the range R1 and R2 is equal. (b) Assume the angle of projection of one of the projectile is θ, then its value can be calculated from expression for R as R = v02 sin 2θ / g θ = ½ sin-1 (Rg/v02) The angle of projection for the other projectile is 90º - θ. Solution:(a) The quantities sin 2 (45º - θ) and sin 2 (45º + θ) can be written as sin 2 (45º± θ). Also, sin 2 (45º± θ) = sin (90º± 2θ) = cos (± 2θ) = cos (2θ) It is important to note that cos (± 2θ) = cos (-2θ) = cos (θ), and therefore the terms sin 2 (45º - θ) and sin 2 (45º + θ) are equal. This proves the range R1 is equal to range R2. (b) Substitute 30 m/s for v0, 20.0 m for R and 9.8 m/s2 for g in equation θ = ½ sin-1 (Rg/v02), θ = ½ sin-1[(20.0 m) (9.8 m/s2) / (30 m/s)2] = 6.28º Round off to two significant figures,θ = 6.3º The other possible elevation angle of projectile is: ϕ = 90º – θ Substitute θ = 6.3º, ϕ = 90º – 6.3º = 83.7º Therefore, the first possible elevation angle of projectile is 6.3º, whereas the other possible elevation angle of projectile is 83.7º.

Newton’s Laws of Motion:Newton's laws of motion are of fundamental importance in classical physics. Newton gave three laws connected with motion and are, popularly, known as Newton‟s laws of motion. (a) First Law (Law of Inertia):-

Everybody continues in its state of rest or of uniform motion in a straight line unless it is compelled by some external force to change that state.

(b) Second Law:-

The rate of change of momentum of a body is directly proportional to the impressed force and takes place in the direction of the force. So, F= dp/dt This results force acting a body F is equal to the mass of the body m times acceleration of the body a. So, F = ma

(c)Third Law:-

To Every action there is an equal and opposite reaction. FAB = - FBA The first two laws relate to the type of motion of a system that results from a given set of forces. The third law studies the basic nature of all forces i.e. they act in action - reaction pair. "Newton's laws of Motion" is one of the easiest and important chapters of Mechanics in the Physics syllabus of IIT JEE, AIEEE and other engineering examinations. These laws are not new to any aspirant as he is using these laws even in previous classes. The examples based on this are very easy and can be seen even in day to day life. The chapter is important not only because it fetches 2-3 questions in most of the engineering examination but also because it is prerequisite to the other chapters of Mechanics. Types of Forces:-

If we want to produce motion in a body, already at rest, or if we want to destroy the motion of a moving body, we have to apply „effort‟ known as force in the form of a pull or push. Force:-

Force is defined as that pull or push which produces or tends to produce, destroys or tends to destroy motion in a body, increases or decreases the speed of the body or changes its direction ofmotion. Therefore, “Force" is an external or internal agent present to "influence" the natural state of motion of an object. Thus, this is an influence (force) needed to change the natural state of body; that is of rest or of uniform motion. Force is the basic cause of motion.

Classification of Forces:There are different types of forces in our universe. Based on the nature of the interaction between two bodies, forces can be classified into two main categories.

(a)

Contact Forces:-

These are the forces which come into play due to actual contact between the sources and the object. Basically, it is the force that act between the bodies in contact with each other. For example, Normal Reaction, Friction etc. (i) Tensional Force (T):When a string, thread or wire is held taut, the ends of the string or thread (or wire) pull on whatever bodies are attached to them in the direction of the string. This force is known as Tension. If the string is massless, then the tension T has the same magnitude at all points throughout the string. The direction of tension is always from the point of attachment to the body.

(ii) Spring Force:Force in an extended (or compressed) spring is proportional to the magnitude of extension (or compression). i.e. F α x, in magnitude, but opposite in direction. So, F = -kx, where k is a positive constant, also known as the spring constant of the spring; and x is the compression or elongation from the natural length. (iii) Normal reaction:-

When a body exerts a force on another, the second provides a reaction which acts perpendicular to the surface of 2nd body. (iv) Friction:-

It is a force that acts between bodies in contact with each other along the surface of contact and it opposes relative motion (or tendency of relative motion) between the two bodies.

(v) Air Resistance (Fa):Applicable when motion takes place through air. This force becomes appreciable for bodies moving at high speeds.

(vi) Weight (W):It is a field force, the force with which a body is pulled towards the center of the earth due to its gravity. It has the magnitude mg, where m is the mass of the body and g is the acceleration due to gravity.

(b)

Non-Contact Forces (Field Forces or Action at a Distance Forces):-

Forces which come into existence without any physical contact between the bodies. These forces are due to some inherent characteristics of the body.

Forces that act between bodies separated by a distance without any actual contact. For example, Tension, Spring, Weight etc. Examples:(i) Gravitational Forces (Fg):- This is due to the gravitation attraction between two bodies. If we deal with cases of attraction due to earth, this force is always directed downwards. (ii) Electrical Forces (Felec):- These forces are due to the charges present on the two bodies. The direction of these forces depends upon the type of charges on the two interacting bodies. (iii) Magnetic Forces (Fmag):- Forces which come into play between two bodies due to their magnetic characteristics. Their direction also depends upon the nature of magnetic polarity acquired by the body. In mechanics we shall only deal with gravitational forces from this category.

Newton’s First Law of Motion:-

To study Newton‟s first law of motion, the concept of equilibrium should be clear to us. Whenever a number of forces act on a body and they neutralize each other‟s effect, the body is said to be in equilibrium. In such a case there is no change in the state of rest or of motion. If however, the system of forces have a resultant, the state of rest or that of motion undergoes a change. This is explained by Newton‟s first law of motion. It states that,” Every body continues in its state of rest or of uniform motion in a straight line unless it is compelled by some external force to change that state. Therefore, every object persists in its natural state of motion i.e. continues to be at rest or moves in a straight line with uniform (constant) velocity, in the absence of a net external force acting (impressed) on it. It can be easily deduced from the statement of change in the state of motion. It is directly related to a frame of reference about which we have discussed earlier. To mark the point here, we can discover that by viewing objects from different frame of references the natural state of motion as perceived by different observers will be obviously different (can only be same if the frames are truly equivalent). Therefore, the change in state will also depend on the choice of reference frame. Finally, the amount of acceleration produced in a body (or change in velocity) will depend on our choice of reference frames.

Law of Inertia:-

Inertia is the property of all bodies by virtue of which they are unable to change their state of rest or of uniform motion in a straight line without the help of an external force. In other words inertia can also be termed as a resistance to change the state of motion of a body. Inertia can be classified into following three categories. (a) Inertia of Rest:It is the property of a body by virtue of which it is unable to change its state of rest without the help of an external force. (b) Inertia of Motion:It is the property of a body by virtue of which it is not able to change its speed without the help of an external force. (c) Inertia of Direction:It is the property of a body by virtue of which it is unable to change its direction of motion without the help of an external force. Qualitative definition of force from first law:Newton‟s first law states that there cannot be any change in the state of rest or that of motion of a body unless some external force acts upon it. In other words force is an agent which is capable of producing any change in state of rest or that of motion (including direction). This provides a qualitative definition of force. Some Conceptual Questions:Question 1:-

A car moving at constant speed is suddenly braked. The occupants, all wearing seat belts, are thrown forward. The instant the car stops, however, the occupants are all jerked backward. Why? Is it possible to stop an automobile without this „jerk‟?

Solution:Newton‟s first law states that, without any external force, if a body is at rest, it will remain at rest and if the body is moving with constant velocity, it will continue to do so. When the car is suddenly braked, due to the inertia, the occupants in the car will tend to move in the forward direction of car. When the car stops the sit belt in the car will produce backward momentum on the occupants. Since the all the occupants wearing seat belts, therefore the occupants are all jerked backward. Yes, it is possible to stop an automobile without this jerk. This can be done by slowing down the car a little longer time. Question 2:Why do you fall forward when a moving bus decelerates to a stop and fall backward when it accelerates from rest? Subway standees often find it convenient to face the side of the car when the train is starting or stopping and to face the front or rear when it is running at constant speed. Why? Solution:Newton‟s first law states that, without any external force, if a body is at rest, it will remain at rest and if the body is moving with constant velocity, it will continue to do so. When the moving bus decelerates, due to inertia, the body will tend to move in the forward direction of the bus. Therefore, you fall forward when the moving bus decelerates to a stop. Again, when the bus accelerates from rest, due to inertia, the body will tend to maintain the rest position. That is why; you fall backward when the bus accelerates from rest. The tendency of a body to remain at rest or in uniform linear motion is called inertia. The reference frame to which it applies is called inertial frames. When the train is starting or stopping, there is no net force acting on the observer. So the observer is in the inertial frame. That is why; it is easy to face the side of the car because both the frame (one is observer in the car and the other one is the outside of the car) are in rest. But when the car is running at constant speed, the observer in the car does not remain at rest. The frame is not an inertial frame. That is why; it is easy to face the front or rear when the car is running at constant speed because the car is in the forward direction. Question 3:A mass 'M' is lying (figure shown below) on a table which is at rest (w.r.t. the table on which it is kept). Explain its state with the help of Newton's First Law of motion.

Solution:Since 'M' is lying on a table, there is no external force acting on it (forget about gravity just for the immediate discussion). As per Newton's first law of motion it will keep on lying at rest with respect to table for infinite time. Here, comes out a very important, intrinsic (that is inherent) property of a body which is that it retains its state of motionlessness (as well as of motion, if it is in motion) which is termed as INERTIA of an object. This is present in all materialistic bodies in this universe.

Newton’s Second Law of Motion:Momentum:-

Momentum of a body is defined as the amount of motion contained in a body. Quantity of motion or the momentum of the body depends upon, (a) mass of the body. (b) velocity of the body. Therefore momentum of a body of mass „m‟ and velocity „v‟ will be,

Quantitative definition:Momentum of a body is equal to the product of its mass and velocity. Momentum is a vector quantity and possesses the direction of velocity. Units:S.I:- kg m s-1 C.G.S:- g cm s-1 Momentum can be put into following two categories. Dimension:[MLT-1] (a) Non-relativistic momentum:According to classical physics (or non-relativistic physics) which is based upon the concepts of Newton‟s laws of motion, mass of a body is considered to be a constant quantity, independent of the velocity of body. In that case momentum

is given by,

. Thus, momentum of a body is a linear function of its velocity. (b) Relativistic momentum:In accordance to Einstein‟s special theory of relativity, mass of a body depends upon the relative velocity „v‟ of the body with respect to the observer. If „m0‟ is the mass of body observed by an observer at rest with respect to body, its relativistic mass „m‟ is given by,

Therefore, momentum of a body according to the concepts of theory of relativity is given by,

Thus, relativistic momentum is not a linear function of v.

Frame of Reference:A system of co-ordinates whose axes can be suitably chosen is said to be a frame of reference. For location of a point „P‟ we need three co-ordinate x, y and z. For complete identification of an event we must know „t‟ also, i.e., the time of the occurrence. Hence an event in characterized by four co-ordinates (x,y,z,t). A reference frame describing an event in these four co-ordinates is known a space time frame.

Inertial and Non-Inertial Frame of Reference:-

(a) Inertial Frame:A frame of reference either at rest or moving with a uniform velocity (zero acceleration) is known as inertial frame. All the laws of physics hold good in such a frame. An inertial frame is endowed with the following characteristics: (i) All the fundamental laws of physics are valid in inertial frames. (ii) All the fundamental laws of physics assume the same mathematical shape in all inertial frames. (iii) They are isotopic with respect to mechanical and optical experiments (b) Non-Inertial or Accelerated Frame:- It is a frame of reference which is either having a uniform linear acceleration or is being rotated with uniform speed. Newton‟s Second Law:-

The rate of change of momentum of a body is directly proportional to the impressed force and takes place in the direction of the force. Newton‟s first law provides a qualitative definition of the force while second law provides a quantitative definition of the force. Let

be the instantaneous velocity of the body. Momentum

of the body is given by,

According to second law, ∝ (rate of change of momentum) Or,

Or,

Or,

Here „k‟ is the constant of proportionality. Mass „m‟ of a body is considered to be a constant quantity.

or,

The units of force are also selected that „k‟ becomes one. Thus, if a unit force is chosen to be the force which produces a unit acceleration in a unit mass, i.e., F = 1, m = 1 and a = 1. Then, k = 1 So, Newton‟s second law can be written , in mathematical form, as

i.e., Force = (mass) (acceleration) This provides us a measure of the force.

Here, if F = 0 then we find a = 0. This reminds us of first law of motion. That is, if net external force is absent, then there will be no change in state of motion, that means its acceleration is zero. Further we can extend second law of motion, (in fact its decomposition) to three mutually perpendicular directions as per our coordinate system. If components in x, y and z direction are Fx, Fy & Fz respectively, the three acceleration produced when Fx, Fy & Fz act simultaneously) in the body are, Now,

If we add three forces then resultant is called net external force. Similarly,

is called net acceleration produced in the body. Unit of Force:S.I:- Newton [kg.m/sec2] C.G.S:- Dyne [g.cm/sec2] Dimension:[MLT-2] Impulse:-

Impulse of a force is defined as the change in momentum produced by the force and it is equal to the product of force and the time for which it acts. Therefore, a large force acting for a short time to produce a finite change in momentum which is called impulse of this force and the force acted is called impulsive force or force of impulse.

According to Newton‟s second law of motion,

or,

So, Impulse of a force = change in momentum. If the force acts for a small duration of time, the force is called impulsive force.

As force is a variable quantity, thus impulse will be,

The area under F - t curve gives the magnitude of impulse. Impulse is a vector quantity and its direction is same as the direction of

.

