Unit 1.Newton's Laws

UNIT 1 MECHANICS Structure 1.1

Introduction Objectives

1.2

Newton’s Laws of Motion

1.3

Newton’s Law of Gravitation

1.4

Kepler’s Laws

1.5

Motion of Extended Bodies

1.6

Angular Velocity and Angular Momentum

1.7

Moment of Inertia and Radius of Gyration

1.8

Work, Power and Energy

1.9

Gyroscope

1.10 Centroids, Centre of Gravity and Centre of Mass 1.11 Summary 1.12 Key Words 1.13 Answers to SAQs

1.1 INTRODUCTION Mechanics is a physical science dealing with the study of forces and motion of bodies when acted upon by external forces. It essentially deals with the study of bodies at rest or in motion when subjected to external forces. A knowledge of its basic concepts and principles is a must for the design and construction of various types of machines and structures. It provides the opportunity to develop logical thinking, analytical capability, reasoning and judgement which are essential for the solution of great variety of problems. The subject of mechanics of a rigid body is divided into statics and dynamics. Statics is concerned with the study of bodies at rest or in equilibrium under the action of applied external forces, whereas dynamics deals with the study of bodies in motion. Dynamics is further divided into kinematics and kinetics. Kinematics is concerned with the description of motion of objects independent of causes of motion. In kinetics, both the motion and its causes are considered. The physical causes of motion are studied in kinetics, when the relations between the motion of the body and the forces acting upon it are considered. In this unit, you will be introduced to the basic concepts of mechanics.

Objectives After studying this unit, you should be able to •

state and apply the Newton’s Laws of Motion,

•

state and explain Newton’s Law of Gravitation,

•

explain variation of ‘g’ due to rotation of earth, latitude, altitude and depth,

•

state and apply Kepler’s Laws,

•

determine orbital velocity, angular velocity, angular acceleration, etc.,

•

apply theorem of parallel and perpendicular axes to calculate the moment of inertia of a rigid body,

•

explain the concepts of work, power, energy, and

•

determine the centroids centre of gravity and centre of mass.

5

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1.2 NEWTON’S LAWS OF MOTION Newton proposed three basic laws of motion which are commonly known as Newton’s Laws of Motion. These can be enunciated as follows.

1.2.1 Newton’s First Law Every body continues in its state of rest or of uniform motion in a straight line, unless it is compelled to change that state by an external impressed force.

1.2.2 Newton’s Second Law The rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction in which the force acts. Let a mass of body m moving with an initial velocity u (x = 0 in the state of rest) is acted upon by a force f. This force will cause its velocity to change from u to final velocity v in time t. Then, initial momentum of the body is mu and the final momentum of the body after time t is mv. Then the change in momentum in time t would be (mv – mu) = m (v – u) or rate of change of momentum would be m (v − u ) = ma . t

where a = acceleration (rate of change of velocity) =

v−u . t

As per Newton’s second law of motion, rate of change of momentum will be directly proportional to applied force Fαma ⇒

F = K ma

where K is constant of proportionality. If the unit of force is defined such as to produce unit acceleration in a body of unit mass, then 1 = K ×1×1

or

K=1

∴

F=ma Force = Mass × Acceleration

In SI unit, the unit of force is Newton (N). It’s dimension is (MLT– 2].

1.2.3 Newton’s Third Law of Motion To every action there is always an equal and opposite reaction. Let us note that action and reaction which always in pairs act on different bodies. If they acted on the same body, the resultant force would be zero and there could never be accelerated motion. Newton’s third law enters significantly into various situation encountered in our daily life.

1.2.4 The Law of Conservation of Momentum When two or more bodies interact with one another, their total momentum remains constant, provided no external forces are acting. The law of conservation of momentum is made use of extensively in studying processes like collision of particles. Example 1.1

6

A constant retarding force of 60 N is applied to a body of mass 30 kg moving initially with a speed of 18 ms– 1. How long does the body take to stop?

Mechanics

Solution Here m = 20 kg, F = − 60 N (retarding force) Now

F=ma

⇒

a =

Also,

v=u+at

Here

u = 18 ms– 1, v = 0

⇒

0 = 18 + (− 3.0) t

F 60 =− = − 3.0 m/s 2 m 20

⇒

t=6s

Example 1.2 A constant force acting on a body of mass 6 kg changes its speed from 2 ms– 1 to 3.5 ms– 1 in 20 s. The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force? Solution Here u = 2 ms– 1, v = 3.5 ms– 1, and t = 20 s Now

v=u+at 3.5 = 2 + a × 20 a = 0.075 ms– 2

⇒

Mass of the body, m = 6 kg Therefore, force acting on the body, F=ma = 6 × 0.075 = 0.45 N Since the applied force increases the speed of the body, it acts in the direction of motion. Example 1.3 A gun has a mass of 30 tonnes. It fires a bullet whose mass is 450 kg with a velocity of 300 m/s. (a)

Calculate the initial velocity of gun recoil.

