Chapter 3
Engineering Mechanics
Chapter 3 DYNAMICS
Dynamics can be divided into two main branches (a) Kinematics (b) Kinetics In kinematics, motion of particles or rigid bodies is studied without considering the forces that produce or change this motion. In kinetics, motion of particles or rigid bodies is studied with the unbalanced force system that produces or changes this motion.
Kinematics of Rectilinear Motion Motion with constant acceleration
Where u=initial velocity, v=final velocity, s=distance of travel, t=time and a=acceleration Example 1 Velocity of a particle is defined as V=Kx3-x2 6x , where “V” is in m/s and ‘x’ is in ‘m’ compute the acceleration when x=2 and k=1 Solution Given, V= Kx3-x2+6x 6 (
6 )
, (
6)
= (123-22+62) (3122-22+6) = (8-4+12)(12-4+16) =1614=224m/s2 THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,
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Chapter 3
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Example 2 The brakes of a train reduce its speed from 60 to 20 km/h while it runs 200 m. Assuming that there exists constant retarding force, find (a) (b)
How much further the train will run before coming to rest, and How long will it take.
Solution 6
v
/ ,
u
/
as
400 – 6
α
a= (a)
/ For the train to come to rest , v
u
as
6 =225 m Further distance moved = 225 -200 =25 m (b)
v
u
at
0 =20 – 8000 t 6
Motion of Bodies Projected vertically upwards When a body is projected vertically upwards, it is under the effect of the downward acceleration due to gravity, i.e., it moves with retardation. Its velocity, therefore, gradually decreases until it becomes zero the body is then for an instant at rest and immediately begins to fall with a velocity which increases numerically but is negative. Thus, we get
v
u
gs
The following important results can be deduced from these equations: THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,
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Chapter 3
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Time to reach the highest point, Maximum height reached, Time for returning to the starting point time of flight = Example 3 A particle is dropped from the top of tower 200 m high and another particle is projected at the same time vertically upwards from the foot of the tower so as to meet the first particle at a height of 50 m. find the velocity of projection of the second particle. Solution Let the particle meet after seconds. The first particle falls from rest a distance = 200 -50 = 160m. Hence 150
√ If
( )
is the required velocity1 of the second particle, then ( )
Adding Equation (a) and (b), we get 200 = 6
/
Example 4 Two balls are projected simultaneously with the same velocity from the top of a tower, one vertically upwards and the other vertically downwards. If they reach the ground in times t and t respectively, show that √t t is the time which each will take to reach the ground if simply let drop from top of the tower. Solution Let h = height of the tower u = velocity of projection Taking downward direction positive, for the first ball, we have ,
,
, gives
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Chapter 3
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( ) For the second ball, ,
,
,
∴
( )
Eliminating
from equation (a) and (b), we get
If is the time to reach the ground when initial velocity is zero, we get
∴
√
Example 5 The speed of a train increases at a constant rate from zero to and then remains constant for an interval and finally decreases to zero at a constant rate β If d be the total distance described, prove that the total time occupied is (
)
Solution Let t , t and t be the times for acceleration, constant speed and deceleration, then , Total distance travelled = Or t
t
t
Now total time And t
,
∴ 2t THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,
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Chapter 3
d
t
(
) (
=
(
Engineering Mechanics
)
)
Example 6 Two stopping points of an electric tram car are 450m apart. The maximum speed of the car is 20km/h and it covers the distance between the stops in 100seconds. If both acceleration and retardation are uniform and the latter is twice as great as the former, find the value of each of them and also calculated how far the car runs at the maximum speed. Solution Let t , t , t be the times for acceleration, uniform velocity and deceleration. Then
Maximum speed of car = 20 km/h m/s 450 = =(
(
)
∴t
t
t
2t
t
t
t
t
) 6
Also ∴t =2t
and t t
6 s, t
6 6
t
6 6 s, t
s
Distance covered during constant speed =
6
Example 7 A person going Eastwards with a velocity of 4 km/h finds that the wind appears to blow directly from the North. He doubles his speed and the wind seems to come from the North-east. In what direction and with what velocity is the wind blowing?
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Chapter 3
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Solution Let the actual velocity of wind be figure.
km/h along OA making an angle θ with OE, as shown in
N N.E
/
O
W
B
( S
(a)
E
θ
A )
C
When the velocity of the man along OE is 4 km/h the velocity of wind is along ON. Therefore, the relative velocity of wind will be along OS. Therefore, the relative velocity of wind will be along OS. The resultant of 4Km/h and the relative velocity of wind is . Resolving along OE, we get 4=
(b)
When the velocity of man is 8km/h the relative velocity of wind is along OB, whose resultant is again . Resolving perpendicular to OB, we get )
cos( √
(cos
(cos
∴
cos
sin )
√
sin )
√ 6
6
√
/
tan = Example 8 The driver of a car travelling at 72 km/h observe the traffic light 300m ahead of him turning red. The traffic light is timed to remain red for 20s before it turns green. If the driver wishes to pass the light without stopping to wait for it to turn green, determine (a) the required uniform deceleration of the car and (b) the speed with which the driver crosses the traffic light.
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Chapter 3
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Solution
6
/
6 ,
Now 300 = 20 ×20 + / Also
/
=
6
/
Example 9 Two trains A and B of length 400 m each are moving on two parallel tracks with uniform speed 72km/h in the same direction, with A ahead of B. The driver of B decides to overtake A and acceleration by 1ms If after 50 s, the guard of B just brushes past the driver of A, what was the original distance between them? Solution Relative velocity of B w.r.t A = 0 Let d
orginaldistance between the trains in m. Train B
Train A
400 m
400 m
Distance covered by B in 50 s = 400 + d +400 =800 +d Using (
)
Example 10 A stone projected vertically upwards from a point A passes a point B after 3 seconds. If it returns to A after a further interval of 4 seconds, find (i) the height of B above A (ii) the velocity with which the stone passes the point midway between A and B. THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,
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Solution Let
projection maximum height reached Time taken from A to C and back to a A
A
= 3 +4 =7 s
Or Time taken from A to B = 3
= 34.3 × 3 or
AB = 58.8 m AD =
( v
) 6
6
6
Kinematics of Curvilinear motion Motion of projectile Maximum height (h) = Time required to reach maximum height (t) =
∴time of flight = Range (R) =
∴for maximum range,
= 450
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Chapter 3
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Acceleration Analysis Background Material: Rate of Change of Vector A vector quantity has magnitude and direction and hence, the vector can change in time in two ways-In magnitude or direction. Therefore, when we take the time-derivative of a vector ie ( ) () lim , the numerator is change in vector from time t to t + ∆t can have two components. •
•
Rate due to change in Magnitude: This is got by differentiation of the scalar magnitude | | n̂ (n̂ | | is the unit vector in the direction of velocity vector) Ex: For vector ̂ ̂, ̂ Rate due to change in Direction: This is more involved, but it can be shown that a fixed vector V rotating in a rate ω has a rate of change due to direction change as
So, in general we have that for a generic vector changing in magnitude and rotating at rate ω, we have | |
̂
Acceleration Analysis Given in the figure, is a fixed ground Coordinate System (CS) inside of which a rotating CS with origin at O rotates at ω, has angular acceleration of , translates at V0 & accelerates at A0. Inside this rotating CS O, is a point P which translates at OVP & accelerates at OAP locally.