Unit of Impulse:- The unit in S.I. system is kgm/sec or newton -second. Dimension:- MLT1

Problem 1:The Sun yacht Diana, designed to negative in the solar system using the pressure of the sunlight, has a sail area of 3.1 km2 and a mass of 930 kg. Near Earth‟s orbit, the sun could exert a radiation force of 29 N on its sail. (a) What acceleration would such a force impart to the craft? (b) A small acceleration can produce large effects if it acts steadily for a long enough time. Starting from rest then, how far would the craft have moved after 1 day under these conditions? (c) What would then be its speed? (See “The Wind from the Sun,” a fascinating science fiction account by Arthur C.Clarke of a Sun yacht race.) Solution:(a) Given Data:Mass of the yacht Diana, m = 930 kg Force exerted by the sun light, F = 29 N Force acting on the body (F) is equal to the product of mas of the body (m) and acceleration of the body (a). So, F = ma

…… (1)

From equation (1), the acceleration (a) of the body would be, a = F/m

…… (2)

Putting the value of m and a in equation (2), the acceleration such force impart to the craft would be, a = F/m = 29 N /930 kg = (3.1×10-2 N/kg) (1 kg. m/s2 /1 N) = 3.1×10-2 m/s2

…… (3)

Thus acceleration such force impart to the craft would be, 3.1×10-2 m/s2. (b) Given Data:Time, t = 1 day = (1day) (24 h/1 day) (60 min/1 h) (60 s/1 min) = 86400 s Initial velocity, vi = 0 Acceleration, a = 3.1×10-2 m/s2 From equation of motion, we know that, Distance travelled by the body (x) = vi + ½ at2

So, x = vit+ ½ at2

…… (4)

Putting the value of vi, a and t in equation (4), the distance travelled by the craft will be, x = vit+ ½ at2 = 0+½ (3.1×10-2 m/s2) (86400 s)2 =1.1571×108

(Since, a = 3.1×10-2 m/s2 and t = 86400 s)

…… (5)

Rounding off to two significant figures, the distance will be 1.2×108 m. Thus from the above observation we conclude that, the craft have moved1.2×108 m after 1 day under these conditions. (c) Given data: Acceleration, a = 3.1×10-2 m/s2 Time, t = 86400 s Acceleration of an object is equal to the velocity of the object divided by time. a = v/t So, v = at

……(6)

Putting the value of a and t in equation (6), velocity would be, v = at = (3.1×10-2 m/s2) (86400 s) = 2678.4 m/s Rounding off to two significant figures, speed will be 2700 m/s. Thus from the above observation we conclude that, speed will be 2700 m/s. Problem 2:A car travelling at 53 km/h hits a bridge abutment. A passenger in the car moves forward a distance of 65 cm (with respect to the road) while being brought to rest by an inflated air bag. What force (assumed constant) acts on the passenger‟s upper torso, which has a mass of 39 kg? Concept:Force acting (F) on the body is equal to the mass of the body (m) times deceleration of the body (a). F = ma …… (1) Solution:First we have to find out the deceleration (a) of the car. If v0 is the initial speed of car and v is the final speed of the car, then the average speed (vav) of the car will be, vav, = ½ (v+ v0)

…… (2)

To obtain the average speed (vav) while the car is decelerating, substitute 53 km/h for v0 and 0 m/s for v in the equation vav = ½ (v+ v0), vav = ½ (v+ v0) = ½ ((53 km/h)+ (0 m/s)) = (½ ×53 km/h) (1,000 m/1 km) (1 h/60 min) (1 min/60 s) = 7.4 m/s

…… (3)

But average speed (vav) is equal to the rate of change of displacement (x). vav = x/ t

So, t = x/ vav

…… (4)

To obtain the time of deceleration t, substitute 0.65 m for x and 7.4 m/s for vav in the equation t = x/ vav, t = x/ vav = 0.65 m /7.4 m/s = 8.8×10-2 s

…… (5)

Deceleration (a) is equal to rate of change of velocity. So, a = Δ v /t = ((0) - (53 km/h))/ 8.8×10-2 s = (-53 km/h)/ 8.8×10-2 s = ((-53 km/h) (1,000 m/1 km) (1 h/60 min) (1 min/60 s))/ 8.8×10-2 s = (-14.7 m/s)/ (8.8×10-2 s) = -1.7×102 m/s2

……(6)

To obtain the force (F) acting on the passengers upper torso having mass 39 kg, substitute 39 kg for mass mand -1.7×102 m/s2 for deceleration a in the equation, F = ma, F = ma = (39 kg) (-1.7×102 m/s2) = -6630 kg. m/s2 = -(6630 kg. m/s2) (1 N/1 kg. m/s2) = -6630 N

…… (7)

Rounding off to two significant figures, the magnitude of the force will be 6600 N. Problem 3:Workers are loading equipment into a freight elevator at the top floor of a building. However, they overload the elevator and the worn cable snaps. The mass of the loaded elevatorat the time of the accident is 1600 kg. As the elevator falls, the guide rails exert a constant retarding force of 3700 N on the elevator. At what speed does the elevator hit the bottom of the shaft 72 m below? Concept:Force acting on the body (F) is equal to the product of mas of the body (m) and acceleration of the body (a). So, F = ma From equation F = ma, the acceleration (a) of the body would be, a = F/m Weight W of the object is equal to the mass m of the object times of the free fall acceleration g. W = mg In accordance to equation of motion, the distance y travelled by the body will be, y = ut + ½ at2 Here u is the initial velocity, t is the time, and a is the acceleration. When the elevator falls, the initial velocity u will be equal to zero. So, u = 0 Substitute 0 for u in the equation y = ut + ½ at2, y = ut + ½ at2 = 0×t +½ at2

= ½ at2 So the time t will be, t = √2y/a Speed v is equal to the product of acceleration a of the body and time t. So, v = at Solution:To obtain the weight of the elevator W, substitute 1600 kg for mass of the elevator m and 9.81 m/s2 for free fall acceleration g in the equation W = (m) (g), W = (m) (g) = (1600 kg) (9.81 m/s2) =15680 kg,m/s2 = (15680 kg,m/s2) (1 N/1 kg,m/s2) = 15680 N The magnitude of the net force F will be, F = W-R To obtain the magnitude of the net force F, substitute 15680 N for W and 3700 N for retarding force R in the equation F = W-R, F = W-R = (15680 N) – (3700 N) =11980 N Rounding off to two significant figures, the magnitude of the net force F will be 12000 N. To obtain the acceleration a, substitute 12000 N for F and 1600 kg for m in the equation a = F/m, we get, a = F/m = 12000 N/1600 kg = (7.5 N/kg) (1 kg.m/s2/1 N) = 7.5 m/s2 To obtain the time t to fall, substitute -72 m for y and -7.5 m/s2 for a in the equation t = √2y/a, t = √2y/a = √2(-72 m) /(-7.5 m/s2) = 4.4 s To obtain the final speed v at which the elevator hits the bottom of the shaft 72 m below, substitute 7.5 m/s2(only magnitude of a) for a and 4.4 s for t in the equation v = at, we get, v = at = (7.5 m/s2) (4.4 s) = 33 m/s From the above observation we conclude that, the speed at which the elevator hits the bottom of the shaft 72 m below would be 33 m/s.

Newton’s Third Law of Motion:It states that,

“To every action there is an equal and opposite reaction”.

Whenever one force acts on a body, it gives rise to another force calledreaction. A single isolated force is an impossibility. The two forces involved in any interaction between two bodies are called “action” and “reaction”. But this does not imply any difference in their nature, or that one force is the „cause„ and the other is the „effect‟. Either force may be considered as „action‟ and the other „reaction‟ to it. It may be noted that action and reaction never act on same body. Note: The most important fact to notice here is that these oppositely directed equal action and reaction can never balance or cancel each other because they always act, on two different point (broadly on two different objects) For balancing any two forces the first requirement is that they should act one and the same object. (or point, if object can be treated as a point mass, which is a common practice) Few examples on Newton’s third law of motion: (a)

Book kept on a table:-

A book lying on a table exerts a force on the table which is equal to the weight of the book. This is the force of action. The table supports the book, by exerting an equal force on the book. This is the force of reaction, as shown in the below figure. As the system is at rest, net force on it is zero. Therefore, forces of action and reaction must be equal and opposite.

(b) Walking on the ground:-

While walking a person presses the ground in the backward direction (action) by his feet. The ground pushes the person in forward direction with an equal force (reaction). The component of reaction in the horizontal direction makes the person move forward. (c) Process of Swimming:A swimmer pushes the water backwards (action). The water pushed the swimmer forward (reaction) with the same force. Hence the swimmer swims. (d) Firing from a gun:When a gun is fired, the bullet moves forward (action). The gun recoils backwards (reaction).

(e) Fight of jet planes and rockets:The burnt fuel which appears in the form of hot and highly compressed gases escapes through the nozzle (action) in the backward direction. The escaping gases push the jet plane or rocket forward (reaction) with the same force, hence, the jet or rocket moves. (f) Rubber ball re-bounds from a wall:When a rubber ball is struck against a wall or floor it exerts a force on a wall (action). The ball rebounds with an equal force (reaction) exerted by the wall or floor on the ball. (g) It is difficult to walk on sand or ice:This is because on pushing, sand gets displaced and reaction from sandy ground is very little. In case of ice, force of reaction is again small because friction between feet and ice is very small. (h) Driving a nail in to a wooden block without holding the block is difficult:This is because when the wooden block is not resting against a support, the block and nails both move forward on being hit with a hammer. However, when the block is held firmly against a support, and the nail is hit, an equal reaction of the support drives the nail into the block. (i) A tea cup breaks on falling on the ground:Tea cup exerts certain force (action) on ground while the ground exerts an equal and opposite reaction on the cup. Ground is able to withstand the action of cup, but the cup being relatively more delicate breaks due to reaction.

Problem 1:Two blocks, with masses m1 = 4.6 kg and m2 = 3.8 kg, are connected by a light spring on a horizontal frictionless table. At a certain instant, when m2 has an acceleration a2 = 2.6 m/s2, (a) what is the force on m2 and (b) what is the acceleration of m1? Concept:Force acting on the body (F) is equal to the product of mas of the body (m) and acceleration of the body (a). So, F = ma From equation F = ma, the acceleration (a) of the body would be, a = F/m Solution:(a) The net force ∑ Fx on the second box having mass m2 will be, ∑ Fx = m2a2x Here a2x is the acceleration of the second block. To obtain the net force ∑ Fx on the second box having mass m2, substitute 3.8 kg mass m2 and 2.6 m/s2 fora2x in the equation ∑ Fx = m2a2x, ∑ Fx = m2a2x = (3.8 kg) (2.6 m/s2)= 9.9 kg .m/s2 = (9.9 kg .m/s2) (1 N/ 1 kg .m/s2)= 9.9 N From the above observation we conclude that, the net force ∑ Fx on the second box having mass m2 would be 9.9 N. There is only one (relevant) force on the block, the force of block 1 on block 2. (b) There is only one (relevant) force on block 1, the force of block 2 on block 1. By Newton‟s third law this force has a magnitude of 9.9 N. So the Newton‟s second law gives, ∑ Fx = m1a1x = -9.9 N But, m1a1x = (4.6 kg) (a1x)

(Since, m1 = 4.6 kg)

(4.6 kg) (a1x) = -9.9 N So, a1x = -9.9 N/4.6 kg = (- 2.2 N/kg) (1 kg.m/s2 / 1 N) = -2.2 m/s2 From the above observation we conclude that, the acceleration of m1 will be -2.2 m/s2. _______________________________________________________________________________________________ Problem 2:A meteor of mass 0.25 kg is falling vertically through Earth‟s atmosphere with an acceleration of 9.2 m/s2. In addition to gravity, a vertical retarding force (due to frictional drag of the atmosphere) acts on the meteor as shown in the below figure. What is the magnitude of this retarding force?

Solution:Given Data: Mass of the meteor, m = 0.25 kg Acceleration of the meteor, a = 9.2 m/s2 The net force exerted (Fnet) on the meteor will be, Fnet = ma = (0.25 kg) (9.2 m/s2) = (2.30 kg. m/s2) (1 N/ 1 kg. m/s2) = 2.30 N

…… (1)

If g (g = 9.80 m/s2) is the free fall acceleration of meteor, then the weight of the meteor (W) will be, W = mg = (0.25 kg) (9.80 m/s2) = (2.45 kg. m/s2) (1 N/ 1 kg. m/s2) = 2.45 N

…… (2)

The vertical retarding force would be equal to the net force exerted on the meteor (Fnet) minus weight of the meteor (W). So, vertical retarding force = Fnet –W

…… (3)

Putting the value of Fnet and W in equation (3), the vertical retarding force will be, Vertical retarding force = Fnet –W = 2.30 N -2.45 N = -0.15 N

……. (4)

From equation (4) we observed that, magnitude of the vertical retarding force would be, -0.15 N. _______________________________________________________________________________________________ Problem 3:Suppose in figure shown above we put one more block of 5 kg mass adjacent to 10 kg and a force of 150 N acts as shown in the figure below, then find the forces acting on the interface.

Solution:The

combined

acceleration

a = F/((10+5))=150/15=10/sec

of

the

two

bodies

when

treated

as

one

is

2

So each one moves with a = 10m/sec2 keeping their contact established. Here you can feel that due to 150N force the body of 5 kg feels as if it is being pushed by the 10 kg mass. There is force acting on 5kg called R1, to oppose it by third law this body exerts a force R2 on 10kg. The interface is as shown in Figure given below.

Also, third law tells us that R1 = R2 in magnitude and is opposite in direction.

R1 = R2 = R Here since 150 N force acts on the 10kg mass and only r acts on the 5kg mass. For motion in 5kg only R is responsible. We can write the initial equation as: F = 150 = (10 + 5) a 150 = 10a + 5a Here 10a is force experienced by 10kg mass. And 5a is experienced by 5kg mass. R = 5a a = 10m/sec2 So,R = 50N Thus,Net force experienced by 10kg block is (150-R) = 10a 150-R = 1010 = 100 N Therefore, R = 50 Therefore we get R = 50N for both blocks. Hence we find "action and reaction are equal and opposite". Now net force on the body of 10kg mass is 100N & Net force on the body of 5kg mass is 50N and on the interface action and reaction are both equal and also are equal to force experienced by second body. _________________________________________________________________________________________________ Problem 4:An object is hung from a spring scale attached to the ceiling of an elevator. The scale reads 65 N when the elevator is standing still. (a) What is the reading when the elevator is moving upward with constant speed of 7.6 m/s? (b) What is the reading of the scale when the elevator is moving upward with a speed of 7.6 m/s and decelerating at 2.4 m/s2? Solution:Weight of the object (W) when the elevator is standing, W = 65 N (a) We have to find out the scale reading when the elevator is moving upward with a constant speed of 7.6 m/s. Since the elevator is moving upward with a constant speed, therefore there is no acceleration of the system resulting there is no force. Thus the scale reading must be equal to the weight of the object and that will be 65 N. (b) We have to find out the scale reading when the elevator is moving upward with a speed of 7.6 m/s and decelerating at 2.4 m/s2.

The force exerted on the object due to the deceleration at the rate 2.4 m/s2 (a = - 2.4 m/s2) will be, F = ma = (W/g) a

…… (1)

Where W is the weight of the object (W = 65 N) when the elevator is at rest and g is the free fall acceleration of the object (g=9.80 m/s2). Putting the value of W, g and a in equation (1) the force exerted on the object will be, F = (W/g) a = (65 N/9.80 m/s2) (-2.4 m/s2) = -15.92 N …… (2) (Rounding off to two significant figure)

= -16 N

When the elevator is moving upward with a speed of 7.6 m/s and decelerating at 2.4 m/s2, the force would be, = F- (-W) = -16 N –(-65 N) = -16 N+65 N = 49 N

…… (3)

From the above observation we conclude that, the scale reading when the elevator is moving upward with a speed of 7.6 m/s and decelerating at 2.4 m/s2 would be 49 N. Frame of Reference:A system of co-ordinates whose axes can be suitably chosen is said to be a frame of reference. For location of a point „P‟ we need three co-ordinate x, y and z. For complete identification of an event we must know „t‟ also, i.e., the time of the occurrence. Hence an event in characterized by four co-ordinates (x,y,z,t). A reference frame describing an event in these four co-ordinates is known a space time frame.

Inertial and Non-Inertial Frame of Reference:In general we solve the problem of mechanics using inertial frame, which was discussed in chapter two, but as the same time it is possible to solve the same problem using a non-inertial frame. Let us discuss about the difference between these frames. When Newton stated his first law he made a very important distinction. He decreed the absolute equivalence between a state of rest and one f uniform motion and distinguished it specifically and absolutely from that of an accelerated motion. If the environment is completely symmetric then no direction is preferred over another and therefore if a body possesses a initial velocity (which might be zero) it will persist with that velocity. If suppose we say that the velocity will change then we will have to concede that the velocity

changes in a particular direction. But why should it change in one direction and not in the other, since all direction are equally favored. So the only way it can change is to change in all directions. But this is impossible so it will not change at all, i.e. if environment is really symmetric. Therefore if we grant a change in velocity we will also have to grant an irregularity in the environment in the same direction as the change in velocity. The acceleration, he said to be understood as an irregularity and he expressed force as that basic asymmetry in the environment which produces this irregularity. The most important aspect in all this is that force is theoretical construction to explain away the irregularities in motion and is not to be understood as a tangible entity.