(b)

If a resistivity force of 600 kN is applied on gun on an average, calculate the distance travelled by the gun during recoil?

(c)

Also compute the time period of recoil.

Solution Mass of gun (M) = 30,000 kg, Mass of bullet (m) = 450 kg, its Velocity (v) = 300 m/s. Let V is recoil velocity of gun (a)

Then by the principle of conservation of momentum MV + mv = 0 or

V =− = −

mv M 450 × 300 = − 4.5 ms − 1 30000

(i.e. gun velocity is opposite to that of bullet.)

7

(b)

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Resistive force = 600 kN = 600,000 N Retardation, a = R esistive force = 600, 000 = 20 ms − 2 M ass of the gun

We know

v2 = u2 – 2 a s

⇒

0 = 4.5 2 − 2 × 20 × s

⇒

s = 0.506 m

30, 000

Example 1.4 A pile of mass 1000 kg is driven 30 cm into ground by a pile driver of mass 250 kg falling from a height of 2.0 m. Find the average resistance of the ground to penetration of pile, assuming g = 10 ms– 2. Solution Let the velocity of pile driver after falling 2.0 m is v. v=

Then

2 gh =

2 × 10 × 2 = 6.325 m s − 1

. . . (a)

After the pile driver strikes the pile, the common velocity of system is V while its mass is ( M + m ) = 1000 + 250 = 1250 kg . Applying Conservation of Momentum Momentum before impact = Momentum after impact Then

m × v + M × 0 = (M + m) × V

⇒

250 × 6.325 = 1250 V

⇒

V =

250 × 6.325 = 1.265 ms − 1 1250

Since the pile is driven 0.3 m into ground before coming to rest (e.g. final velocity = 0). v 2 = u 2 + 2 as

⇒

0 2 − 1.265 2 = 2 × a × 0.3

or

a = −

(1.265) 2 = − 2.67 ms − 2 (retardation) 2 × 0.3

By Newton’s second law of motion, then F = (M + m) × a = 1250 × 2.67 = 3337.50 N

The resistance of ground R will be the retarding force so created plus the weight of the pile driven system. Hence,

R = 1250 × 10 + 3337.50 = 15837.50 N

SAQ 1

8

(a)

A cannon of mass 20,000 kg fires a shell of mass 100 kg with a muzzle velocity of 800 ms– 1. Calculate the recoil velocity of cannon, average uniform force required to stop in a distance of 400 mm and time period required for recoil.

(b)

Two masses 7 kg and 12 kg are connected at the two ends of a light inextensible string that goes over a frictionless pulley. Find the acceleration of the masses, and the tension in the string when the masses are released.

Mechanics

1.3 NEWTON’S LAW OF GRAVITATION “Every particle of matter in the universe attracts every other particle of matter with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.” M. R. Isaac Newton Two particles of mass m1 and m2 are attracted towards each other with a force whose magnitude ‘F’ is proportional to the product of their masses and inversely proportional to the square of the distance (r) between them. The magnitude of the force of gravitation is given by F =

G m1 m 2

. . . (1.1)

r2

where m1 and m2 are the masses of the two particles, r is distance between the particles and G is the universal constant, and its value is 6.67 × 10– 11 N m2 kg– 2. You must understand that •

The law applies only to the particles or to point masses.

•

Gravitational force between two particles is independent of the presence of other bodies.

•

Gravitational force is an action reaction pair.

The earth exerts the force of gravitation on all bodies on it. The attractive force by which any object tends to be drawn towards the centre of the earth is called the force of gravity. The acceleration of the object is termed the acceleration due to gravity (g), and g = 9.8 m/s2. The force of gravity acting on an object of mass is given by F=mg

. . . (1.2)

Now, you know that F =

GM m

. . . (1.3)

R2

where, M is the mass of the Earth and R, its radius. Equating (1.2) and (1.3), we get the expression for g : g =

GM

. . . (1.4)

R2

Example 1.5 The acceleration due to gravity at the moon’s surface is 1.67 ms– 2. If the radius of the moon is 1.74 × 106 m, calculate the mass of the moon. Use the known value of G. Solution We know that g =

⇒

M =

⇒

M =

GM R2 g R2 G

1.67 × (1.74 × 10 6 ) 2 6.67 × 10 − 11

= 7.58 × 10 22 kg

∴ The mass of the moon is 7.58 × 1022 kg. 9

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Variation of Acceleration due to Gravity The acceleration due to gravity (g) has different values at different places on the surface of the earth. Let us now study its variation (a)

due to altitude,

(b)

due to depth,

(c)

due to latitude, and

(d)

due to rotation of the earth.