Figure : Diagram for Acceleration Analysis OVP =Velocity of Point P as viewed from Origin of Rotating Cord. Sys. O
OAP =Acceleration of Point P as viewed from Origin of Rotating Cord. Sys. O VO =Velocity of Origin of Rotating CS O w.r.t Fixed Ground CS with origin at G AO =Acceleration of Origin of Rotating CS O w.r.t Fixed Ground CS with origin at G THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,
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O
=Angular Velocity of Rotating CS O w.r.t Fixed Ground CS with origin at G
O
=Angular Acceleration of Rotating CS O w.r.t Fixed Ground CS with origin at G
OP =Position vector of a point in w.r.t Rotating CS with origin at O GP =Position vector of a point in w.r.t Ground CS with origin at G GO =Position vector of Origin O of Rotating CS w.r.t Ground CS with origin at G We are interested in acceleration of point P as seen from ground because as coordinate system O is accelerating and rotating system, it is a non-inertial coordinate system and hence Newton’s laws are not applicable for vectors in that frame. Hence, we need to write P in terms of ground coordinate system if we are to apply the same for subsequent force/torque analysis. So, we start by writing down the position of point P as seen from ground coordinate system.
GP = GO + OP ⇒ ⃗
⃗
⃗
Now, differentiating w.r.t time, we get
⃗
⃗
⃗
⃗
⃗
But we know that (velocity of point P as viewed from O) & (velocity of point P as viewed from ground). Now, ⃗ is a vector which has change in magnitude (motion in the frame O with a local velocity of OVP, acceleration OAP) & also has change in direction (rotates at ω & α). So, we can write as follows. ⃗ ⇒ We now have the velocity of point P w.r.t ground G. To get acceleration, further differentiate both the sides.
We now have the velocity of point P w.r.t ground G. To get acceleration, further differentiate both the sides.
Now, we know that
is a rotating vector in the CS at O and hence has magnitude change(
and direction change(rotates ( )
, )⇒
)
⇒
Simplifying the above expression gives as follows (
)
So, we see that we have five terms in total here on the Right Hand Side which is why we call this above formula as the five point acceleration formula. THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,
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Chapter 3
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(
): This is the acceleration of the origin O of rotating CS w.r.t ground CS at G. Term 2( ): This is the acceleration of the point P local to/when viewed from within the rotating CS at O. Term 3( ) This component is the Coriolis acceleration component produced due to velocity inside a rotating frame. Term 4 ( ): This is the Tangential acceleration component which acts in the instantaneous tangential direction of motion of point P. Term 5(ω (ω rop)) This is the Radial/Centripetal acceleration component and acts radially inward towards the point of center of rotation.
Thus, all the five terms of motion are clearly derived from this highly generic method. Newton's law can now be applied to the LHS term which is got by adding up the five acceleration components in the RHS.
Acceleration Analysis in Cylindrical or Path or Curvilinear Coordinates In case of cylindrical coordinates, we have the coordinate directions as shown below. Clearly, for a motion from time , The direction of CS itself changes in & direction whereas remains constant. We proceed with derivation in the same manner as above. P (Projecting Outward) Trajectory of the particle P (
)
O
( ) ( ) (
)
dθ
∆ ( ) ( )
∆ ( ) (
)
( )
(
)
Rate of change
Rate of change
Rotation of CS from t to t+td Figure: Diagram for Acceleration Analysis
⇒(
)
Now, as shown in figure, when the CS rotates by (
)
( )
∆
in a time dt, we see that
( )
Now, note that in the limiting case, as , the direction of ∆ ( ) approaches tangential direction in . Also, as far as magnitude is concerned, THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,
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Chapter 3
|
|
|
( )
(
|
Engineering Mechanics
| |
)⇒
Similarly, from the diagram above, using the same logic, we can prove that However, as
does not change direction in time ̇
Using the above results, we proceed to get as follows.
̇ ̇
Again differentiating, we get ̈
̇
̇
( ̈
)
(
̇
)
̈
This is the final formulae for velocity and acceleration of a moving particle in cylindrical coordinates. Examples 11 Acceleration of a point moving in a rotating frame. Consider the rotating tube shown in figure. It is given that the arm OAB rotates with counterclockwise angular acceleration / and at the instant shown the angular speed / . Also, at the same instant, the particle P is falling down with speed / / and acceleration / / w.r.t. tube. Find the absolute acceleration of the particle at the given instant. Take in the figure. Solution Fixing a ground coordinate system on ground and a rotating coordinate system with origin at O and moving along with the frame OAB, we see that for point P, position vector , But clearly, & are all fixed vectors from geometry which does not change in magnitude (length) with time. So, only change is change in direction. But changes in length because point P has a local velocity when viewed from B & also it changes it direction like the other two vectors.