Now, for constant mass system If force is a tangible entity then the force in all systems on the same body should be same. Let us see if this is true. Consider a body of mass m. We will observe its motion from three different frames. Simulation for Frame of Reference:(i)

Reference frame is at rest:-

The acceleration of the mass will be, say, Therefore the force on it will be

.

.

We will reason that

(ii) Reference frame starts moving with constant velocity vector-v :The acceleration of frame = Thus, acceleration of mass m relative to frame is given by

Force on it will be

inertial and we will reason that

(iii) Reference frame moves with constant acceleration:Let the acceleration of frame be

.

Thus, acceleration of mass relative to frame will be

Let there be force

.

frame on mass we will reason, that

We see that the force is not the same as that in the inertial frames. Therefore we postulate that under observation from an accelerated reference frame we substitute the inertial forces on the body with those same initial forces plus an additional force which numerically equal to the mass of the body under observation times the acceleration of the frame taken in the opposite direction. This force we call as pseudo force. Now, we can work on a problem from an accelerated reference frame by just adding a pseudo force and pretending that nothing has changed. Let us illustrate our point. (a) Inertial Frame:A frame of reference either at rest or moving with a uniform velocity (zero acceleration) is known as inertial frame. All the laws of physics hold good in such a frame.

(b) Non-Inertial or Accelerated Frame:It is a frame of reference which is either having a uniform linear acceleration or is being rotated with uniform speed.

An inertial frame is endowed with the following characteristics:

   

All the fundamental laws of physics are valid in inertial frames. All the fundamental laws of physics assume the same mathematical shape in all inertial frames. Inertial frames are isotopic with respect to mechanical and optical experiments The optical experiments performed in an inertial frame in any direction will always yield the same results.

Is earth an inertial frame of reference?

Earth rotates around its axis as also revolves around the sun. In both these motion, centripetal acceleration is present. Therefore, strictly speaking earth or any frame of reference fixed on earth cannot be taken as an inertial fame. However, as we are dealing with speeds ≈ x 108 ms-1 (speed of light) and speed of earth is only about 3 x 104 m/s, therefore when small time intervals are involved effect of rotation and revolution of earth can be ignored. Furthermore, this speed of earth can be assumed to be constant. Hence earth or any other frame of reference set up on earth can be taken as an approximately inertial frame of reference. On the contrary, a frame of reference which is accelerated or decelerated is a non-inertial frame. Other examples of inertial frames of reference are: (i)

A frame of reference remaining fixed w.r.t. stars.

(ii)

A space ship moving in outer space, without spinning and with its engine cut off.

Apparent weight of a man inside a lift:-

(a) The lift possesses zero acceleration (fig-1): W = mg (b) The lift moving upward with an acceleration a (fig-2): W = mg + ma = mg + mg = 2 mg (c) The lift moving downward with an acceleration a (fig-3): W = mg – ma = mg – mg =0 Conceptual Problem:Problem 1:Suppose that you are standing on the balcony of a tall tower, facing east. You drop an object so that it falls to the ground below; see below figure. Suppose also that you can locate the impact point very precisely. Will the object strike the ground at a, vertically below the release point, at b to the east, or at c to the west? The object was released from rest; the Earth rotates from west to east.

Solution:If one assumes that the factor by which the Earth rotates is negligible during the time the object takes to reach the ground, then the object will hit the ground at point a and one will not have to concern about theCarioles Effect. In case the factor is non-negligible, and the Earth moves from West to East, the object will hit the ground at point c. The situation is similar to the perceived leftward displacement of the air moving to a low pressure point from north to south. To the person standing on the balcony of a tall tower, a psuedo force has acted on the object in a direction from east to west. ____________________________________________________________________________________________________ Problem 2:What is the distinction between inertial reference frames and those differing only by a translation or rotation of the axes? Solution:One can distinguish the inertial frame of reference against the translation of the axes when the translation or rotation occurs nonuniformly. If the translation of one frame of reference relative to the other is such that observer in one frame measures some acceleration of the other, then the observer can draw a distinction between his frame and the inertial frame of reference. It is important to note that the rotating frame will always be non-inertial because to account for the rotation there must be a change in velocity vector. Therefore the inertial reference frame will always be distinct from the rotation frame. However the translation can be distinguished from the inertial frame of reference only when the translation occurs at a uniform velocity. _____________________________________________________________________________________________________ Problem 3:-

A passenger in the front seat of a car finds himself sliding toward the door as the driver makes a sudden left turn. Describe the forces on the passenger and on the car at this instant if the motion is viewed from a reference frame (a) attached to the Earth and (b) attached to the car. Solution:(a) From the reference frame attached to the Earth, when the car takes a left turn, the inertia of the passenger would maintain the state of motion of the passenger (in accordance with Newton‟s first law of motion). This causes the passenger to move in the direction of velocity vector of car before car took a turn. Therefore, to the observer at Earth frame, the passenger has the initial velocity equal to the velocity of the car before it takes the turn. The observer also see the car taking a turn, and accounts for an acceleration for the same. Thus, the observer would expect frictional forces to exist between the passenger and the seat.

Given that the passenger has moved towards the door when the car took a left turn, it is obvious that the passenger moves to the right. Therefore the frictional force would be in a direction opposite to the motion of the passenger, that is, towards the left. If you assume that the frictional force between the passenger and seat is given by f , mass of passenger by m whereas the acceleration by a. Then the equation depicting the motion of passenger relative to the observer in Earth‟s frame is: f = ma Therefore the magnitude of frictional force between the passenger and the seat is equal to the magnitude of the deceleration of the passenger. (b) According to the observer in car‟s frame, the car experiences a centripetal acceleration when it turns to left. Therefore a psuedo force will act on the passenger, accelerating him in the direction of the seat. But there is a friction between the seat of the car and the passenger, which decelerate the passenger in a direction opposite to the direction of its motion. Thus the magnitude of net force acting on the passenger is the difference between magnitude the centrifugal force and magnitude the frictional force. If you assume that the centrifugal force acting on passenger is F, the frictional force between the passenger and the seat is f , the mass of passenger is m whereas the acceleration is a. Then the equation depicting the motion of passenger relative to the observer in car‟s frame is: F – f = ma Therefore one can see that the perceived acceleration for the observer in car differs from that of the observer on Earth. __________________________________________________________________________________________________

Problem 4:Do you have to be concerned with the carioles effect when playing tennis or golf? If not, why not?

Solution:No, one does not have to concern with the Carioles Effect when playing tennis or golf because the factor by which the Earth rotates during the time the ball goes from its source to the destination, is very small and can be neglected. Therefore the player can take his/her shot as if the destination of the ball would be at the same position as it was at the time the shot was taken.

Friction:-

Whenever the surface of a body slides over another, each body experiences a contact force which always opposes the relative motion between the surfaces. This contact force is called frictional force. Intermolecular interaction arising due to elastic properties of matter is the cause of frictional force.This force acts tangentially to the interface of two bodies. Cause of friction:Old view:Earlier it was thought that roughness of the two surfaces causes friction in the figure because it can be easily seen that smoother the surfaces, lesser is the friction. Interlocking of irregularities of the two surfaces causes hindrance to sliding. This, however, is not the current view. Current view:The current view is a slight deviation from the old view. Earlier we thought that interlocking of irregularities of surfaces was causing friction. Now, it is though that due to irregularities, the common surface area which is in actual contact of the two surfaces, is much less than the total overall area in contact. In one experiment, it came out to be 1/10,000th of the apparent area. Thus, while the total interactive (action and reaction) forces between the two surfaces remain the same, the pressures at the points of contact are extremely high and cause the humps to flatten out (undergoing plastic deformation) until the increased area of contact

enables the upper solid to be supported. It is thought that at the points of contact, small, cold-welded joints are formed by the strong adhesive forces between molecules which are very close together. These have to be broken away before one surface can move over the other. Thus the force of friction is found to depend upon the following factors.

(i) The nature of two surfaces with the surfaces are being pressed together. (ii) Normal force with the surfaces are being pressed together. (iii) Actual area of contact Types of Friction:There are four types of friction. (a) Static Friction (b) Kinetic Friction

(c) Rolling Friction

(d) Fluid Friction

(a)Static friction:-

Static friction is the force of friction between two surfaces so long as there is no relative motion between them. It is always equal to the applied force. The static frictional forces are incorporated in the following inequality.

The magnitude of static friction fs (static frictional force) has a maximum value fs,max that is given by, fs,max = µsN

Here µs is the coefficient of kinetic friction and N is the normal force. So, coefficient of static friction, µs= fs,max /N Static friction is always equal to the apllied force. It will be observe that value of static friction increases to certain maximum value, beyond which if the applied force is increased body starts moving. This maximum value of force of friction is called limiting friction. Limiting friction is the maximum value of force of friction between two surfaces so long as there is no relative motion between them. (b) Kinetic friction:Kinetic friction is the force of friction which comes into play between two surfaces when there is some relative motion between them. The magnitude of force of kinetic friction fk (kinetic frictional force) is proportional to the normal force N.

So,

Here µk is the coefficient of kinetic friction. Thus, coefficient of kinetic friction, µk = fk/N Laws of limiting friction:(a) The direction of force of friction is always opposite to the direction of motion. (b) The force of limiting friction depends upon the nature and state of polish of the surfaces in contact and acts tangentially to the interface between the two surfaces. (c) The magnitude of limiting friction „F‟ is directly proportional to the magnitude of the normal reaction R between the two surfaces in contact, i.e., F∝R (d) The magnitude of the limiting friction between two surfaces is independent of the area and shape of the surfaces in contact so long as the normal reaction remains the same. (c) Rolling friction:Force of friction which comes into play, between two surfaces, while one is rolling over the other is called rolling friction. Rolling friction is similar to kinetic friction. So,

Here µr is the coefficient of rolling friction and N is the normal force. (d) Fluid friction:Fluid friction is the opposing force which comes into play when a body moves through a fluid. Cause and direction of rolling friction:-

A wheel of radius R rolling without sliding on a flat surface will experience a resistance due to the very small local deformation that takes place, which is sometimes elastic sometimes inelastic, i.e., a kind of ridge is formed in front of the wheel as shown exaggerated in the figure. This gives rise to the force FR, whose line of action passes through the center C of the wheel and P the horizontal force necessary to force the wheel to topple over the point M, the total clockwise torque acting on the wheel about M must be more than or marginally more than the total anticlockwise torque about M. F = Frictional force Therefore, P × R cos θ > mg × R sin θ P > mg tan θ The value of "tan θ" is called the coefficient of rolling friction (μg). This value does not depend upon R. If the two surfaces are absolutely rigid, then no ridge will be formed and q will be zero i.e., coefficient of rolling friction will be zero. Typical values are μR = 0.006 for steel and 0.02 - 0.04 for rubber tiers on concrete surfaces. Rolling friction is very small compared to the sliding friction. In case of pure rolling μR = 0.



The value of angle of friction and that of angle of repose are same and the tangents of both of them is equal to the co-efficient of friction.



When a body slides down an inclined plane, whose angle of inclination with the horizontal is equal to angle of repose, it moves with uniform velocity.



Sliding friction and rolling friction are independent of velocity.



Fluid friction depends upon velocity. It increases with an increase in velocity.



Force of limiting friction does not depend upon the size and shape of surface in contact.



Coefficient of kinetic friction is less than coefficient of static friction i.e.,µk ?µs Co-efficient of Friction:According to the law of limiting friction, F∝ R Or F = μR

............ (1)

where μ is a constant of proportionality and is called the coefficient of limiting friction between the two surfaces in contact. From (1),

Hence coefficient of limiting friction between any two surfaces in contact is defined as the ratio of the force of limiting friction and normal reaction between them. The value of μ depends on (i) nature of the surfaces in contact i.e., whether dry or wet; rough or smooth; polished or not polished. (ii) material of the surfaces in contact. For example, when two polished metal surfaces are in contact, μ ≈ 0.2, when these surfaces are lubricated, μ ≈ 0.1. Between two smooth wooden surfaces, μ varies between 0.2 and 0.5. Obviously, μ has no units. When a body is actually moving over the surface of another body, we place F by Fx, the kinetic friction, and μ and μk. Therefore,

μk is then called the coefficient of kinetic or dynamic friction. As Fk < F, therefore, μk is always less than μ i.e. coefficient of kinetic or dynamic friction is always less than the coefficient of limiting friction. Table gives the values of coefficient of limiting/kinetic friction between some pairs of materials: S.No.

Surface in contact

Coefficient of limiting friction

Coefficient of kinetic friction

1.

Wood on wood

0.70

0.40

2.

Wood on leather

0.50

0.40

3.

Steel on Steel (mild)

0.74

0.57

4.

Steel on Steel (hard)

0.78

0.42

5.

Steel on Steel (greased)

0.10

0.05

Angle of Friction:-

The angle made by the resultant reaction force with the vertical (normal reaction) is known as the angle of the friction. Now, in the triangle OAB, AB/OB = cotθ So, OB = AB/ cotθ = AB tanθ Or, tanθ = OB/AB =f/N So, tanθ = f / N = µs Angle of Repose:It is the angle which an inclined plane makes with the horizontal so that a body placed over it just begins to slide of its own accord.

Consider a body of mass m resting on an inclined plane of inclination θ . The forces acting on the body are shown – Ff being the force of friction. If friction is large enough, the body will not slide down. Along x: mg sin θ – f = 0 Along y: N –mg cosθ = 0 i.e. N = mg cos θ and f = mg sin θ Thus, f ≤ µsN gives, mg sin θ ≤ µs mg cosθ So, tan θ ≤ µs. This signifies, the coefficient of static friction between the two surfaces, in order that the body doesn‟t slide down. When θ is increased, then tan θ > µ . Thus sliding begins, and the angle θr = tan-1µ. This angle is known as the angle of repose. Methods of Reducing Friction:Friction can be reduced if we try to remove the cause of friction. (a) By rubbing and polishing (b) By lubricants (c) By converting sliding into rolling friction (d) By streamlining Problem:A horizontal bar is used to support a 75-kg object between two walls, as shown in the below figure. The equal forces F exerted by the bar against the walls can be varied by adjusting the length of the bar. Only friction between the ends of the bar and the walls supports the system. The coefficient of static friction between bar and walls is 0.41. Find the minimum value of the forces F for the system to remain at rest.

Concept:-

The diagram below shows the forces involved in the system: The magnitude of frictional force between the walls and the bar is, f = µsN Here µs is the coefficient of static friction between the walls and the bar , and N is the normal force exerted by the wall on the bar. If the bar is at rest, the horizontal equilibrium is maintained and the sum of the horizontal forces on the bar must be zero, that is

F–N=0 F=N Substitute F = N in equation f = µsN , f = µsN =µs F To account for the vertical equilibrium of the block, the sum of the vertical forces must be zero,

f+f–W =0 f+f=W 2f = W It is important to note that the term on the left hand side of the above equation account for the fact that the frictional force exist at both the ends of the bar and acts in a direction opposite to the direction of weight. The weight of the block is calculated by multiplying the mass of the block with acceleration due to gravity, W = mg Substitute W = mg and f = µsF in equation 2f = W, 2f = W 2(µsF) = W F = mg/2µs This equation can be used to calculate the magnitude of force exerted by the bar on the wall that will balance the system. Solution:To calculate the magnitude of force F, substitute 75 kg for m , 9.81 m/s2 for g and 0.41 for µs in equation F = mg/2µs , F = mg/2µs = (75 kg) (9.81 m/s2)/2(0.41) = (897.2 kg.m/s2) (1 N/1 kg.m/s2) = 897.2 N Round off to two significant figures, F= 897.2 N Therefore, the magnitude of the force exerted by the bar on the wall, to balance the system is 897.2 N.