P
ω, ̇
/
O
/
ℓ = 2
A ℓ
2ℓ
B
ℓ f = t Figure:2Problem 1 Figure
ft
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t
Chapter 3
Engineering Mechanics
So, ⇒ ( ) . Same way, on further differentiating, we / get acceleration ( ) ( ( ) ) . / / / Now, we have to represent all vectors in same CS. Here, we can write ̂, / ̂ ,̂ ,̂ ̂, and upon simplifying, we get the answer as ̂
̂&
/
6̂
=- ̂ Substitute these values / & 66 ̂ ̂ /
Example 12 A particle moves along the spiral as shown in figure, and defined by the equation r= 12t and o=2nt. Determine the velocity and the acceleration of the particle when t=0 < t=0.3 see Y
0 O
X
Solution Given, R= 12t and = 2nt
We know, for r- System, = 12+12t 2 =12(1+2t) m/s Acceleration a= ar+a [
(
) ]
*
+
= -(12t2)2 + 0+ 2122 a=48-5762t2 m/s2 V= 12(1+2t)m/s a=48-5762t2 m/s2 THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,
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Example 13 Find the maximum horizontal range of a cricket ball projected with a velocity of 15m/s. If the ball has a range of 20m, find the least angle of projection and the least time taken. Solution Horizontal range = For maximum horizontal range, α ∴R
m
Now Sin α α
6 or 6 6
∴ Least angle of projection =
6
Least time of projection = The value of the maximum range (
r
) (
)
Example 14 A particle is projected from a point on an inclined plane with a velocity of 30m/s. The angle of projection and the angle of plane are and to the horizontal respectively. Find the range and time of flight of the particle. Solution Range on an inclined plane *sin(
)
)
Here For the range to be maximum α–β Or 2 ×5 Thus, this corresponds to maximum range on the given inclined plane. Maximum range, r
u
(
)
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Chapter 3
=
(
)
(
=
6
Time of flight, T =
Engineering Mechanics
)
6 (
(
)
)
= Example 15 A projectile is launched as shown in the figure. (i) Compute the velocity of the projectile if R = 250m (ii) Find the flight time of the projectile Solution
u
(i) R =
]
Sin
600
sin ] A
300
Given, -
R
R = 250m (range of the projectile) B
G = 9.81 m/s2 (Acceleration due to gravity) Angle of the projectile = 600 Β
Angle of inclination,
0
U = velocity of projectile = ? ∴
sin(
∴u
6
)
sin
]
m/s
(ii) Flight time, (
T= =
(
)
)
= 10.09sec. u = 42.89m/s T = 10.09sec
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Example 16 A rope, to which a weight is attached, passes around a pulley 50cm in diameter. The angular acceleration of the pulley is 18 rad/s . If the pulley is initially at rest, find(a) the time required for the weight to attain a velocity of 15m/s, (b) the number of revolutions through which the pulley rotates during that period and (c) the total acceleration of a point on the rim of the pulley 0.5 second after it was at rest. Solution (a) ω
6
ω
ω
/
αt
60= 0+18t
(b) θ
ω t =
(
)
Number of revolutions = (c) Tangential acceleration α
rα
cm /s
Normal acceleration α
rω
After 0.5 s, we have ω
ω
αt
rad/s
∴
cm/s
Total acceleration, = √(
) (
)
√ cm/s
Example 17 A ball is thrown from the top of a tower 60m high with velocity of ms at an elevation of above the horizontal distance from the foot of the tower to the point where it hits the ground. (g = 10 ms ) THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,
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/ O
60 m
A
R
B
Solution Refer the figure sin 6
sin
6 or
6
or t
t √
√ 6
6
Hence, R = cos = 20 ×
√
√
cos
6
6 = 79.67 m
Example 18 A shell is required to be projected so as just to pass over the wall of a fort horizontally. If the wall is 20m high and 60 m distant from the gun, find the angle and the velocity of projection. Solution The shell grazes the wall horizontally. Therefore, the height of the wall is the maximum height attained.
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With reference to figure ( ) Also
6
or
R = 120m
Now 120 =
( )
Dividing (1) by (2) tan 6 Or tan ( )
6
From (1), 20 = u 6
/
Kinetics of Rectilinear Motion Frames of Reference , ,
G
Figure: Diagram for Acceleration/Force Analysis THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,
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=Velocity of Point P as viewed from Origin of Rotating Cord. Sys. O =Acceleration of Point P as viewed from Origin of Rotating Cord. Sys. O =Velocity of Origin of Rotating CS O w.r.t Fixed Ground CS with origin at G =Acceleration of Origin of Rotating CS O w.r.t Fixed Ground CS with origin at G =Angular Velocity of Rotating CS O w.r.t Fixed Ground CS with origin at G =Angular Acceleration of Rotating CS O w.r.t Fixed Ground CS with origin at G OP =Position vector of a point in w.r.t Rotating CS with origin at O GP =Position vector of a point in w.r.t Ground CS with origin at G GO =Position vector of Origin O of Rotating CS w.r.t Ground CS with origin at G As seen in acceleration analysis, the acceleration experienced by a point P moving in a rotating and accelerating coordinate system, as seen from the global frame of reference G, is given by(5term acceleration formula) (
)
Now, we know that Newton's law is F = ma or Force F required to make an object of mass m move at acceleration a is ma.Therefore, standing in the frame G, we see acceleration given by previous formula & therefore, if we multiply both sides of the equation by mass of particle P, i.e., m, we get force as follows. (
(
))
However, from the rotating frame of reference O, if we write the Newton's law, we see that it because an observer in O frame only sees the)AP acceleration for the point P. However, there can be only one force required to move the object from any frame. So, which one is correct value, or FG? Clearly, the observer in frame O is applying a lesser force as per Newton's law but that's because he is neglecting many acceleration components including Coriolis acceleration components & hence, FO is wrong. So, the conclusion is observer in frame O cannot employ Newton's law to estimate force whereas observer in frame G can. Frames where Newton's law can be applied (G) are called Inertial Frames of Reference & those frames where Newton's law cannot be applied(O) are called Non-Inertial Frames of Reference. Simply put, any frame accelerating or rotating w.r.t a stationary frame is a Non-Inertial Frame and any frame stationary or moving at a constant velocity w.r.t a stationary frame is a Inertial Frame. Clearly, between an Newton's Law equations for Inertial & Non-inertial frames, the difference comes in 4 terms. These 4terms are 4 forces not in strict physical sense as such forces don't physically exist but come up in the equations. Hence, these terms are called Fictitious Forces or Phantom Forces. They are as follows. 1.
Inertial Force: The term & ( ), caused by translational & rotational acceleration of frame of reference O are fictitious forces. THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,
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Chapter 3
2.
3.
Engineering Mechanics
)caused by velocity of particle P inside the Coriolis force: The term ( accelerating frame O is a fictitious force. A major physical manifestations of this force include cyclones/hurricanes. ( ))due to rotation of the frame O is a fictitious Centripetal Force: The term ( force.
( ) ( ) ( )) Further, we see that So, ( from the above equation, we can conclude that, if we indeed want to write force balance equation in frame O, we should take into consideration, the fictitious forces & draw them in opposite direction in the Free Body Diagram (FBD). Finally, if we also consider the term as a fictitious inertial force and draw it in opposite direction, this is equivalent to writing an equation (
)
(
)
(
(
))
Which is the form∑ which is the equation for equilibrium/statics. Thus, we see that considering each fictitious force in opposite direction of motion in the FBD, we can convert a dynamics problem to statics problem. This famous principle is called the D'Alembert's Principle. D’ Alembert’s Principle It was pointed out first of all by D’Alembert that on the line of equation of static equilibrium, equation of dynamic equilibrium can also be established by introducing inertia force in the direction opposite to acceleration in addition to the real forces acting on the system. According to Newton’s second law of motion, F = ma where Or Now
is the inertia force Example represents the D’ Alembert’s principle, which may be
stated as follows: When different forces act on a system such that it is in motion in a particular direction, the algebraic sum of all the forces acting on the system in the direction of the motion, including the inertia force taken in opposite direction to motion is zero. Thus in general F Where
, ,
ma
or ∑F
∑ ma
]
Where ∑ indicates the sum of all forces acting on the body in the direction of motion.