Free body diagram It is to be noticed here that during previous examples, we were using a concept called FBD implicitly which can now be brought to you conscious attention.

It is like this. Whenever one attempts a problem involving forces and acceleration (say of dynamics or statics) one must show all forces and acceleration (possible acceleration may be unknown also) on each part of the system treating that part separately (it is called dividing system into possible subsystems). By all forces we mean external as well as internal forces (internal forces refer to mutual reactions). Now each system is ready to get treatment of laws of motion e.g. acceleration, velocity. Coming to actual situation one should first identify all the component involved in the system, say mass m1, mass m2 .........., pulley 1, pulley 2, etc. Now separate them from others by cutting the string contacts (sort of imaginary separation) In effect, make them free (that's where comes the name free body) from other components and at the same time show all the forces acting on it, external as well as internal, arising due to separation from other parts. These are called mutual interaction forces (one of this type we have seen in the illustration of IIIrd law) and show all possible acceleration. The diagram thus obtained is called a free body diagram Illustration : Let us draw FBD for various given systems [Here we are assuming that all the surfaces strings and pulleys are ideal that is we are neglecting their masses & any friction present] FBD (1) : Block of mass M is resting on a frictionless rigid surface

There are only two forces in the system in figure above. mg which is the weight of the block, and is acting on the surface. R2 = mg through its centre (since the body is symmetric). So, here itself from FBD we can see that net external force on the block is zero. That is why, it is stationary on the surface.

FBD (2) : Draw the free body diagram of the block shown in figure 1.12.

Where R is reaction from the surface (R = Mg) FBD (3) : Draw free body diagrams of both the blocks (figure 1.13 a), Assuming a reaction of magnitude 'R' is present at the interface.

Now a question comes Can 'R' have a direction opposite to what is shown here? Answer is, of course it can have,

[before we proceed further let me point out that once you make calculations you will find that values of R will automatically come negative, which tells us that earlier direction were the correct ones]. FBD (4) : Suppose situations is as shown in figure above that a light inextensible string pulls a block of mass M on a frictionless rigid surface.

Here string is acting as a force transmitting element It will experience a tension T in it. Let us cut it (imagine) at then the situation is as shown in figure shown below.

(We join it again then net force should become zero at that point. Since t, and T are oppositely directed their sum will come out to be zero). So tension t is responsible for dragging mass of block. Therefore free body representation is

Caution : What happens if rope is not massless? FBD (5) : Draw free body in case of system shown (figure shown below).

Suppose T is tension in the string, then by cutting it at 1-1' and 2-2' we can draw FBD's as shown in figure shown below.

R1 and R2 are reactions from plane, and from Newton's law they are equal to the weights of respective blocks.

FBD (6) : Draw FBD where one of the block is resting on an inclined plane and rope goes over a frictionless massless pulley (Figure given below).

FBD becomes

Here the noticeable fact is that R2 is perpendicular to the inclined plane and mg is perpendicular to horizontal plane. Since Mg is always directed downwards. R2 is due to the component of Mg, which is acting, perpendicular to inclined plane. (Here, the rule is that reaction force is always normal to the surface, which provides the reaction). Here for "ease" we can resolve force Mg in two components. One parallel to inclined plane and other on perpendicular to inclined plane. As shown in figure below.

.·. free Body, of the block on the inclined plane can be represented by figure shown below.

Finally at the end of this FBD, I want to point out that T2 will always be equal to T2for a continuous homogeneous massless inextensible string passing over a massless frictionless pulley. FBD (7) : Suppose both the block are on inclined planes as shown in figure given below.

For FBD cut the strings at 1-1' and 2-2' and separate the two blocks and pulleys as subsystems.

FBD (8) : If pulley is hanging from a rigid support say roof and masses are connected as shown in figure given below.

FBD's become.

Once we are comfortable in drawing FBD then we can proceed to write equation of motion, using Newton's II Law. But prior to that, it is necessary to briefly introduce you with the concept of equilibrium of bodies.

Equilibrium: An object is said to be in equilibrium if the vector sum of all the forces acting on the body (externally applied + forces arising due to mutual interaction;) is zero.

That is

1+

2 +......+

N=0

But since it is impractical to apply it, as it is, to problems, therefore we resolve all forces in the three directions X, Y & Z equilibrium and if some of the forces is zero in each direction then the body is said to be in equilibrium. (to be more specific - translator equilibrium). F1x + F2x + ............ + Fnx = 0 F1y + F2y + ............ + Fny = 0 F1z + F2z + ............ + Fnz = 0 Enquiry : How to apply equations of motion to any problem? Writing Down Equations of Motion: Once we have made a free body diagram then we can write equations of motion for each part of the system, for which we have drawn FBD. To write down equations of motion for a sub-system for which we have already drawn the FBD the requirement is to choose two direction (if required we will need three directions) in which we shall work to reduce the complexity say, direction X and Y

which in most of the cases is natural. In other cases, where we are dealing with

inclined planes we can fix our coordinate axis x-y in any desired orientation which reduces our trouble of finding out components of the forces which are acting in various directions and by experience one knows that it is better to choose X axis parallel to the inclined plane and Y axis perpendicular to it. For inclined plane

is the right choice of coordinate axis, tilted as per

inclination. Now once can go for writing F = ma equations in two directions for any ith sub-system. ∑Fx = miax

∑Fy = miay Let us write down equation of motion for the FBD's we have already drawn. Refer to FBD (2):

Here figure is given below.

∑Fy = R - Mg It will be zero since we know that on the surface the body can't move upwards or downwards. .·. R - mg = 0

=> R = Mg and ∑Fx = F

And this alone should accelerate it ∑ F = Ma, => a = F/m Refer to FBD (3): Here Figure is given below.

for the smaller block ∑Fy = R1 - mg = 0 ∑Fx = F-R For the Large Block ∑Fx = R

.·. R1 = mg .·. F-R = m a1 ∑Fy = R2 - Mg = 0 .·. R2 = Mg .·. R = Ma2

Since both Blocks move as on system, their accelerations will be equal a1 = a2 & R = Ma

.·. F - R = ma from here we can solve for R & a.

Refer to FBD (4): Here figure is given below.

For block ∑Fy = R - Mg = 0 .·. R = mg ∑Fx = T = Ma .·. a = T/m Refer to FBD (5): Here Figure is given below. Here also, since system is fully connected and we assume that string is not loose, then the acceleration in different parts will be equal. That is, Blocks m and M as well as the string will move with the same acceleration 'a'.

Equation for smaller Block ∑Fy = R1 - mg = 0 .·. R1 = mg ∑Fx = T = ma ............ (1) For Block - 'M'

∑Fy = R2 - Mg = 0

.·. R2 = Mg ∑Fx = F - T = Ma ......... (ii) solving (i) & (ii) we get T & a. Refer to FBD (6): Here Figure is given below. Since, system is fully connected by light inextensible string therefore acceleration in all the parts will be equal.

For Block 'm' R1 - mg = 0 .·. R1 = mg T1 = ma. For Block 'M' (Refer to the diagram made with mg cos α & mg sin α components. R2 - Mg cos α = 0 (since it is not moving to the plane) R2 = Mg cos α and is parallel to the plane Mg sin α - T2 = Ma

............ (ii)

It is already explained that T1 = T2 = T .·. solving (i) & (ii) we get 'T' & 'a' for the system. Refer to FBD (7) : Here Figure is given below. T1 = T2 = T (similar to the analysis done in FBD (6))

For block m1 perpendicular to inclined plane T1 - m1g cos α = 0 R1 = m1g cos α. Parallel to the plane. T-m1g sin α = m1a ............. (1) For block m2 perpendicular to plane R2 - m2g cos β = 0 R2 = m2g cos β and m2 g sin β - T = m2 a ...... (ii) Solving (i) & (ii) we get 'a' & 'T. Refer to FBD (8) : Here Figure is given below.

In Y-direction : For block m1 T - m1g = m1a For Block m2 m2g - T = m2 a From (i) & (ii) we can solve for 'T' & 'a'. On pulley 2T force is acting downwards. Therefore the pulley experiences a force F = 2T. Now you are well conversant with making FBD's and writing down equations of motion for the given system. Therefore, now we can write down some definite steps to follow

for any given problem, which can considerably reduce your diversion or confusion or say the possibility of getting trapped into complexity of a problem. (of course for that you have to use your brain all the time with these steps). (i)

Draw the fully connected clear diagram.

(ii)

Define subsystems, that is parts of the system on which you will work to get your answers.

(iii)

Draw free body diagrams for all possible subsystems.

(iv)

Resolve forces as well as accelerations in x y & z direction. (or perpendicular and || to the plane whenever require).

(v)

Write, ∑Fx, ∑Fy, ∑Fz equations with the physical constraints appearing in the problem.

(vi)

Eliminate some variables and get the required one(s).

Centripetal Force If a body is moving with a constant speed in a circle, as seen from an inertial frame, it is continuously towards the centre of rotation with magnitude vr/r (known as centripetal acceleration), where v is the speed of the particle and r is the radius of the circular path. According to Newton's second law, this body will experience net force directed towards the centre called the centripetal force. Therefore, net force acting on the body towards the centre = mv2/r, where m is mass of body. Centrifugal force is a pseudo force acting on the body from a rotating frame. Illustration: A bob of mass m is suspended form a inextensible, massless describe a horizontal uniform circular motion as shown in figure 1.42. about a vertical. Analyze the dynamics of this system.

Solution: The path of the system described above is shown in figure 1.42. Let the radius of circular path of bob is r equal to l sin θ and tension in string is T. The string makes an angle θ with the vertical. Consider the bob is at A. We can draw the free body diagram of bob at a as shown in figure 1.43. The force acting on the bob is it's weight mg and tension T of the string. Tenstion T is resolved in two components T cos θ and T sin θ as shown in figure 1.43. we can write the equation of motion

T cos θ = mg

T sin θ = mv2/r

Banking of Roads:-

Perhaps you have noticed that when a road is straight, it is horizontal too. However, when a sharp turn comes, the surface of the road does not remain horizontal. This is called banking of the roads.

Purpose of banking:-

Case I: If coefficient of friction, μ = 0:What we really wish is that even if there is no friction between the tyres and the road, yet we should be able to take a round turn. In the given figure Vertical N cos θ component of the normal reaction N will be equal to mg and the horizontal N sin θ component will provide for the necessary centripetal force. [Please note that as we are assuming μ to be zero here, the total reaction of the road will be the normal reaction.] Frictional forces will not act in such a case.

Thus, N cos θ = mg

.......... (i)

2

N sin θ = mv /r

.......... (ii)

Dividing equation (ii) by (i), we get tan θ = v2/rg

where θ is the angle of banking. Case - II : If coefficient of friction, μ ≠ 0:In the above figure shows a section of the banked road and the view of the vehicle form the rear end. vehicle-form-the-rear-end The total forces acting are, N1 and N2 = normal reactions f1 and f2 = frictional forces mg = weight r = radius θ = angle of banking Let N = Resultant of N1 and N2. f = Resultant of f1 and f2. Let us resolve all the forces horizontally and vertically. As the vehicle has Equilibrium in vertical direction. so, N cos θ + f sin θ = mg

............ (i)

The resultant of horizontal components i.e., (f cos θ + N sin θ), however, this becomes the net external force acting on the vehicle in the radially inward direction of the round-turn. This thus provides for the necessary centripetal force (mv2/r). Therefore, f cos θ + N sin θ = mv2/r

............ (ii)

Further, if μ is the coefficient of friction, we have f = μN

............ (iii)

These are the three basic equations from which, we can find out whatever we want to find out.'

Putting (iii) in (i) gives N cos θ = μN sin θ + mg =>N(cos θ - μsin θ) = mg => N = mg/cos θ - μsin θ

...............(iv)

Putting (iii) and (iv) in (ii) gives μ × mgcos θ/(cos θ - μsin θ) + mgsinθ/(cos θ - μsin θ) = mv2/r =>

μ mgr cos θ + mgr sin θ = mv2 cos θ - μmv2 sin θ

Thus, tan θ = (v2 - μrg)/(rg + μv2)

............... (A)

or

v2 = rg(μ+ tanθ)/(1-μtanθ)

(A)

Gives the angle of banking for the maximum velocity v and (B) gives the value of the maximum velocity which the vehicles

................ (B)

should be allowed on a road banked at an angle θ. Notes : 1. The value of μ is the minimum value of μ required. The value of v is the maximum allowable velocity. Therefore, the best angle of banking θ so that there is absolutely nil wear and tear due to frictional force for the given values of v and r can be determined by putting μ = 0 in this formula. If we put μ = 0 in formula (A), we get tan θ = v2/rg Further, for zero frictional wear and tear, the velocity for the given values of θ and r will be v = √rgtanθ. Skidding:-

Let us consider the situation in the figure. You are cycling fast on road I. You then want to take a turn to go to road II. However, due to big leak of mobile oil from some truck, the portions of the roads within the area ACBD have become slippery. You do not know about it. You are cycling fast. When you reach the line AC, you turn the handle mounted on the front wheel towards road II. What will happen? Will you be able to take the turn? No, you won't be. Although your front wheel is aligned to go towards road II, you still continue to go straight to road III. This is called skidding. You will skid.

The tendency to slip transverse (transverse means across) to the intended line of run, is called skidding. Thus as soon as you turn the handle of your cycle (or the driving wheel in a motor car), skidding will try to occur. If there is enough friction, this start of skidding brings into action a frictional force between the road and the bottom surface of the wheels of the vehicle. This frictional force then provides for the necessary centripetal force required to negotiate the turn. If friction is not enough, skidding will start which will not let you take the turn in a normal way. Skidding will also cause additional friction wear and tear of the tyres of your vehicle. How to avoid skidding? Let us consider the situation given in the figure. Let r is the radius of turn which you have to take. N1 and N2 are normal reactions mg the weight and F1, F2 the frictional forces on the inside and outside wheels.

So, (N1 + N2) = mg Thus, F1 + F2 = μ(N1 + N2) = μgm

......... (i) ......... (ii)

This must be greater than or equal to the centripetal force required. Therefore, μmg > mv2/r or, v < √μrg or, Vmax = √μrg

If the velocity of the vehicle is more than √μrg, it will skid. Overturning:You may have seen overturned trucks lying on the road. Such heavily loaded trucks! Who could have overturned them!! Overturning occurs on the roads when the trucks try to change directions, take sharp turns. Overturning occurs, more after, in case of vehicles which have greater height or whose center of gravity are much high up from the surface of roads. Let us first consider why over turning would take place at all. Suppose a heavily loaded truck is going straight. Suddenly it takes a sharp turn towards its left. Now what actually happens is that while the upper portion of the truck still tends to go straight because of

its inertia (Newton's first law of motion), the lower portion starts going towards left because you have turned the driving wheel accordingly. Thus, if the inertial forces on the upper portion are much height, they provide so much torque on the truck at its center of gravity that overturning takes place. Thus overturning always takes place by lifting off the inner wheels from the ground on the curved path. Limiting case when a four wheeler just begins to overturn on a plain horizontal road:Let, Mg = weight N = Total normal reaction = N1 + N2 F = Total friction force v = velocity r = radius of the round 2a = distance between inner and outer wheels. G = center of gravity h = height of center of gravity from Earth. Frictional force F will provide for the necessary centripetal force.