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Example 19 An elevator weighs 25 kN and is moving vertically downwards with a constant acceleration. Write the equation for the elevator cable tension. Starting from rest it travels a distance in this time. Neglect all other resistance to motion. What are the limits of cable tension? Solution Here W = 25kN For downwards motion of elevator, using D’Alembert’s principle, we get T=W– T=W( Now
) ,
,
35 = 0 + / T = 2(
)
6 kN
Maximum tension T = 25 =(
)
6
Example 20 A collar ‘P’ is free to slide along the smooth shaft ‘Q’ mounted in the vertical frame as shown in the figure below Determine the horizontal accretion ‘a’ of the frame necessary for maintain the collar in a fixed position on the shaft.
P
a
Q
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Solution: F B D of the Collar ‘P’ a
x ma y mg
N
Consider mass is “m’ Acceleration of ‘P’ & ‘Q’ same, ∴ ∑ N cos -mg = 0 1 & ∑ ma-Nsin = 0 2 From eq 1 & 2 we, can write
a= gtan Example 21 A load W is used to raised by a rope from rest to rest through a height h and the greatest tension which the rope can safely wear is W. Show that the least time in which the ascent can be made is √(
)
Solution The time required will be least if the motion is first with maximum possible acceleration and then with maximum possible retardation. The maximum possible acceleration is obtained when the tension in the rope, is , and maximum retardation is when this tension is zero. For the first part when the load is going up, we have by D’Alembert’s principle T –W -
( ) For the second part
a =-g Let t
time for accelration
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t
Engineering Mechanics
time for retardation
v
maximum velocity
Then acceleration =
(
)
Retardation = Total distance moved =
(
Now t t
(
t (
)
(
)
Or
t
)
t
√ (
)
,
(
)
(
)
(
(
)
) nh )
Example 22 A system of weights connected by spring, passing over pulleys A & B is shown in Fig below. Find the acceleration of the three weights assuming weightless springs and ideal conditions for pulleys. Solution Pulley A has two weights 150N and 60 + 40 = 100N acting on it. Let T 1 be the tension in the string for the downward motion of N weight using D Alembert’s principle, we have T1 – 150 + 150/g =0 T1 – 100 -10 a1 /g = 0 Solving these, a1= 1.962
N
N 6 N
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For pulley B, let T2 be the tension in the rope and a2 be the acceleration of the 60N downwards. Again using D Alembert Principle, we get T2- 60 + 60a2/g =0 T2- 40 - 40a2/g=0 Solving, a2=1.962 m/s2 Hence the acceleration of 150N and 60 N weights is 1.962 m/s2 downward and the acceleration of 40 N weight is also 1.962 m/s2 upwards. Example 23 Two blocks A and B are connected to each other by a spring and a string. The string passes over a frictionless pulley as shown in Figure below. Block B slides along the vertical side of a stationary block C and the block A slides along the vertical side of C, both with same uniform speed. The coefficient of friction between the surfaces of the block is 0.2. Force constant of the spring is 1960N/m2. If the mass of block A is 2 Kg, calculate the mass of block B and the energy stored in the spring. Solution Free body diagram is shown below B
T R
F C
A
T B A
R= mB g T=F= R = mB g
-------------------(1)
For Block A =T- mAg
-------------------(2)
From (1) & (2) mAg = mB g mB = 10 kg T= mAg=19.6N Extension of spring = T/k = 19.6/1960= 1/100 m Energy stored in the spring = ½ kx2 = 0.098 J THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,
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Example 24 Two weights 800 N and 200 N are connected by a thread and move along a rough horizontal plane under the action of a 400N applied to the first weight of 800N as shown in Figure below. The coefficient of the friction between the sliding surfaces of the weights and the plane is 0.3. Determine the acceleration of the weights and the tension in the threads D Alembert’s principle Solution Using D Alembert’s principle, we have 400 - W2 a /gT - W1 a/g –
R2 –T =0 R1 =0
Solving, a = 0.981 m/s2 and T = 80N
Kinetics of Curvilinear Motion Central force motion Centrifugal force = Where r = radius of the path = angular velocity v = linear speed g = acceleration due to gravity Moment of momentum (angular momentum)of the whole body
Iω
Where I = mk , k being the radius of gyration.
Moment of Inertia We know that Newton’s laws of motion define the equations that govern motion of bodies in macro scale. As far as translation is concerned, • First Law: At its crux defines Inertia or the resistance offered by body to change in state(rest or motion). For translation, inertia is same as mass of body & is the resistance offered to change of translation state of the body. • Second Law: This defines the relation between inertia and force required to induce an momentum change in the body as ( ) • Third Law: This proposes the action and reaction pairs of Forces. Now, if we extend this to rotation, the concept analogous to these are generated as follows. • First Law: At its crux defines Moment of Inertia or the resistance offered by body to change in state(rest or motion). For rotation, moment of inertia is the resistance offered to change of rotation state of the body. • Second Law: This defines the relation between moment of inertia and the torque required to induce a change in angular momentum in the body as ( ) THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,
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• Third Law: This proposes the action and reaction pairs of Torques. The moment of the inertia force on a particle around an axis multiplies the mass of the particle by the square of its distance to the axis and forms a parameter called the moment of inertia (denoted by I). The moment of inertia of an object is defined by the distribution of mass around an axis. It depends not only on the total mass of the object but also on the square of the perpendicular distance from the axis to each element of mass. This means the moment of inertia increases rapidly as masses are distributed more distant from the axis. For example, consider two wheels that have the same mass, one that is the size of a bicycle wheel and one that is half the size. The larger wheel has four times the moment of inertia even though it is only twice the diameter.
Figure 1: Computation of Moment of Inertia • • •
For system of particles, moments of inertia of individual particles sum to define the ∑ moment of inertia of a body rotating about an axis ie For rigid bodies moving in a plane, such as a compound pendulum, the moment of inertia is a scalar got by integrating over the body mass ie I ∫ r dm ∫ r dV For rigid bodies moving in three dimensions, such as a spinning top, the moment of inertia becomes a matrix, also called a tensor with 9 components (6 independent components). For easiness, moment of inertia is written in lumped form as where k is called radius of gyration.
Theorems for computing Moment of Inertia for compound bodies The main theorems which will help in computing Moment of Inertias are as follows. • Superposition: The moment of inertia of the body is additive. That is, if a body can be decomposed (either physically or conceptually) into several constituent parts, then the moment of inertia of the whole body about a given axis is equal to the sum of moments of ∑ inertia of each part around the same axis ie •
Perpendicular Axis: If Ix, Iy, Iz are moments of inertia around three perpendicular axis passing through the bodies center of mass, then each of them cannot be greater than the sum of two others: For example . Here the equality holds only if the body is flat/planar body (negligible z values ie z ≈ 0)). When body is planar, ∫ ∫( ) ( ) ) ≈ )⇒ ∫ ∫ ∫( ∫( , thereby proving the above identity.