Therefore, F = mv2/r

............... (i)

When the vehicle just begins to overturn, the inner wheels will just begin to lift off from the ground. Their pressure on ground will become zero, so the reaction N1 on the inner wheels will become zero. Thus, N1 = 0 N1 + N2 = N2 = mg Let us take moments about G N2 × a = F × h Putting (i) and (iii) in (iv) gives mg × a = mv2/r × h, or v i.e., vmax = √rga/h If speed goes beyond it, the vehicle will overturn. Minimum μ required to prevent overturning We know frictional force is μN. =>

F = μN = μmg

Putting equation (v) in (i) gives

............ (v)

μmg = mv2/r, Or μ = μmin = v2/rg F = mv2/r = mω2r

Centripetal force is not a new kind of force. It is the radial component of the net force acting on the particle moving along a circle. Centrifugal force is a type of pseudo force used by an observer moving in a circle. Numerically, it is equal to the centripetal force but is oppositely directed if observer and the body both are moving on same circle as a single unit.

   

The banking angle is the angle at which the vehicle is inclined about its longitudinal axis with respect to the plane of its curved path. If the force of friction is not strong enough, the vehicle will skid. Even if there is very little force of friction the vehicle can still go round the curve with no tendency to skid. Roads are banked because of the inertia of vehicles driving on the road. Problem 1:A curve has a radius of 50 meters and a banking angle of 15º. What is the ideal, or critical, speed (the speed for which no friction is required between the car's tires and the surface) for a car on this curve? Solution:-

Here, radius of curve, r = 50 m banking angle, θ = 15º free-fall acceleration, g = 9.8 m/s2 We have to find out the ideal speed v (the speed for which no friction is required between the car's tires and the surface) From the free-body diagram for the car:Fnet = Fcentripital mg tanθ = mv2/r v2 = rg tanθ v = √rg tanθ

= √(50 m) (9.8 m/s2) (tan 15º) = 11 m/s If the car has a speed of about 11 m/s, it can negotiate the curve without any friction. ____________________________________________________________________________________________ Problem 2:A 1200 kg automobile rounds a level curve of radius 200 m, on a unbanked road with a velocity of 72 km/hr. What is the minimum co-efficient of friction between the tyres and road in order that the automobile may not skid. (g = 10 m/s2) Solution:In a unbanked road, the centripetal force is provided by the frictional force. So, ffriction = mv2/r But flimiting friction > ffriction or, μmg = ffriction or μmg = mv2/r So, μmin = v2/gr = (20×20)/(10×200) = 0.2. The bank angle is the angle at which the vehicle is inclined about its longitudinal axis with respect to the plane of its curved path. If the force of friction is not strong enough, the vehicle will skid. even if there is very little force of friction the vehicle can still go round the curve with no tendency to skid. Roads are banked because of the inertia of vehicles driving on the road. ________________________________________________________________________________________ Problem 3:A turn of radius 100 m is being designed for a speed of 25 m/s. At what angle should the turn be banked? Solution:-

Here, radius of turn, r = 100 m speed of the car, v = 25 m/s free-fall acceleration, g = 9.8 m/s2 We have to find out the bank angle, θ. From the free body diagram of the car, Fnet = Fcentripetal mg tanθ = mv2/r tanθ = v2/rg θ = tan-1(v2/rg)

= tan-1 [(25 m/s)2/ (100 m) (9.8 m/s2)] = 33º So, the banking angle should be about 33º.

PROBLEMS Problem 1:A 26-ton Navy jet as shown in the below figure requires an air speed of 280 ft/s for lift-off. Its own engine develops a thrust of 24,000 lb. The jet is to take off from an aircraft carrier with a 300-ft flight deck. What force must be exerted by the catapult of the carrier? Assume that the catapult and the jet‟s engine each exert a constant force over the 300-ft takeoff distance.

Concept:Force acting on the body (F) is equal to the product of mas of the body (m) and acceleration of the body (a). So, F = ma Weight W of the object is equal to the mass m of the object times of the free fall acceleration g. W = mg So the mass m of the object would be, m = W/g To obtain force F in terms of weight W, substitute W/g for m in the equation F = ma, F = ma = (W/g) (a) Time t taken by a body to travel a distance x with average velocity vav will be, t = x/ vav The deceleration a is equal to the rate of change of velocity, a = Δv/Δt Solution:To find the time t for the plane to travel 300 ft, substitute 300 ft for x and 140 ft /s for vav in the equation t = x/vav, t = x/ vav = (300 ft)/(140 ft/s) = 2.14 s To obtain the acceleration a, substitute 280 ft/s for Δv and 2.14 s for Δt in the equation a = Δv/Δt,

a = Δv/Δt = (280 ft/s)/(2.14 s) = 130 ft/s2 To find out the net force F on the plane, substitute 52,000 lb for W, 130 ft/s2 for a and 32 ft/s2 for free fall acceleration g in the equation F =(W/g) (a), F =(W/g) (a) = (52000 lb)( 130 ft/s2)/( 32 ft/s2) = 2.1×105 lb The force exerted by the catapult Fc is equal to the difference of the net force F on the plane and the thrust develop by the own engine T. So, Fc = F-T To obtain the force exerted by the catapult Fc, substitute 2.1×105 lb for F and 24,000 lb for T in the equation Fc= F-T, Fc = F-T = (2.1×105 lb) - (24,000 lb) = (2.1×105 lb) - (2.4×104 lb) =1.86×105 lb From the above observation we conclude that, the force exerted by the catapult Fc would be 1.86×105 lb. _________________________________________________________________________________________________________ ___________ Problem 2:A 77-kg person is parachuting and experiencing a downward acceleration of 2.5 m/s2 shortly after opening the parachute. The mass of the parachute is 5.2 kg. (a) Find the upward force exerted on the parachute by the air. (b) Calculate the downward force exerted by the person on the parachute.

Concept:Force acting on the body (F) is equal to the product of mas of the body (m) and acceleration of the body (a). So, F = ma From equation F = ma, the acceleration (a) of the body would be, a = F/m Weight W of the object is equal to the mass m of the object times of the free fall acceleration g. W = mg Solution:(a) The net force Fnet on the system is equal to the sum of force exerted by the person and force exerted by the parachute. So, Fnet = (mpe+mpa) (a)

Here, mpe is the mass of person, mpa is the mass of parachute and a is the downward acceleration. To obtain the net force Fnet on the system, substitute 77 kg for mpe , 5.2 kg for mpa and -2.5 m/s2 for a in the equation Fnet = (mpe+mpa) (a), Fnet = (mpe+mpa) (a) = (77 kg + 5.2 kg) (-2.5 m/s2) = (-210 kg,m/s2) (1 N/1 kg,m/s2) = -210 N The weight W of the system will be, W = (mpe+mpa) (g) To obtain the weight W of the system, substitute 77 kg for mpe , 5.2 kg for mpa and 9.81 m/s2 for free fall acceleration g in the equation W = (mpe+mpa) (g), W = (mpe+mpa) (g) = (77 kg + 5.2 kg) (9.81 m/s2) = (810 kg,m/s2) (1 N/1 kg,m/s2) = 810 N If P is the upward force of the air on the system (parachute) then, P = Fnet +W = (-210 N)+ (810 N) = 600 N From the above observation we conclude that, the upward force exerted on the parachute by the air would be 600 N. (b) The net force Fnet on the parachute will be, Fnet = mpa a Here, mpa is the mass of parachute and a is the downward acceleration. To obtain the net force Fnet on the parachute, substitute 5.2 kg for mpa and -2.5 m/s2 for a in the equationFnet = mpa a, Fnet = mpa a = (5.2 kg)(-2.5 m/s2) = (-13 kg.m/s2) (1 N/1 kg,m/s2) = -13 N The weight W of the parachute will be, W = (mpa) (g) To obtain the weight W of the system, substitute 5.2 kg for mpa and 9.81 m/s2 for free fall acceleration g in the equation W = (mpa) (g), W = (mpa) (g) = ( 5.2 kg) (9.81 m/s2)= (51 kg,m/s2) (1 N/1 kg,m/s2) = 51 N If D is the downward force of the person on the parachute then, D = - Fnet-W+P = -(-13 N)-(51 N)+(600 N) = 560 N _________________________________________________________________________________________________________ _________ Problem 3:A 1400-kg jet engine is fastened to the fuselage of a passenger jet by just three bolts (this is the usual practice). Assume that each bolt supports one-third of the load. (a) Calculate the force on each bolt as the palne waits in line for clearance to take off. (b) During flight, the palne encounters turbulence, which suddenly imparts an upward vertical acceleration of 2.60 m/s2 to the palne. Calculate the force on each bolt now. why are only three bolts used? See below figure.

Concept:Force acting on the automobile (F) is equal to the product of mas of the automobile (m) and acceleration of the auto mobile (a). So, F = ma Weight of the automobile (W) is equal to the product of mass of the automobile (m) and acceleration due to gravity on the surface of the earth (g = 32 ft/s2). W = mg Solution:(a) First we have to find out the weight W of the engine. To obtain the weight of the engine W, substitute 1400 kg for mass m and 9.81 m/s2 for g in the equation W =mg, W = mg = (1400 kg) (9.81 m/s2) = 1.37×104 kg.m/s2 = (1.37×104 kg.m/s2) (1 N/1 kg.m/s2) = 1.37×104 N As each bolt supports 1/3 of this force, thus the force F on a bolt will be, F = 1.37×104 N/3 = 4600 N From the above observation we conclude that, the force on each bolt would be 4600 N. (b) To find out the force f on each bolt, first we have to find out the upward force Fup on the bolt. Again to obtain the upward force Fup on the bolt, we have to obtain the net force Fnet on the engine. To obtain the net force Fnet on the engine, substitute 1400 kg for mass of the jet engine m and 2.60 m/s2 for acceleration a in the equation Fnet = ma, Fnet = ma =(1400 kg) (2.60 m/s2) =3.64×103 kg.m/s2 =(3.64×103 kg.m/s2) (1 N/1 kg.m/s2) =3.64×103 N So the upward force Fup from the bolts will be equal to the sum of net force Fnet on the engine and weight of the engine W. Fup = Fnet + W To obtain the upward force Fup from the bolts, substitute 3.64×103 N for Fnet and 1.37×104 N for W in the equation Fup = Fnet + W, Fup = Fnet + W = (3.64×103 N) + (1.37×104 N) = 1.73×104 N The force per bolt f will be equal to the 1/3 of the upward force Fup from the bolts.

So, f = Fup/3. To obtain the force per bolt f, substitute 1.73×104 N for Fup in the equation f = Fup/3, f = Fup/3 = (1.73×104 N)/3 = 5800 N From the above observation we conclude that, the force on each bolt would be 5800 N. _________________________________________________________________________________________________________ _________ Problem 4:A research ballon of total mass M is descending vertically with downward acceleration a as shown in below figure. How much ballast must be thrown from the car to give the ballon an upward acceleration a, assuming that the upward lift of the air on the ballon does not change?

Solution:Let us consider initially mass of the system (balloon) is, M. So the force (f) acting on the system having downward acceleration, a will be, f = -Ma

…… (1)

And the weight of the system (W) will be, W = Mg

…… (2)

Where, g is the free fall acceleration of the system. Therefore the upward force acting on the system (F) will be, F = W + f = Mg+(-Ma) F = Mg-Ma

…… (3)

Again let us consider m is the mass of single ballast which is thrown from the balloon. Now mass of the system is, M-m. So the force (f1) acting on the system having upward acceleration, a and mass, M-m will be, f1 = (M-m) a

…… (4)

And the weight of the system (W1) having mass M-m will be,

W1 = (M-m) g

…… (5)

Where, g is the free fall acceleration of the system. To give the balloon an upward acceleration a, the upward force (F) must be equal to the addition of the force (f1) acting on the system having upward acceleration, a and mass, M-m and the weight of the system (W1) having mass M-m will be. So the equation will be, f1+W1 = F

…… (6)

(M-m) a + (M-m) g = Mg-Ma Ma-ma +Mg-mg = Mg-Ma ma + mg = Ma+ Mg –Mg + Ma m(a+g) = 2Ma m = 2Ma/(a+g)

…… (7)

From equation (7) we observed that, 2Ma/(a+g) mass of ballast must be thrown from the car to give the balloon an upward acceleration a. _________________________________________________________________________________________________________ __________ Problem 5:A child‟s toy consists of three cars that are pulled in tandem on small frictionless rollers as shown in below figure. The cars have masses m1 = 3.1 kg, m2 = 2.4 kg, and m3 = 1.2 kg. If they are pulled to the right with a horizontal force P = 6.5 N, find (a) the acceleration of the system, (b) the force exerted by the second car on the third car, and (c) the force exerted by the first car on the second car.

Solution:Given Data: Mass of the first car, m1 = 3.1 kg Mass of the second car, m2 = 2.4 kg Mass of the third car, m3 = 1.2 kg Horizontal force on the car, P = 6.5 N Acceleration a body is equal to the force exerted on the body divided by mass of the body. (a) The acceleration of the system is equal to the total horizontal force acting on the system divided by total mass of the system. So, a = F/(m1+ m2+ m2) = (6.5 N)/(3.1 kg+2.4 kg+1.2 kg) = (6.5 N)/(6.7 kg) = (0.97 N/kg) (1 kg. m/s2/1N) = 0.97 m/s2

Therefore the acceleration of the system would be 0.97 m/s2. (b) Force exerted by the second car on the third car would be, F32 = (mass of third car) (total acceleration of the system) = m3 a = (1.2 kg) (0.97 m/s2) = (1.164 kg. m/s2) (1 N/ 1 kg. m/s2) = 1.164 N Rounding off to two significant figures, the force exerted by the second car on the third car would be 1.2 N. (c) Force exerted by the first car on the second car would be, F32 = (sum of mass of second car and third car) (total acceleration of the syste) = (m2+ m3) a = (2.4 kg +1.2 kg) (0.97 m/s2)= (3.492 kg. m/s2) (1 N/ 1 kg. m/s2) = 3.492 N Rounding off to two significant figures, the force exerted by the first car on the second car would be 3.5 N. _________________________________________________________________________________________________________ __________ Problem 6:A landing craft approaches the surface of Callisto, one of the satellite (moons) of the planet Jupiter as shown in the below figure. If an upward thrtust of 3260 N is supplied by the rocket engine, the craft descends with constant speed. Callisto has no atomsphere. If the upward thrust is 2200 N, the craft accelerates downward at 0.390 m/s2. (a) What is the weight of the landing craft in the vicinity of Callisto‟s surface? (b) What is the mass of the craft? (c) What is the acceleration due to gravity near the surface of Callisto?