•
Parallel Axis: If the objects moment of inertia around a certain axis passing through the center of mass is known (ICM), then the parallel axis theorem or Huygens Steiner theorem provides a convenient formula to compute the moment of inertia of the same body around THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,
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a different axis (Id), which is parallel to the original and located at a distance d from it. The formula is only suitable when the initial and final axis are parallel & is given by Id = ICM + Md2 Examples 25 1.
Given 2 masses - m@2r & 2m@r from axis of rotation, to find the moment of inertia, we see that this is a particle system and hence have Inet = £\ Ij = 2m(r)2 + m(2r)2 = 6mr2.
2.
Given a disk of radius R & mass M, we can divide the continuum into discrete rigid bodies and take integral as follows. I = L,r2dm = f„ 7rn fR nr2(rd6dr^L) = ±MR2. However, note that in case of a loop/hoop it has same mass and radius,
Figure: Computation of Moment of Inertia for a disc
The Moment of Inertia / = MR2 because in that case the whole of the mass is distributed at a larger distance from the rotation/central axis. 3.
For a given right circular cone with radius R, mass M and height H about the vertical axis, we have as follows. We divide the cone into stacked discs of diminishing radius as shown and do a summation of individual I’s for getting the total moment of inertia of the compound body. ) ∫ ( ) ⇒ ∫ ( (
) On
Figure: Computation of Moment of Inertia for a cone
Simplifying, we get Further, for any axis passing through the apex of cone, parallel to base (3rd figure in above diagram), we proceed as follows. Here, a common mistake people make is to employ the perpendicular axis theorem but the point to be noted is that it can be used only for planar bodies. So, we again break up the bodies into smaller bodies and do integration as follows. We see that the moment of inertia about an axis parallel to base but passing through the elemental disc shown is given by (obtained by applying perpendicular axis THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,
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theorem for the disc which is a planar body). Now, applying the perpendicular axis theorem for getting the moment of inertia about the axis parallel to base passing through the apex, we get IdApex = Id + dm(H - h). Finally, integrating this, we get as follows. (
∫ (
) )
(
gives
(
∫ (
) )(
) which on solving
)
Similarly we can find the moment of inertias for any given shape. I for some common shapes are given below. No. Shape I 1 Rod of length L and mass in (Axis of rotation at the end of the rod) 2
Rod of length L and mass m (Axis of rotation at the center of the rod)
3
Sphere (hollow) of radius r and mass m
4
Ball (solid) of radius r and mass m
5
Thin rectangular plate of height h and of width w and mass m
6
Ellipsoid (solid) of semi-axis a, b, and c with axis of rotation a and mass m
( (
) )
Example 26 The drum shown in Fig. (a) has a radius of gyration of 30 cm and weighs 1.8 kN. It is supported by means of small hubs which rest in bearings. A weight of 1 kN is attached to one end of a rope, the other end being wrapped around the drum. Neglecting friction in the bearings, determine the acceleration of the weight, the angular acceleration of the drum, and the tension in the rope. Solution Moment of inertia of drum, (
)
6 ω
37.5 cm
37.5 cm α
1.8kN
R W=1.8kN
T T 1kN
(a)
a
T 1kN (b)
The free body diagram is shown in Fig. (b) Let T be the tension in the rope Using D’ Alembert’s principle, we have for the 1kN weight, THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,
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Chapter 3
∑
(
Engineering Mechanics
)
1000 – T-
(a)
And for the drum ∑
(
)
∴
6
Now a = r
(b)
= 0.375
..
(c)
Substituting Eq. (c) in (a), we get 1000 – T – ∴
6
Substituting in Eq. (b), we get 0.375 (1000 – 38.226 )- 16.5 375 – 14.335 - 16.5 6
=0
/ 6 6
6
/ 6
Example 27 Two weights and of 8 kN and 5 kN are attached at the ends of a flexible cable. The cable passes over a pulley of 100 cm diameter. The weight of the pulley is 600 N with a radius of gyration of 50 cm about its axis of rotation. Find the torque which must be applied to the pulley to arise the 8 kN weight with an acceleration of 1.5 / . Neglect friction in the pulley bearing. Solution Let T and T be the tensions in the cable on the weights W and W sides respectively and be the torque applied to the pulley to accelerate the 8000 N weight with 1.5 / as shown in Fig. Using D’ Alembert’s principle, we have for T
=
,
(a)
and for (b) For the pulley, we have THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,
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∑
Or M
(
)
(c )
Also a = r
(d)
From Eq. (d), we get / ∴
∴
(
6
)
(
)
α 50 cm
600N
a
a 8kN W1
5kN W2
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Example 28 A step pulley and mass system is shown in the figure along side. Consider pulley is mass less & frictionless, string is non extendable
r1 r2
m1
m2
Find the m2>m1 and m2 moving downward Solution F.B.D is, m1a1 S1
S2 a1
a1
m 1g
m 2g m2a2
From the F.B.D, we can write,.. S1-m1a1-m1g & S2-m2a2-m2g
2
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For the rotation of ‘O’ of the step pulley, we can write n1 0 n2
x1=r10 x2 = r20
∴
we can write,
Considering equilibrium at that instant S1r1=S2r2 From eg 2,3,4, we can write
From eq 1 & 5 we can write (
)
(
)
Kinetic of system with variable mass From newton’s 2nd Law of motion, (
)
Where m = mass of the system v = Velocity of the system F = Force Consider mR= mass of rocket THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,
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mf= mass of the fuel vR = Velocity of rocket vf = Velocity of the fuel u
Rocket Fuel (u-v) (
)
Now vr=v And vf= -(u-v) u=Velocity of burn fuel with respect to rocket
(
)
(
)
(
)
∫
∫( (
m0 = Initial mass of the system (
)
(
) )
)
Friction Friction is the "evil" of all motion and always resist relative motion between two bodies. Friction is actually a force that appears whenever two things rub against each other. Although two objects might look smooth, microscopically, they're very rough and jagged, as this picture shows: 1 Mass
2
1 2
Figure: Microscopic Source of Friction THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,
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As they slide against each other, their contact is anything but smooth. They both kind of grind and drag against each other. This is where friction comes from. But friction is not all bad. In fact, it has a lot to do with life as we know it here on Earth. Without it, we wouldn't be able to walk, sit in a chair, climb stairs, or use a mouse to surf the web. Everything would just keep slipping and falling all over the place. When we walk, we try to push Earth backward, but friction opposes this relative motion by a forward reaction on the body which pushes us forward making us walk. Another important example of friction is rolling. Friction is usually distinguished as being either static friction (the frictional force opposing placing a body at rest into motion) and kinetic friction (the frictional force tending to slow a body in motion). In general, static friction is greater than kinetic friction. The force due to kinetic friction is generally proportional to the applied force. However, there is no strong consistent theory for friction and so, based on experimental observations, an empirical result is developed whereby a coefficient of friction is defined as the ratio of frictional force to the normal force on the body. We'll carry out a thought experiment. Take a mass (say 1Kg) and start applying a force on it horizontally. Plot the frictional force and force applied against each other as shown below. ,
/ f =friction force F= Applied Force f a F
F m
m F Figure: Friction thought Experiment
Apply force of say 0.1N. The body does not move. So, body is in equilibrium ⇒ ∑ ⇒ Same continues till some critical . Note that in this whole region till , the frictional force is same as applied force and hence a line is traced as shown in figure.