Concept:Force acting on the automobile (F) is equal to the product of mas of the automobile (m) and acceleration of the auto mobile (a). So, F = ma So mass m of the body will be, m = F/a Weight of the automobile (W) is equal to the product of mass of the automobile (m) and acceleration due to gravity on the surface of the earth (g = 32 ft/s2). W = mg So, g = W/m

Solution:(a) As the craft descends with constant speed, so the net force will be equal to zero. Thus the thrust balances weight. As the upward thrust of 3260 N is supplied by the rocket engine, therefore the weight of the landing craft in the vicinity of Callisto‟s surface will be 3260 N. (b) An upward thrust of 3260 N is supplied by the rocket engine; the craft descend with constant speed. If the upward thrust is 2200 N, the craft accelerates downward at 0.390 m/s2. So the net force F will be, F = 2200 N – 3260 N = -1060 N To obtain the mass m of the craft, substitute -1060 N for F and -0.390 m/s2 for acceleration a (negative sign due to downward direction) in the equation m = F/a, m = F/a = (-1060 N)/(0.390 m/s2) = (2720 N/(m/s2)) (1kg.m/s2 /1 N) = 2720 kg From the above observation we conclude that, the mass m of the craft will be 2720 kg. (c) To find out the acceleration due to gravity g near the surface of Callisto, substitute 3260 N for W and 2720 kg for m in the equation g = W/m, g = W/m = 3260 N/2720 kg = (1.20 N/kg) (1 kg.m/s2 /1 N) = 1.20 m/s2 From the above observation we conclude that, the acceleration due to gravity g near the surface of Callisto would be 1.20 m/s2. _________________________________________________________________________________________________________ _______ Problem 7:Two blocks are in contact on a frictionless table. A horizonatl force is applied to one block, as shown in the below figure. (a)If m1 = 2.3 kg, m2 = 1.2 kg, and F = 3.2 N, find the force of contact between the two blocks. (b) Show that If the same force F is applied to m2 rather than to m1, the force of contact between the blocks is 2.1 N, which is not the same value derived in (a). Explain.

Concept:Force acting on the body (F) is equal to the product of mas of the body (m) and acceleration of the body (a).So, F = ma From equation F = ma, the acceleration (a) of the body would be, a = F/m Solution:(a) The acceleration a of the two block will be, a = F/(m1+m2)

Here F is the horizontal force, m1 is the mass of the block 1 and m2 is the mass of the block 2. The net force Fnet on block 2 is from the force of contact, and will be, Fnet = m2a = F m2/(m1+m2)

(Since, a = F/(m1+m2))

To obtain the force Fnet of contact between the two blocks, substitute 3.2 N for F,1.2 kg for m2 and 2.3 kg form1 in the equation Fnet = F m2/(m1+m2), Fnet = F m2/(m1+m2) = (3.2 N) (1.2 kg)/(2.3 kg+1.2 kg) = 1.1 N From the above observation we conclude that, the force Fnet of contact between the two blocks would be 1.1 N. (b) The acceleration a of the two block will be, a = F/(m1+m2) Here F is the horizontal force, m1 is the mass of the block 1 and m2 is the mass of the block 2. The net force Fnet on block 1 is from the force of contact, and will be, Fnet = m1a = F m1/(m1+m2)

(Since, a = F/(m1+m2))

To obtain the net force Fnet on block 1, substitute 3.2 N for F,1.2 kg for m2 and 2.3 kg for m1 in the equationFnet = F m2/(m1+m2), Fnet = F m1/(m1+m2) = (3.2 N) (2.3 kg)/(2.3 kg+1.2 kg) = 2.1 N From the above observation we conclude that, if the same force F is applied to m2 rather than to m1, the force of contact between the blocks would be 2.1 N. _________________________________________________________________________________________________________ _________ Problem 8:A block of mass M is pulled along a horizontal frictionless surface by a rope of mass m, as shown in the below figure. A horizontal force

is applied to one end of the rope. Assuiming that the sag in the rope is negligible, find (a) the acceleration of rope and

block, and (b) the force that the rope exerts on the block.

Concept:Force acting on the body (F) is equal to the product of mas of the body (m) and acceleration of the body (a). So, F = ma From equation F = ma, the acceleration (a) of the body would be, a = F/m Solution:(a) Treat the system as including both the block and the rope.

Thus the mass of the system will be M+m. Here M is the mass of the block and m is the mass of the rope. As there is one horizontal force which is acting to one end of the rope, so,

In accordance to Newtons second law, the horizontal force P will be, P = (M+m) ax Here ax is the horizontal acceleration of the rope and block. From equation P = (M+m)ax, the horizontal acceleration ax of the rope and block will be, ax = P/(M+m) From the above observation we conclude that, the horizontal acceleration ax of the rope and block would be P/(M+m). (b) Now consider only the block. The horizontal force does not act on the block, instead there is the force of the rope on the block. We will assume that, the magnitude of the force is R, and this the only relevant force on the block. Thus the net force on the block will be,

In this case Newton‟s second law would be written as, R = Max Here the horizontal acceleration ax of the block is equal to the acceleration of the block and rope system andM is the mass of the block. To obtain the force R that the rope exerts on the block, substitute P/(M+m) for acceleration ax in the equation R = Max we get, R = Max= m [P/(M+m)] = [M/(M+m)] P From the above observation we conclude that, the force R that the rope exerts on the block would be [M/(M+m)] P. _________________________________________________________________________________________________________ _________ Problem 9:A light beam from a satellite-carried laser strikes an object ejected from an accidently launched ballistic missile; see below figure. The beam exerts a force of 2.7×10-5 N on the target. If the “dwell time” of the beam on the target is 2.4 s, by how much is the object displaced if it is (a) a 280-kg warhead and (b) a 2.1-kg decoy? (These displacements can be measured by observing the reflected beam.)

Concept:Force acting on the automobile (F) is equal to the product of mas of the automobile (m) and acceleration of the auto mobile (a). So, F = ma From the above equation F = ma, the acceleration a will be, a = F/m In accordance to equation of motion, the distance y travelled by the body will be, y = ut + ½ at2 Here u is the initial velocity, t is the time, and ais the acceleration. Since the object ejected from an accidentally launched ballistic missile, so the initial velocity u will be equal to zero. So, u = 0 Substitute 0 for u in the equation y = ut + ½ at2, y = ut + ½ at2 = 0×t +½ at2 = ½ at2 Solution:(a) To find the displacement y which is the object displaced, first we have to find out the acceleration a. To obtain the acceleration a, substitute 2.7×10-5 N for F and 280 kg for m in the equation a = F/m, a = F/m = (2.7×10-5 N)/280 kg = (9.64×10-8 N/kg) (1 kg.m/s2 /1 N) = 9.64×10-8 m/s2 To find the displacement y which is the object displaced from the original trajectory, substitute 9.64×10-8m/s2 for a and 2.4 s for t in the equation y = ½ at2, y = ½ at2 = ½ (9.64×10-8 m/s2) (2.4 s)2 = 2.8×10-7 m From the above observation we conclude that, the displacement y which is the object displaced from the original trajectory would be 2.8×10-7 m. (b) To find the displacement y which is the object displaced, first we have to find out the acceleration a.

To obtain the acceleration a, substitute 2.7×10-5 N for F and 2.1 kg for m in the equation a = F/m, a = F/m = (2.7×10-5 N)/2.1 kg = (1.3×10-5 N/kg) (1 kg.m/s2 /1 N) = 1.3×10-5 m/s2 To find the displacement y which is the object displaced from the original trajectory, substitute 1.3×10-5 m/s2for a and 2.4 s for t in the equation y = ½ at2, y = ½ at2 = ½ (1.3×10-5 m/s2) (2.4 s)2 = 3.7×10-5 m From the above observation we conclude that, the displacement y which is the object displaced from the original trajectory would 3.7×10-5 m. _________________________________________________________________________________________________________ ________ Problem 10:What strength fishing line is needed to stop a 19-lb salmon swimming at 9.2 ft/s in a distance of 4.5 in.? Concept:We have to calculate the force acting on the salmon. Force (F) acting on the body is equal to the mass of the body (m) times acceleration (a) of the body. F = ma In terms of weight (W = mg) the above equation (F = ma) will be, F = ma = (ma) (g)/g = (mg)a/g = Wa/g Solution:Initial speed (vi) of the salmon is, vi = 9.2 ft/s Final speed (vf) of the salmon is, vf = 0 ft/s So the average speed of the salmon will be, vav = (vi + vf) /2 = (0 ft/s+9.2 ft/s)/2 =

= 4.6 ft/s

Time required (t) to stop the salmon is equal to the distance travelled (x) by the salmon divided by the average velocity (vav) of the salmon. So, t = x/ vav To obtain the time required (t) to stop the salmon, substitute 4.5 in for x and 4.6 ft/s for vav in the equation t= x/ vav, t = x/ vav =(4.5 in)/(4.6 ft/s) =(4.5 in×0.0833 ft/1 in)/(4.6 ft/s) = (0.38 ft) /(4.6 ft/s) = 8.3×10-2 s The deceleration a of the salmon will be, a = Δv/Δt. To find out the deceleration a of the salmon, substitute 9.2 ft/s for Δv and 8.3×10-2 s for Δt in the equation a= Δv/Δt, a = Δv/Δt = (9.2 ft/s)/(8.3×10-2 s) =110 ft/s2 To find out the force on the salmon, substitute 19-lb for W, 110 ft/s2 for a, and 32 ft/s2 for g in the equation F= Wa/g, F = Wa/g = (19-lb) (110 ft/s2)/ (32 ft/s2) = 65.31 lb Rounding off to two significant figures, the force will be 65-lb.

From the above observation we conclude that, the force required to stop the 19-lb salmon swimming at 9.2 ft/s in a distance of 4.5 in would be 65-lb. _________________________________________________________________________________________________________ _________ Problem 11:A jet plane starts from rest on the runway and accelerates for takeoff at 2.30 m/s2. It has two jet engines, each of which exerts a thrust of 1.40 ×105 N. What is the weight of the plane?

Solution:Given Data: At takeoff time, acceleration of the jet plane, ax = 2.30 m/s2 Thrust of each jet engine, Fx = 1.40×105 N Jet plane consist two jet engines. So the total force exerts by the get plane will be, ∑ Fx = 2(1.40×105 N) = 2.80×105 N

…… (1)

Force exerted on a body is equal to the mass of the body times acceleration of the body. So, ∑ Fx = max

…… (2)

So from equation (2), m = ∑ Fx /ax

…… (3)

Putting the value of ∑ Fx and ax in equation (3), mass of the jet plane will be, m = ∑ Fx /ax = 2.80×105 N/2.30 m/s2 = (1.22 ×105 N/m/s2) (1 kg. m/s2 / 1 N) = 1.22 ×105 kg

……. (4)

Weight of a body is equal to the mass of the body times (m) acceleration due to gravity (g) on that surface. W = mg …… (5) We know that that, the value of acceleration due to gravity (g) on the surface of earth is 9.81 m/s2. So putting the value of m and g in equation (5), weight of the jet plane would be, W = mg = (1.22×105 kg) (9.81 m/s2) = (11.9682×105 kg. m/s2) (1 N/1 kg. m/s2)

= 1.19682×106 N

…… (6)

Rounding off to three significant figures, weight of the jet plane would be 1.20×106 N.

Conservation of Linear Momentum:It states that, “In an isolated system (no external force), the algebraic some of the momenta of bodies, along any straight line, remains constant and is not changed due to their mutual action and reaction on each other”. This can be verified by a following simple experiment. Consider a body „A‟ of mass „m1‟ moving with a velocity

strike against another body „B‟ of mass m2,moving with velocity in same

direction as shown in the below figure. Two bodies remain in contact with each other for small time „?t‟. They get separated and move with velocities

Let

and

after collision.

be the force exerted by „A‟ upon „B‟ and

be its reaction. Since the system is isolated, i.e., no external force is

there,

So, …... (1) This is in accordance with Newton‟s third law of motion that „action and reaction are equal and opposite‟. Considering the momenta of the bodies before and after collision. Body A

Body B

Momentum of A before collision =

Momentum of A before collision =

Momentum of A after collision =

Momentum of B after collision =

Change in momentum of A =

Change in momentum of B =

Time taken for the change of momentum =?t

Time taken for the change of momentum =?t

Rate of change of momentum of A (=Force on A) =

Rate of change of momentum of B (=Force on B)=

So,

So, Substituting for

and

in equation (1),

Or, Thus, the total momentum of the system before collision is equal to the total momentum of the system after collision.

This verifies the law of conservation of momentum. It may be noted that the conservation of momentum is closely connected with the validity of Newton‟s third law of motion, since we have used equation (1) [which is nothing but third law] to prove it. Alternative Method:According to Newton‟s second law of motion,

Since

(momentum of body),

Incase of an isolated system,

Thus,

or, Therefore, momentum (in vector form) of an isolated system remains constant. This is in accordance with the law of conservation of momentum. IMPORTANT NOTE:-

While applying law of conservation of momentum to a system following consideration must be kept in mind: (a) The system must be isolated. (b) While finding the algebraic sum of momenta it must be ensured that all of them are along a particular straight line. Applications of conservation of momentum:-

Following few examples with illustrate the law of conservation of momentum. (a) Recoil of gun:-

A gun and a bullet constitute one isolated system. On firing the gun, bullet moves out with a very high velocity experiences a recoil. It moves in the opposite direction as shown in the below figure. Velocity „ calculated by the application of law of conservation of momentum.

. The gun

‟ of the recoil gun can be

Before Firing

After Firing

Momentum of bullet = 0

Momentum of bullet =

Momentum of gun = 0

Momentum of gun =

Total momentum of the system = 0

Total momentum of the system =

Here „m‟ and „M‟ are the masses of bullet and gun respectively. According to the law of conservation of momentum, momentum before collision and after collision must be same.

or, or,

Negative sign indicates that direction of motion of gun is in opposite direction. (b) Rocket and jet plane:Fuel and oxygen is burnt in the ignition chamber. As hot gases escape from a rear opening, with some momentum, the rocket moves in the forward direction with the same momentum.

(c)

Explosion of a bomb:-

Momentum of a bomb before explosion is zero. After explosion different fragments fly in various directions. It will be observed that their momenta, when represented by the slide of a polygon, from a closed polygon, indicating that net momentum after explosion is also zero. Thus, if the bomb exploded into two fragments, they must move in opposite directions.

(d) A man Jumping from a boat:When a man jumps from the boat to the shore, the boat is pushed backward. It can, exactly, be explained as in the case of recoil of gun. Some Conceptual Questions:-

Question 1:Figure below shows a popular carnival device, in which the contestant tries to see how high a weighted marker can be raised by hitting a target with a sledge hammer. What physical quantity does the device measure? Is it the average force, the maximum force, the work done, the impulse, the energy transferred, the momentum transferred, or something else? Discuss your answer.

Answer:The device will measure impulse. The impulse of the net force acting on a particle during a given time interval is equal to the change in momentum of the particle during that interval. Since the contestant is hitting the target with a sledge hammer the change in momentum is large and the time of collision is small, therefore it signifies that the average impulsive force will relatively large. Suppose two persons bring the harmer from the same height, but they are hitting with different forces. The person who hits with greater force for the short time interval the impulse will be more and this results the height of the mark will be more. Thus the device will measure impulse. _________________________________________________________________________________________________________ _______ Question 2:Can the impulse of a force be zero, even if the force is not zero? Explain why or why not? Answer:Yes, the impulse of a force can be zero, even if the force is not zero.