Apply force F till the body just start moving. Then in this limiting case = F . Now, in this limiting case, let us denote frictional force f N where N=normal reaction=mg in this case. So, maximum frictional force is in this limiting static condition & is f = N.
Further increasing F, we see that body now moves with some acceleration. So, by Newton's Law, we have F f ma. But, now on, we see that remains constant. Let us denote f N. Note that there is a kink in plot which shows that (f Dynamic Friction Force) (f Static Friction force)
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Hence, we conclude that static friction force is not constant but dynamic friction force is always constant. The study of friction is called tribology.
Rolling Friction Rolling is direct consequence of friction. However, the mechanism differs slightly. As shown below, the microscopic surface irregularities act like microscopic gear tooth which mesh with each other, there by the contact force appear as friction forces as shown below. So, clearly, when we take the free body diagram, we see that ∑ (also in pure rolling, , ) . frictional force induces rolling in Force/Torque free motion. But again, the difference in mechanism can easily also help conclude that magnitude of rolling friction is only 1 or 2 orders lesser than sliding friction & just as in sliding case, we have frictional force due to rolling Coefficient of rolling friction. Example: A truck in neutral rolling down the incline has friction force pointing up the incline/backwards. At point of contact, surface irregularities act like Gear teeth meshing each other
O
C Figure: Rolling Friction Mechanism in Force/Torque Free Case However, in case of forced/torque motion the case changes, as shown below. Here, when the body is generating a Torque for motion, this causes the microscopic tooth to push back at ground just like walking. Consequently, by Newton's third law, ground pushes back with a reactive frictional force as shown above which causes friction.
, T O T , C Figure: Rolling Friction Mechanism in Forced/Non-zero Torque Case Example: A truck moving up an incline in gear has friction force direction forward/up the incline
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Impulse and Momentum From Newton’s second law of motion
Or For a finite period of time, integrating, we get ∫ If
∫ is constant, above equation may be integrated, giving
Where
indicates vector difference of two momentum terms.
If the forces are variable, they must be given as a function of time and should be similarly integrated. Forces that cannot be expressed mathematically as a function of time may be plotted on a force-time curve, in which the area under the curve is equal to the left side of the equation. Linear impulse of a force is defined as Ft and linear momentum is defined as mv. Thus, it may be expressed as follows: Ft = mv – mvo The resultant impulse of the external forces acting upon a body is equal to the change of momentum of the body. Both impulse and momentum are vector quantities. The units of impulse and momentum are Ns. Conservation of Linear Momentum If the sum of the external forces acting on any system of mutually attracting and impinging bodies resolved in any direction is always zero, the total momentum of the system in that direction remains constant during the motion. Suppose a system consists of only two bodies that are in contact for a period of time t. Let there be no external forces acting on this system According to Newton’s third law of motion, if one body exerts a force on a second body, the second body exerts an equal and opposite force on the first since only a pair of equal and opposite forces act on the system, the resultant impulse acting on the system must be zero. It follows, then, that the net change of momentum must also to be zero. Let the two bodies have masses and with velocities and respectively, before coming into contact with each other, and velocities and at the end of the period of contact. Then according to the conservation of linear momentum, we have |
|
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Example 29 A gun weighing 80kN shoots a 250N projectile in a horizontal direction with a muzzle velocity of 500m/s. The bullet weighs 150N and is assumed to move with one-half of the muzzle velocity of the projectile. Find the initial recoil velocity of the gun. If the recoil is resisted by a constant force, find this force if the recoil distance is 80cm. Solution Let v be the recoil velocity in m/s Total momentum = Since all portions of the system are at rest initially, using the principle of conservation of momentum, we have 8
6 /
The negative sign indicates that the gun moves in a direction opposite to that of the projectile. Now using the work-energy relation, ( (
)
∆
, we have
) ]
= 21kN
Example 30 A car of mass 750kg is moving East with a velocity of 30km/h and collides with a second car of mass 1200kg moving NW@ W with a velocity of 45km/h. Find the velocity of the two cars after impact, assuming that they remain constant. Solution Applying the principle of conservation of momentum, we have (
|
)
| |
6
From the vector diagram shown in Fig. we have =(
)
(
6
)
6
cos
= 133.125 + 766.847 - 541.924 = 448.048 21.17 km/h tan
=
/√ /√
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N
27.692
11.538
E
= = 6 66 Or
=
6 66 West of North
Therefore, the direction of the motion of the cars after impact is
N W.
Example 31 A body of mass 6kg moving with a velocity of 3m/s meets a body of mass 4kg moving(a) in the same direction, (b) in opposite direction, with a velocity of 1.5m/s. If they coalesce into one body, find the velocity of the compound body and the loss of kinetic energy. Solution (a) Let
be the velocity after coalescing, then (6
6
)
18 + 6 = 10 ∆
(
)
= (6
)
= (
(6
6)
(b) (6
)(6
( )(
)
,
)
2.7N.m )
12 = = 1.2m/s ∆
(6
) = 24.3N.m
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Example 32 A shot of mass m penetrates a thickness s of a fixed plate of mass M. Prove that if M is free to move, the thickness it penetrated is (
)
Solution Let be the velocity of the shot and F be the average resistance to penetration. When the plate is fixed, using ∆ we get
Let be the thickness penetrated when the plate is free to move, then using the principle of conservation of momentum, we have (
)
and using the principle of work and energy, we have (
)
,
= =
( ( (
) )
)
(
)
Collision of Elastic Bodies If two bodies suddenly collide, an impulsive force, or impact, is set up between them. When the direction of each body is along the common normal at the point where they touch, the impact is said to be direct. When the direction of motion of either or both, is not along the common normal at the point of contact, the impact is said to be oblique. If the pressure exerted on the surfaces of contact coincides with the line joining the mass centers of the bodies, the impact is central. If such is not the case, it is eccentric. For a very short period of time after the two bodies come in contact, the mass centers continue to approach each other. This is known as the period of deformation. During this period the intensity of the force between the surfaces increases. For an instant at the end of the period of deformation, the mass centers are moving with the same velocity. If the bodies are elastic, the impulsive forces causes centers to begin separating and, after a second short interval, the surfaces of the bodies are no longer in contact. This second short period is known as the period THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,
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is known as the period of restitution. Time of impact is the sum of the period of deformation and period of restitution. The time of impact is very small. For this reason, the resultant impulse of the external forces acting on the system during this time must be small and can be neglected. On the bodies of this assumption, the sum of the momentum before impact is equal to the sum of the momentum after impact, i.e, the conservation of momentum holds, thus for direct central impact, we have
Coefficient of Restitution For direct central impact Newton verified experimentally that the relative velocity after impact is in a constant ratio to the relative velocity before impact. If the bodies collide obliquely, the same fact holds for their compound velocities along the common normal at the point of contact. This ratio is known as the coefficient of restitution, and is denoted by e. Thus
In which the proper sign of the four velocities must be included. The value of e lies between zero and one. It is zero for perfectly inelastic bodies and one for perfectly elastic bodies. Example 33 A ball of mass 4kg moving with a velocity of 1.5 m/s is overtaken by a ball of mass 6kg moving with a velocity of 3m/s. (a) in the same direction as the first, (b) in the opposite direction. If e = 1/5, find the velocity of the two balls after impact. Find also the loss of energy in the first case. Solution Here
= 4kg,
= 1.5m/s,
= 6kg,
= 3 m/s, e = 1/5 = 0.2
(a) ∴
6
∴
6 =6
. . . . . . . . (a)
Also 0.2 = = 0.75
. . . . . . . . . .(b)
Solving Eqns. (a) and (b), we get 2.5
= 5.25 = 2.1 m/s = 2.85m/s
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Loss in kinetic energy = (
)
= (
(
)
6
=
)
(
6
6
)
N.m
(b) 6
6
Also, 0.2 = = 0.9 ∴ 2.5 m/s m/s Example 34 A ball weighing 1kg and moving with a velocity of 3m/s impinges on a smooth fixed plane in a direction making 6 with the plane. Find its velocity and direction of motion after impact, the coefficient of restitution being 0.5. Find also the loss in kinetic energy and the impulse on the plane due to the impact. Solution: Here
v = 3m/s, m = 1kg α
, e = 0.5
Now
(sin
cos
= 9(sin
)
cos
= 9(0.25 + 0.25
)
) = 3.9375
1.984m/s Also cot
cot
cot
= 0.866
Loss in kinetic energy ,
(
)
( Impulse =
(
) = 2.583N.m ) cos
= 1(1 + 0.5) cos = 1.5
√
= 3.879kg. m/s
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Chapter 3
Engineering Mechanics
Example 35 A solid cylinder, as shown in Fig. is supported on frictionless bearings. Find the angular velocity of the cylinder and the tension T in the rope, 8s after the system is released from rest. 60cm
2kN
T 600N
Solution The impulse-momentum equation for the 600N load is (
) (
(600 – T) 8 =
)
( )
The angular impulse-momentum equation for the cylinder is: ( (
) ( 6) (
6) 6
) ( )
Now
6
∴ Eq.(a) becomes (6
)
6
6
6
( )
Comparing Eqs. (b) and (c), we get ∴ T = 375N rad/s 6
= 29.43m/s
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Chapter 3
Engineering Mechanics
Conservation of Angular Momentum According to this principle, if a system of two rotating bodies are brought into contact for a short time period, and no external torque is applied to the system during this time, the resultant angular impulse on the system is zero. Suppose the two bodies have moments of inertia and and angular velocities and repectively, before coming, into contact, and angular velocities and at the end of the period of contact. Then the principle of conservation of angular momentum may be stated as
Example 36 The 2.5kN body shown in Fig. rolls freely on the inclined plane and has a centroidal radius of gyration of 0.6r. Determine the velocity of each body and tension T, 4s after the system has been released from rest.
T
r
1.5kN
2.5kN O
F=µR R
Solution
=
( 6 )
66
For the rolling body, =
(
)
∴(
sin 66
)
66 (
) . . . . . . . (a)
For the translating body, ( ( 6
)
) . . . . . . . . (b) THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,
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Chapter 3
Now Or
Engineering Mechanics
= 2r =
∴ Eq. (a) becomes Or
= 86.65
( )
Adding Eqs. (b) and (c), we get 3500 = 239.55 ∴
= 14.61m/s .305m/s 66
6
N Example 37 A pile hammer weighing 12kN drops from a height of 75 cm on a pile weighing 6N. If the pile penetrates a distance of 4cm for every blow, find the resistance of the ground assuming it to the uniform and the impact to be plastic. Solution Velocity of hammer on impact 6m/s
√
√
Let be the velocity of the pile and hammer after impact, then applying the principle of conservation of linear momentum, we have 6
6
(
6
)
= 2.557m/s Let
be the resistance of the ground. Using the principle of work end energy
(
)
6
(
∆
, we get
)
6 N
Example 38 A cricket ball of mass 150g is moving with a velocity of 12m/s and is hit by a bat so that the ball is turned back with a velocity of 20m/s. The force of blow acts for 0.01s on the ball. Find the average force exerted by the bat on the ball. Solution gram,
= 12 m/s
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Chapter 3
Engineering Mechanics
m/s, = 0.01s Change in momentum (
)
(
) = 4.8kg m/s
Force = Rate of change of momentum =
= 480N
Example 39 A body of mass 8kg is moving 2m/s without any external force on it. An internal explosion suddenly breaks the body into two equal parts and 16 joules of translational kinetic energy is imparted to the system by the explosion. Neither part leaves the line of original motion. Determine the speed and direction of each part after the explosion. Solution Let
and
be the velocities of the disintegrated parts (Fig.) 2m/s
8kg
4kg
4kg
Applying the principle of conservation of momentum, ( or
)
=4
. . . . . . (1)
Now applying law of conservation of energy. 6
(
Or
)
= 16
. . . . . . (2)
Substituting the value of (
)
from (1) in (2),
= 16
16+
= 16
2
=0
Or = 4m/s And
=0
Work and Energy Work: If a force acts on a body and causes it to move some distance, work is said to be done by the force. Thus, work is a measure of accomplishment. Therefore, work done by a constant force is equal to the product of the force and the displacement of its point of application in the direction of the force. It is measured in Nm. THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,
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Chapter 3
Engineering Mechanics
Energy: The capacity to do work is called energy. It is measured in N.m. Potential Energy: This is the energy which a body possesses because of its position. Kinetic Energy: This is the energy which a body possesses because of its velocity. Power: The rate of doing work is called power. Work Done by a Force The work done by a force is equal to the product of the force, F, and its displacement, s, provided the force is constant and the displacement, of the body is in the same direction as the force. Denoting work by ,
, we have
= F.s
Relation between work and change of kinetic energy Net work = change in kinetic energy ,
∆
Where
represents kinetic energy. This equation represents the principle of work and energy.