Impulse of a force is defined as the change in momentum produced by the force and it is equal to the product of force and the time for which it acts. The impulse of a force can be zero, if the net force acting on the particle during that time interval is constant. Since the force is constant (both magnitude and direction), so change in momentum produced by the force will be zero. Therefore impulse of the force will be zero. From the above observation we conclude that, impulse of a force can be zero, even if the force is not zero. _________________________________________________________________________________________________________ _________ Question 3:Explain how conservation of momentum applies to a handball bouncing off a wall. Answer:Law of conservation of linear momentum states that, in an isolated system (no external force), the algebraic sum of momenta of bodies, along any straight line, remains constant and is not changed due to their mutual action and reaction on each other.

The momentum of particle (p) is equal to the mass of particle (m) times the velocity of particle (v). So,p = mv

…… (1)

Let us consider m is the mass of the ball and v is the velocity of the ball when the ball is collides with wall. So using equation (1), the momentum of the ball before collision (p1) will be, p1= mv

…… (2)

After collision, when the ball re bounces, the velocity of the ball will be, -v. So again using equation (1), the momentum of the ball after collision (p2) will be, p2= -mv

…… (3)

Conservation of linear momentum states that, the algebraic sum of momenta of bodies, along any straight line, remains constant and is not changed due to their mutual action and reaction on each other. p1 + p2 = 0 So, mv + (-mv) = 0

…… (4)

From equation (4) we observed that, linear momentum of the hand ball is conserved. _________________________________________________________________________________________________________ Question 4:-

Give a plausible explanation for the breaking of wooden boards or bricks by a karate punch. (See “Karate Strikes.” by Jearl D. Walker, American Journal of Physics, October 1975, p.845.)

Answer:In the process, breaking of wooden boards or bricks by a karate punch, the collision between the hand and brick is only for a few milliseconds. Because the applied external force is large and the time of collision is small therefore the average impulsive force is relatively large. Thus when you break a wooden board or bricks by a karate punch you have to apply large force for the minimum time which is impulse. Therefore the impact force on the brick or wooden boards will be high. Some Solved Problems:-

Problem 1:A 75.2-kg man is riding on a 38.6-kg cart travelling at a speed of 2.33 m/s. He jumps off in such a way as to land on the ground with zero horizontal speed. Find the resulting change in the speed of the cart. Concept:Momentum of the body p is equal to the mass of the body m times velocity of the body v. So, p = mv In accordance to the principle of conservation of energy, the final momentum of the system is equal to the initial momentum of the system. Consider the initial momentum of the man is pi,m, initial momentum of the cart is pi,c, final momentum of the man is pf,m and final momentum of the cart is pf,c. We define, pf,m = mmvf,m pf,c = mcvf,c pi,m = mmvi,m pi,c = mcvi,c Here, mass of the man is mm, mass of the cart is mc, initial velocity of the man is vi,m and cart is vi,c, and final velocity of the man is vf,m and cart is vf,c. Solution:So applying conservation of momentum to this system, the sum of the initial momentum of the man and cart will be equal to the sum of the final momentum of the man and cart. pf,m + pf,c = pi,m + pi,c Substitute, mmvf,m for pf,m, mcvf,c for pf,c, mmvi,m for pi,m and mcvi,c for pi,c ijn the equation pf,m + pf,c = pi,m + pi,c, pf,m + pf,c = pi,m + pi,c mmvf,m + mcvf,c = mmvi,m + mcvi,c

vf,c- vi,c = (mmvi,m - mmvf,m)/ mc Δvc = (mmvi,m - mmvf,m)/ mc To obtain the resulting change in the speed of the cart Δvc, substitute 75.2 kg for mm, 2.33 m/s for vi,m and 0 m/s for vf,m in the equation Δ vc = (mmvi,m - mmvf,m)/ mc, Δvc = (mmvi,m - mmvf,m)/ mc = (75.2 kg) (2.33 m/s) – (75.2 kg) (0 m/s)/(38.6 kg) = 4.54 m/s As the sign of the change in the speed of the cart Δvc is positive, this signifies that, the cart speed increases. From the above observation we conclude that, the resulting change in the speed of the cart Δvc would be 4.54 m/s. _________________________________________________________________________________________________________ __ Problem 2:A space vehicle is travelling at 3860 km/h with respect to the Earth when the exhausted rocket motor is disengaged and sent backward with a speed of 125 km/h with respect to the command module. The mass of the motor is four times the mass of the module. What is the speed of the command module after the separation? Concept:Momentum of the body p is equal to the mass of the body m times velocity of the body v. So, p = mv In accordance to the principle of conservation of energy, the final momentum of the system is equal to the initial momentum of the system. Consider the initial momentum of the motor is pi,m, initial momentum of the command module is pi,c, final momentum of the motor is pf,m and final momentum of the command module is pf,c. Initial momentum of the motor is equal to the initial momentum of the command module. So, initial momentum of the motor will be, pi,m = mmvi,c and final momentum of the motor pf,m will be, pf,m = mm(vf,c – vrel) Here, speed of the motor relative to the command module is vrel. Again, pf,c = mcvf,c pi,c = mcvi,c Here, mass of the motor is mm, mass of the command module is mc, initial velocity of the motor is vi,m and command module is vi,c, and final velocity of the motor is vf,m and command module is vf,c. Solution:So applying conservation of momentum to this system, the sum of the initial momentum of the motor and command module will be equal to the sum of the final momentum of the motor and command module. pf,m + pf,c = pi,m + pi,c Substitute, mm(vf,c – vrel) for pf,m, mcvf,c for pf,c, mmvi,c for pi,m and mcvi,c for pi,c in the equation pf,m + pf,c = pi,m+ pi,c, pf,m + pf,c = pi,m + pi,c

mm(vf,c – vrel)+ mcvf,c = mmvi,c + mcvi,c (mm + mc)vf,c - mm vrel = (mm +mc) vi,c (mm + mc)vf,c = mm vrel + (mm +mc) vi,c So, vf,c = [mm vrel + (mm +mc) vi,c]/ (mm + mc) As, mass of the motor is four times the mass of the module, thus, mm = 4 mc To obtain the speed of the command module vf,c, substitute 4 mc for mm, 125 km/h for vrel and 3860 km/h forvi,c in the equation vf,c = [mm vrel + (mm +mc) vi,c]/ (mm + mc), vf,c = [mm vrel + (mm +mc) vi,c]/ (mm + mc) = [4 mc(125 km/h) + (4 mc+mc) (3860 km/h)]/ [4 mc+ mc] = 3960 km/h From the above observation we conclude that, the speed of the command module vf,c would be 3960 km/h.

Work, Energy and Power:-

Work, Energy and power are some of the more important topics of mechanics in classical physics. A car moving, a rocket taking off, an aero plane flying, are a few of the events which take place regularly and can give us insight into the concept of Work and Energy. These are some of the common sense notions of work and energy - which can however be precisely defined and measured in physics. These definitions and measurement can be used consistently to describe and predict the behavior of bodies. Work is related to energy as they are inter-convertible and also the sums of them are conserved in Newtonian Physics. Power is a step further which defines the rate of change of energy flow.The examples based on this are very easy and can be seen even in day to day life. Work:-

Work is said to be done if a force, acting on a body, displaces the body through a certain distance and the force has some component along the displacements. Thus, work is done when the point of application of a force moves.It is defined as the product of magnitude of displacement and the component of the force along the displacement.

Power:-

The rate at which work is done is called power.

Energy:-

Energy is the ability of the body to do some work. The unit of energy is same as that of work. The chapter is important not only because it fetches 3-4 questions in most of the engineering examination but also because it is like a linking pin to the previous and other forthcoming chapters of Mechanics.

Work:-

Work is said to be done if a force, acting on a body, displaces the body through a certain distance and the force has some component along the displacements. Thus, work is done when the point of application of a force moves. Work Done by a Constant Force:Consider a body experiencing a force F in a direction inclined at an angle θ with the positive direction of x-axis. Let the body be displaced from A to B through a distance s. First Definition:Work done is defined as the product of force and displacement. Let

be a constant force acting on a body, and let the body undergo a displacement

work done by the force

is defined as,

, as shown in Figure given below, then the

W =

.

= (F cosθ)s = F(s cosθ) = Fs cosθ

Here θ is the angle between F and s. Work done by the force is positive if the angle between force and displacement is acute (0˚ a E = GM/(4ar2 ) [z+(a2-r2)/z2 ](r-a)(r+a) = GM/r2 We see that the shell may be treated as a point particle of the same mass placed at its centre to calculate the gravitational field at an external point. Case II: P is outside the shell, r > a E = GM/(4ar2)[z+(a2-r2)/z2 ](a-r)(a+r) = 0 We see that the field inside a uniform spherical shell is zero. Gravitational field outside a solid sphere The sphere can be thought of as composed of many shells from radius = 0 to radius = a. The point P is at a distance r from the centre of all these concentric shells. =>

E = G/r2[ΔM1 + ΔM2 +...] E = GM/r2

Gravitational field inside a uniform solid sphere of radius 'R' To find the field at a point P inside the sphere at a distance r < R form the centre, let us consider a sphere of radius r.

Consider a point P on the surface of the shaded sphere. Since this point is inside the shells having radii larger than r, they do not contribute to the field at P. Shells that are less than radius 'r', contribute to the gravitational field at P. The mass of the sphere of radius r is

.·.

M' = (M.4/3 Π r3)/(4/3 Π R3 ) = (Mr3)/R3 EP = (GM')/r2 = (GM r)/R3

The adjacent graph shows the variation of E due to a solid sphere of radius R with the distance r from its centre.

E=

GM/r2 3

E = (GM/R ) r

(r > R) (r < R)



This result holds good for the earth if it is assumed to be a uniform solid sphere.



As by definition, g = Fg/m and also E = Fg/m , so g = E, i.e. acceleration due to gravity and gravitational intensity E at a point are synonymous. Illustration: Two concentric shells of masses M1 and M2 are present. Calculate the gravitational force on 'm' due to M1 and M2 at points P, Q and R.

Solution: Field at P, EP = 0 => F=0 Field at Q, due to M2 will be zero but there will be field due to m1, EQ = (GM1)/b2 =>

F = (GM1m)/b2

Field at R, is the sum of fields due to M1 and M2, ER = G(M1+M2 )/c2 Acceleration due to Gravity (g) The Earth attracts a mass m on its surface by a force F given as: F = GMe m/Re2 , where Me is the mass of the Earth and Re its radius. This force imparts an acceleration to the mass m, which is known as acceleration due to gravity (g). By Newton's second law, acceleration = Force/ Mass , => g = F/m = GMe/Re2 Illustration: Find the value of g at the surface of Earth? Radius of Earth = 6.37 × 106 meter. Mass of Earth = 6 × 1024 kg. Solution:

g = GM/R2 G = 6.67 × 10-11 Nm2 kg-2 M = 6 × 1024 kg R = 6.37 × 104 m

As

...... (1)

Put all values in (1), we get g = 9.8 m/s2

Ans.

Variation of Acceleration due to Gravity (g) i)

Due to altitude

Consider a mass m at a height h from the surface of the earth. Now, the force acting on the mass due to gravity is F=G

, where M is the mass of the earth and R is the radius of the earth.

If the acceleration due to gravity at the given height is g', then mg' = G

=>

,

g' = G (Expanding binomially and neglecting the higher order terms).

(ii)

Due to depth:

If a particle of mass m is kept at a depth 'd' form the surface of earth, then gravitational force exerted on the particle of mass 'm'. (GM' m)/(R-d)2 = GM(R-d)/R3

where M' = mass of earth within radius of (R - d) .·.

M' = M/R3(R-d)3 = (GM/R3) R(1-d/R) = g(1-d/R)

(iii) Due to rotation of the earth: Consider a body at a point with altitude θ, on the surface of the earth. Let R = radius of the earth and ω = angular velocity of the earth about its own axis. F.B.D. of the body with respect to an inertial frame is shown in the adjacent figure.

Acceleration of the body with respect to the earth's centre O is (Rcos θ)ω2 directed towards the axis of rotation (i.e. the centripetal acceleration). From Newton's second law in the radial direction

or or

mg - Fn = m(R cos θ) ω2 cos θ Fn = m[g - Rω2 cos2 θ] Fn/m = g' = g - Rω2 cos2 θ

where g' is the apparent value of the acceleration At poles, θ = 90o => g' = g At the equator, θ = 0o => g' = g - Rω2 Figure given below has illustrated the variation of g with the distance of separation from Earth's centre.

Illustration: Find the value of g at a height equal to the radius of Earth. Solution: g = GM/(R+h)2 and g0 = GM/R2 (At the surface of Earth). g/g0 = R2/(R+h)2 = R2/4R2 ·.·h = R => g = g0/4 = 9.8/4 = 2.45 m/s2. Caution:

Ans.

Here h = R, so we cannot apply g' = g(1-2h/Re)

Illustration: At what angular velocity Earth should rotate, for the weight of an object at the equator to be zero? What would be the duration of a day in this case? Radius of Earth = 6.4 × 106 m g0 = 9.8 m/s2 Solution: For the weight to be zero, the value of g should be zero. That is Here, g' = g0 - Reω2 = 0 or ω = √(g0/Re ) = √(9.8/(6.4×106 )) = 1.2 × 10-3 rad/s The duration of one day will be equal to the time period of rotation T = 2Π/ω = 2Π/1.2*10-3 sec. Orbital Velocity When a satellite revolves in an orbit around a planet, it requires centripetal force to do so. This centripetal force is provided by the gravitational force between the planet and the satellite. If the mass of the satellite is m and that of the planet is M, then the gravitational force between them at a height h above the surface of the planet is

F = GMm/(R+h)2', where R is the radius of the earth. If the speed of the satellite in its orbit is (R + h), then the required centripetal force is mv 2/(R+h). .·.

mv2/(R+h) = (G Mm)/(R+h)2

or

v = √(GM/(R+h))=√(GM/R(1+h/R)) = √(gR/(1+h/R))

If the height is very small compared to the radius of the earth, then v = √gR If the time period of the satellite is 24 hrs rotating in the same sense as the rotation of the earth and if the plane of the orbit is at right angle to the polar axis of the planet (earth), then the satellite will always be above a certain place of the earth. This kind of a satellite is called geostationary satellite. Illustration: An artificial satellite of mass 100 kg is in circular orbit at 500 km above the earth's surface. Take the radius of the earth as 6.5 × 106 m. (a)

Find the acceleration due to gravity at any point along the satellite path.

(b)

What is the centripetal acceleration of the satellite?

Solution: Here, h = 500 km = 0.5 × 106 m R = 6.5 × 106 m r = R + h = 6.5 × 106 + 0.5 × 106 = 7.0 × 106 m (a)

g' = g(R/R+h)2 = 9.8((6.5 * 106)/(7.0 * 106)) = 8.45 m/s2

(b)

Centripetal force on the satellite, F = mv2/R

.·.

Centripetal acceleration, a = F/m = v2/r = (√(gR2/r))2/r = (gR2)/r2 =g R2/(R+h)2 = 8.45 m/s2

Gravitational Potential Energy

When two or more bodies interact with each other, some work has to be done in assembling them together in their respective places. Total work done in assembling bodies together, in their respective places, is called gravitational potential energy of the system. Potential energy of a system of two masses is defined as the amount of work done in bringing these two masses from infinity to their respective places.

In the right side figure, as the bag moves upwards, kinetic energy decreases, and gravitational energy increases. At the highest point (B) Kinetic energy is zero, and Gravitational potential energy is highest. As the bag did not get to, it start falling from point B downwards due to gravity. It starts falling slowly (kinetic energy is low) and then speeds

At

point

up

A,

bag

is

at

full

speed

and

kinetic

energy is

downwards.

highest,

whiles

gravitational

energy

is

nearly

lost.