Conservative/Non-Conservative Force Fields and Energy Balance We know work done is the dot product between force F & displacement s ie W=F.s. Further, assume we have a body of mass m being acted upon by a spatially varying force F(s) which is a function of displacement (Ex: Spring force which is ks where s is the displacement from equilibrium). Then, assume that the body moves from position s to a small change in position s + ds, during which time we can assume that F(s) remains constant. Then, the work done to achieve this motion is dW = F(s).ds Now, assume that during this motion, the velocity changes from v to v+dv. By Newton’s law, if a(s) denotes the acceleration of the body during this motion,
F(s) = ma(s)⇒ dW But we know
ma(s) ds (
⇒
)⇒
Then, if we move from position 1 to 2, then the total work to be done for this is ∫
∫
∫
= Change in Kinetic Energy
Further, suppose for a given force function F(s), we can find some V(s) such that F(s) = — ∆V(negative gradient of some function V), then we get as follows.
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Chapter 3
∫
=Change in Function V between points
∫
⇒
Engineering Mechanics
⇒
⇒
The function V is usually called Potential Energy Function. So, if we are able to find a potential function V which satisfies the above condition, clearly, then the quantity (V + KE) ie Sum of Potential and Kinetic Energies is conserved. This brings us to 2 main classifications of forces. •
Conservative Forces: Those force fields F(s) for which a function V can be found out such that F(s) = –∆V In these cases, KE+PE=Constant. Examples: Gravitational force mg whose potential function is mgh(Gravitational Potential Energy), spring force kx whose potential energy function is
•
Non-Conservative Forces: Those force fields F(s) for which there is no potential function which satisfies F(s) = —∆V exists Examples Friction force, Damping force etc. In this case,KE PE Constant
Therefore, in a generic scenario with both conservative and non-conservative forces, one has to apply energy work balance and not energy conservation which in turn comes out of the following form.
Power = (F ) v is the velocity of the point where the force F is acting. is the angle between the directions of the force and the velocity. If both are in the same direction then . One metric horse power = 735.5 watts Work of the Elastic force: If a prismatic bar of area of cross section A, length and elastic constant E is stretched then the work of elastic force can be calculated by treating it as a spring of stiffness k.
Principle of work and Energy Work energy principle: The work done by a force acting on a particle during its displacement is equal to the change in kinetic energy of the particle during that displacement. ) & are Net W.D. by the force for displacing a body from (1) to (2) ( ) W.D. by a force for displacing a body from (1) (2) is positive (+ve) and from (1) ⟵ (2) is negative ( ve). U
(
Principle of conservation of energy: means that the sum of the potential energy and the kinetic energy of a particle (or of a system of particles) remains constant during the motion under the action of conservative forces. KE PE KE PE This principle cannot be applied where frictional force is involved. THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,
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Chapter 3
Engineering Mechanics
Work of the gravity force (
)
(
is positive upwards
is negative upwards.
)
Force exerted by the spring is not a constant force but it varies linearly with the displacement from the undeformed position. U
(
)U
∫ du ∫ F dx sign indicates that Force and displacement are in opposite directions. ∫
If a particle of mass m is moving with velocity , it’s kinetic energy k E is given by
Let v and and .
be the velocities of the particle at points 1 and 2 and the corresponding distance be
U ( (K E )
(P E )
) (K E )
(P E )
W.D. by the springs +ve ⟸ W.D. by the system (springs) W.D. on the system ve. ⟸ W.D. on the system (springs) When spring is stretched W D by force is –ve ie work is done on the spring { When the spring returns towards undeformed position W D is ve(or)work is done by the spring Example 40 Block A in Fig. weights 500 N and block B weights 200 N. The coefficient of friction under B is 0.2. Find the velocity of each block after A has moved 6 m from rest. Also find the tension T in the cord attached to B. Solution By inspection of the weights, we observe that A will move downwards while B will move up the inclined plane. Also B will travel by half the amount of A. C T T T R T B W
T A
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Chapter 3
Now R =W cos F= Also 500
Engineering Mechanics
6
∆ 6
6 (
=
)
(
)
3000 – 300 – 103.92 = 2596.08 =
(250
)
Now ∴
(
6
)
m/s = 9.62 m/s To find tension in the cord, we apply (T –
6 –
)
∆ (
)
to the block ]
(T – 134.64)3 = 235.84 T = 213.25 N Work Done by a Couple or Torque Let a couple Fr act on a body so that the body starts rotating. As it rotates through a small angle d , the work done by the force is Fds = Fr d When the body rotates through the angle , the total work done is, ∫
d
Relation between Work and Kinetic Energy for Rotation Consider a rigid body rotating about an axis 0 with an angular velocity w as shown in a Fig. The particle of mass dm in this body has a velocity v = rw normal to the radial line r. The kinetic energy of the particle is,as
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Chapter 3
Engineering Mechanics
W
O Rigid body Therefore, ( ∴
(
∫
=
) )
∫
rotating about a fixed axis.
Where
is the mass moment of inertia with respect to the axis of rotation.
∫
Example 41 A 1.6 m diameter cylinder as shown in Fig. starts from rest and rolls freely towards the right because of the action of the 400 N weight. Determine the velocity of the cylinder and of the weight after the weight has descended 8 m. Neglect friction and weight of the pulley. 80 cm 40 cm
1.5kN R
400N
Solution (
W=
)
(
)
= THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,
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Chapter 3
(
) = 14.68 kg.
(
400 320 = 20.39 66.26
Engineering Mechanics
)
6 (
)
+45.87
= 320
2.2 m/s ∴
= 5.5rad/s m/s
Example 42 A bullet moving at the rate of 60 m/s is fired into a thick target which penetrates to the extent of 15 cm; if fired into a target 7.5 cm thick with equal velocity, with what velocity would it emerge supposing the resistance to be uniform and same in both cases. Solution Let mass of the bullet be = m kg Force of resistance = F Then, using,
∆
F
(6 )
F=
, we get
12000mN
Let v be the velocity of emergence, then 12000 m
(6 )
12000
]
6 6 m/s
√ Example 43
A machine gun bullet weighing 3 N is fired with a velocity of 500 m/s. What is the kinetic energy of the bullet? If it can pernetrate 30 cm in a block of wood, what is the average resistance of the wood? What will be the exit velocity of the bullet if fired into a similar block of wood 15 cm thick? Solution Kinetic Energy, (
=
)
38226 N.m
Let F be the average resistance of the wood. Applying ∆
, we have
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Chapter 3
(
F
Engineering Mechanics
) (
)
]
= 12742 N 12742 (
(
)
]
)
250000 – v = 353.5 m/s
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