When the bag of gold coins hits the ground, kinetic energy is converted into heat and sound by the impact. It is worth noting that the higher the gravitational energy of an object moving downwards, the lower the kinetic energy, and the lower the kinetic energy of an object moving upwards, the higher its gravitational energy. Gravitational potential energy of a system:Change in gravitational potential energy of a system is defined as the -ve of the work done by the gravitational force as the configuration of the system is changed.

Uf - Ui = Wgr Change in gravitational potential energy of two point masses m1 and m2 as their separation is changed from r1 to r2 is given by U(r2) - U(r1) = Gm1m2 [1/r1 - 1/r2] If, at infinite separation, gravitational potential energy is assumed to be zero, then the gravitational potential energy of the above two point mass system at separation r,

The below figure shows that the energy of the dart/gun system is initially present in the form of the elastic potential energy (PEs) and gravitational potential energy (PEg). The springs of the dart gun are compressed which accounts for the elastic potential energy. Furthermore, the dart is initially elevated at a height of 1-meter above the ground which accounts for the gravitational potential energy. The presence of these two initial forms of energy are shown by the PEg and PEs bars of the bar chart. Once projected, the dart no longer has elastic potential energy since the springs of the dart are no longer compressed. However, the dart does have a large amount of kinetic energy (energy of motion) since it is now moving at a high speed as it leaves the dart. The dart also has gravitational potential energy since it is still elevated to some height above the ground. As the dart ascends towards its peak, it is continuously slowing down under the influence of the downward force of gravity. During this ascent, there is a transformation of the mechanical energy from the form of kinetic energy (energy of motion) to gravitational potential energy (the stored energy of vertical position). At the peak, there is only a small amount of kinetic energy (the dart still has a horizontal motion) and a large amount of gravitational potential energy (the dart is at its highest vertical position). Finally, as the dart descends to the ground, the force of gravity speeds it up. As it falls, there is an increase in kinetic energy (due to the gain in speed) and a decrease in gravitational potential energy (due to a loss in vertical position).

Gravitational Potential:Gravitational field around a material body can be described not only by gravitational intensity vector but also by a scalar function, the gravitational potential V. The gravitational at any point may be defined as the potential energy per unit mass of a test mass placed at that point. V = U/m (where U is the gravitational potential energy of the test mass m). Thus, if the reference point is taken at infinite distance, the potential at a point in the gravitational field is equal to amount of work done by the external agent per unit mass in bringing a test mass from infinite distance to that point. The expression for the potential is given by

With the above definition, the gravitational potential due to a point mass M at a distance r from it is

Potential is a scalar quantity. Therefore, at a point in the gravitational field of a number of material particles, the resultant potential is the arithmetic sum of the potentials due to all the particles at that point. If masses m1, m2, ......, mn are at distances r1, r2, r3, ......, rn then potential at the given point is V = -G(m1/r1 + m2/r2 + m3/r3 +..........) The field and the potential are related as, E = -dV/dr Gravitational potential due to a shell (i) at a point outside the shell is: -GM/r (r>R) (ii) at a point on the surface of the shell is: -GM/R (iii) at a point inside the shell is: -GM/R Gravitational potential (V) due to a uniform solid sphere (i) Outside of the sphere at a distance r form the center, V = -GM/r. (ii) Inside the sphere at a distance r from the center, V = -GM/R3 (R2/2 - r2/6) Relation between Gravitational Potential and The Gravitational Field Strength Gravitational field strength at a point is given by E = GM/r2

We can also describe the gravitational field of a body by a scalar function called the potential. Let us begin with a test particle of mass m0 at an infinite separation from the body and move the test particle toward the body until their separation is r, where the potential energy is U(r). Now the gravitational potential V at that point is defined as V(r) = U(r)/m0. As we can calculate the radial component of gravitational force from U(r) according to the relation F = -dU/dr => m0E = -m0 dv(r)/dr Hence, we can find the radial component of the gravitational field intensity from V(r) According to the relation, E = -dV/dr.

Escape Velocity Escape velocity on the surface of earth is the minimum velocity given to a body to make it free from the gravitational field, i.e. it can reach an infinite distance from the earth. Let ve be the escape velocity of the body on the surface of earth and the mass of the body to be projected be m. Now, conserving energy at the surface of the earth and infinity, 1/2 mve2 - GMm/R = 0 => ve = √2GM/R.

Binding Energy Binding energy of a system of two bodies is the amount of minimum energy needed to separate the bodies to a large distance. If two particles of masses m1 and m2 are separated by a distance r, then the gravitational potential energy of the system is given by U = Gm1m2/r Let T amount of energy is supplied to the system to separate the bodies by a large distance. When the bodies are separated by a large distance, gravitational potential energy of the system is zero. For minimum T, conserving energy for initial and final positions, T+U=0 => T - Gm1m2/r = 0 or T = Gm1m2/r Hence, binding energy of a system of two particles separated by a distance r is equal to T = Gm1m2/r, where m1 and m2 are the masses of the particles. Gravitational Self Energy of a Uniform Sphere

Consider a sphere of radius R and mass M uniformly distributed. Consider a stage of formation at which the radius of the spherical core is r. Its mass will be 4/3 πr3ρ at that time, where r is the density of the sphere. Let an additional mass be added so that the radius of the core be increased by dr in the form of a spherical layer over the core. The mass of this layer will be 4πr2dr. ρ The mutual gravitational potential energy of the above mass and the spherical core of radius r, dU = (-G(4/3πr3ρ)(4πr2dr.ρ))/r = 16/3 π2ρ2Gr4dr Hence, total energy evolved in the formation of the spherical body of radius R i.e., self energy.



Gravitational potential at any point is a scalar quantity.



Its unit, in C.G.S. system, is erg g-1 and in S.I. it is J kg-1.



With an increase in „r‟, it becomes less negative i.e., it increases as we move away from the source of gravitational fleld.



Maximum value of gravitational potential is zero and this happens only at infinite distance from the source.

Problem 1:What is the gravitational Potential energy of the Moon-Earth system, relative to the potential energy at infinite separation? Solution:M = Mass of Earth = 5.98 × 1024 kg m = mass of Moon = 7.36 = 1022 k d = mean separation between Earth and moon = 3.82 × 108 m. So, gravitational potential energy of the Moon-Earth system (U) = -GMm/d Put all values U = -7.68 × 1028 J ______________________________________________________________________________________ Problem 2:A projectile is fired vertically from the Earth's surface with an initial velocity of 10 km/s. Neglecting atmospheric retardation, how far above the surface of the Earth would it go? Take the earth's radius as 6400 km. Solution:Let the projectile go up to a height h. Then the law of conservation of mechanical energy gives 1/2 mv2 - GMm/R = -GMm/R+h h = v2R2/2GM-v2R = v2R2/2gR2-v2R where g is the acceleration due to gravity on the surface of the Earth. = ((104)2 (6.4×106)2)/(2(9.8)(6.4×106)2-(104)2 (6.4×106)) _____________________________________________________________________________________ Problem 3:-

Two spherical bodies of masses 2M and M and of radii 3R and R, respectively, are held at a distance 16R from each other in free space. When they are released, they start approaching each other due to the gravitational force of attraction. Then, find: (a)

the ratio of their acceleration during their motion.

(b)

their velocities at the time of impact.

Solution:(a) Due to the mutual attraction, the masses attract each other. If the acceleration are a1 and a2, the net external force on the system = 0 =>

cm

= 0 => m1a1 - m2a2 = 0

or a1/a2 = m2/m1 = ½

(b) Taking both the bodies as a system, from conserving momentum of the system, m1v1 - m2v2 = 0 => m1/m2 = v2/v1 = 2 Now, conserving the total mechanical energy, we have 1/2 (2M) V12 + 1/2MV22 - G(2m)M/4R = - G(2m)M/16R and solving it we get. v1 = √GM/8R and v2 = 2√GM/8R Note:



The velocities and acceleration are w.r.t. the inertial reference frame (i.e. the center of mass of the system).

Problem 4:At a point above the surface of earth, the gravitational potential is -5.12 × 107J/kg and the acceleration due to gravity is 6.4 m/s2. Assuming the mean radius of the earth to be 6400 km, calculate the height of this point above the earth's surface. Solution:Let r be the distance of the given point from the centre of the earth. Then, Gravitational potential = -GM/r = -5.12 × 107 J/kg

...... (1)

and acceleration due to gravity, g = GM/r2 = 6.4 m/s2 Dividing (1) by (2), we get r = 5.12*107/6.4 = 8 × 106 m = 8000km

...... (2)

Thus, height of the point from earth's surface = r - R = 8000 - 6400 = 1600 km ______________________________________________________________________________________ Problem 5:The mass of Jupiter is 318 times that of earth, and its radius is 11.2 times the earth's radius. Estimate the escape velocity of a body from Jupiter's surface. [Given: The escape velocity from the earth's surface is 11.2 km/s.]

Solution:Hence, MJ = 318 Me; RJ = 11.2 Re; ve = 11.2 km/s We know, vJ = √2GMj/Rj and ve = √2GMe/Re so, vJ/ve = √(Mj/Me χ Re/Rj) => vJ = ve √(Mj/Me χ Re/Rj) vJ = 11.2{318Me/Me χ Re/11.2Re}1/2 = 11.2(318/11.2)1/2 = 59.7 km/s ___________________________________________________________________________________ Problem 6:Find the escape speed from a point at a height of R/2 above the surface of earth. Assuming mass of earth as M and its radius as R. Solution:Conserving mechanical energy of a point mass m which is to escape, and earth system we have, -GMm/(R+R/2) + 1/2mv2 = 0 => v = √(4GM/3R) ________________________________________________________________________________________ Problem 7:The escape velocity of a body on the surface of the Earth is 11.2 km/s. A body is projected away with twice this speed. What is the speed of the body at infinity? Ignore the presence of other heavenly bodies. Solution:If v is the velocity of projected and v' is the velocity at infinity, then we have by energy conservation principle. ½ mv2 - GMm/R = 1/2mv'2 + 0 Here v = 2ve Thus, (1/2) . 4ve2 - GM/R = 1/2v'2 => 2ve2 - GM/R = 1/2v'2 Now ve = √2GM/R => 2ve2 - v2e/2 = 1/2v'2 or, v'2 = 3 ve2 or, v' = √3 ve = √3 × 11.2 km/s = 19.4 km/s.

Satellites and Planetary Motion

It was Copernicus who, first of all, introduced the idea that the central body of our planetary system was Sun rather than Earth. Kepler later on confirmed by putting it on a solid mathematical back ground. Kepler announced two of his laws in 1609 and the third one 10 years later. The laws can be stated as below: (a) Kepler‟s first law (law of elliptical orbit):- A planet moves round the sun in an elliptical orbit with sun situated at one of its foci. (b) Kepler‟s second law (law of areal velocities):- A planet moves round the sun in such a way that its areal velocity is constant. (c) Kepler‟s third law (law of time period):- A planet moves round the sun in such a way that the square of its period is proportional to the cube of semi major axis of its elliptical orbit. T2 ∝ r 3 Here R is the radius of orbit. T2 = (4π2/GM) r 3 Earth and its Satellite:Consider a satellite of mass m revolving in a circular orbit around the Earth, which is located at the centre of its orbit. If the satellite is at a height h above the Earth's surface, the radius of its orbit r = Re + h, where Re is the radius of the Earth. The gravitational force between Me & m provides the centripetal force necessary for circular motion, i.e., GMem/(Re+h)2 = mv2/(Re+h) Or v2 = GMe/(Re+h) or v = √GMe/(Re+h) Hence orbital velocity depends on the height of the satellite above Earth's surface. Time period T of the satellite is the time taken to complete one revolution. Therefore, T = 2Πr/v = 2Π(Re + h)√(Re+h)/GMe or T2 = 4Π2(Re+h)3/GMe where r = Re + h If time period of a satellite is 24 hrs. Then, r = [GMeT2/4Π2]1/3 = 42400 km and h = 36000 km. This gives the height of a satellite above the Earth's surface whose time period is same as that of Earth's. Such a satellite appears to be stationary when observed from the Earth's surface and is hence known as Geostationary satellite. For a satellite very close to the surface of Earth i.e. h mE dSE ω2 = GmsmE/dSE2 mS = dSE2.ω2/G = 4Π2dSE2/GT2 =(4×(3.14)2×(1.49×1011)2)/(6.67×10-11×(365×24×60×60)2) =1.32 × 1019 kg. Problem 2:A Saturn year is 29.5 times the earth year. How far is Saturn from the sun (M) if the earth is 1.5 × 108 km away from the sun? Solution:It is given that TS = 29.5 Te; Re = 1.5 × 1011 m Now, according to kepler's third law TS2/Te2 = Rs3/Re3 RS=Re(TS/Te)2/3=1.5×1011 ((29.5Te)/Te)2⁄3=1.43×1012 m =1.43×109 km Problem 3:A planet of mass m moves along an ellipse around the sun so that its maximum and minimum distances form the sun are equal to R and r, respectively. Find the angular momentum of this planet relative to the center of the sun. Solution:According to Kepler's second law, the angular momentum of the planet is constant. So, mv1R = mv2r or v1R = v2r If the mass of the Sun is M, conserving total mechanical energy of the system at two given positions we have,

-GMm/R + 1/2 mv12 = -(G M m)/r + 1/2 mv22 So, GM[1/R - 1/r] = v12/2 + v22/2 or GM[(r-R)/Rr] = v12/2 + v22R2/2r2

Thus, v12= (2GM(R-r) r2)/Rr(R2-r2 ) = 2GMr/R(R+r) Now, angular momentum = mv1R = m√2GMR/(R+r)

Fluid Mechanics A flowing liquid may be regarded as consisting of a number of layers one above the other.Fluid Mechanics is of fundamental importance as it talks both about the fluids both at rest and in motion.

Steady Flow (Stream Line Flow) It is the flow in which the velocity of fluid particles crossing a particular point is the same at all the times. Thus, each particle takes the same path as taken by a previous particle through the point. Steady, streamline or laminar flow

The flow of fluid is said to be streamline if the velocity at any point in the fluid remains constant with time (in magnitude as well as in direction) in this case energy needed to drive the fluid is used up in overcoming the viscous force between its layers. All particles passing through a point in a steady flow follow the same path. The paths are known as streamlines> An example is water slowly flowing through a pipe. Turbulent flow When the motion of a particle varies rapidly in magnitude and direction, the flow is said to be turbulent. In other words, when the velocity exceeds beyond the critical velocity, the paths and velocities of liquid change continuously and haphazardly. This flow is called turbulent flow. An example is water coming out of a fountain. Critical velocity If in the case of a steady flow of fluids the velocity of flow is gradually increased it is found that the motion remains steady (streamline or laminar) upto a certain limit. If the velocity of flow crosses a certain limit the fluid particles do not follow the path of the preceding one and the motion becomes turbulent. The maximum velocity upto which fluid motion remains steady is called critical

velocity. According to Reynold, in case of motion of fluids in narrow tubes, critical velocity depends on the density r and coefficient of viscosity η of the fluid as well as radius of the tube. i.e. Vc = η/rρ

or Vc = R η/ρr

Here R is a dimensionless constant called Reynold's number. For steady flow R < 2000. For 2000