Effect of Height (Altitude) on Acceleration due to Gravity Let us consider the earth is a sphere of radius R and mass M. Let g be the value of acceleration due to the gravity at the point A on the surface of earth as shown in Figure 1.1. g′

B

h A

R

o

Figure 1.1

We can write g =

GM

. . . (1.5)

R2

If g′ be the acceleration due to gravity at the point B, which is at a height h above the surface of the earth, then g′ =

GM

. . . (1.6)

(R + h)2

From Eqs. (1.5) and (1.6), we may write g′ R2 = g ( R + h)2

g′ 1 = 2 g h⎞ ⎛ + 1 ⎜ ⎟ R⎠ ⎝

or

g′ ⎛ h⎞ = ⎜1 + ⎟ g R⎠ ⎝

or

[Q Expanding the term ⎛⎜ 1 + h ⎞⎟ R ⎝

⎠

−2

−2

by using Binomial theorem, we have

h g′ 2h . Usually h is quite small as =1− + term containing higher power of R g R compared to R, and hence the term containing higher powers of or 10

g′ ⎛ 2h ⎞ = ⎜1 − ⎟ g R ⎠ ⎝

h are neglected.] R

2h ⎞ ⎛ g ′ = g ⎜1 − ⎟ R ⎠ ⎝

or

. . . (1.7)

Mechanics

[Note : Eq. (1.7) is used to find value of g at a height, when h is very small as compared to the radius R of the earth.] Example 1.6 A body weighs 80 N on the surface of the earth. What is the gravitational force on it due to the earth at a height equal to one-third of radius of the earth? Solution Here,

mg = 80 N, h =

R 3

We know, g′ = g

or

R2 ( R + h)2

mg ′ = mg

R2 (R + h)2

= 80 ×

R2 R⎞ ⎛ ⎜R + ⎟ 3⎠ ⎝

2

= 80 ×

9 = 45 N 16

[Note : We have used Eq. (1.6) because h is not negligible as compared to the radius of the earth.] Effect of Depth on Acceleration Due to Gravity Let us consider that the earth is a sphere of radius R and mass M as shown in Figure 1.2. A y B

R

R−y

O M′

M

Figure 1.2

We already know that the value of acceleration due to gravity at point A, on the surface of the earth is given by g =

GM

. . . (1.8)

R2

If ρ is the density of the material of earth, then M =

g =

4 π R3 . ρ 3 G.

4 π R3 ρ 3 R2 11

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g =

4 πG Rρ 3

. . . (1.9)

Let us now consider that g′ be acceleration due to gravity at point B at a depth of y below the surface of the earth. A body at the point B will experience force only due to the portion of the earth whose radius is OB = R – y. There will be no effect on acceleration due to gravity due to the outer shell whose thickness is y. If M′ is the mass of the portion, whose radius is R – y, then we can write g′ = M′=

But,

GM′ (R − y)2 4 π ( R − y )3 . ρ 3

. . . (1.9a)

Substituting M′ in the Eq. (10.9a), g′ =

4 π G (R − y) ρ 3

. . . (1.10)

From Eq. (1.9) and (1.10), we get g′ R − y = g R y⎞ ⎛ g ′ = g ⎜1 − ⎟ R⎠ ⎝

or

We may calculate that the acceleration due to gravity decreases with depth. At the centre of earth, y = R. So R⎞ ⎛ g ′ = g ⎜1 − ⎟ = 0 R⎠ ⎝

Therefore, at the centre of earth, the acceleration due to gravity becomes zero. Example 1.7 Find the percentage decrease in weight of a body, when taken 32 km below the surface of the earth. Take radius of earth = 6400 km. Solution We know that y⎞ ⎛ g ′ = g ⎜1 − ⎟ R ⎝ ⎠

⇒

32 ⎞ ⎛ g ′ = g ⎜1 − ⎟ 6400 ⎠ ⎝

= g −

1 g 200 g 200

⇒

g − g′ =

⇒

g − g′ 1 = g 200

If m is the mass of the body, then mg and mg′ will be the weight of the body on the surface of the earth, and at a depth of 32 km below the surface of the earth, then ′ % decrease in weight = m g − m g × 100 mg

=

12

g − g′ 1 × 100 = × 100 = 0.5% g 200

Effect of Latitude on Acceleration due to Gravity

From Eq. (1.4) it is clear that the value of g varies inversely as the square of the earth’s radius. •

Do you know that the Earth is not a perfect sphere? It is bulged out at the equator and flattened at the poles.

•

Therefore, the radius of the Earth is not constant. It is greatest at the equator and the least at poles.

•

Since g varies on the radius, and the radius varies from the pole to the equator, its value changes along the surface of the earth. g is least at the equator, since R is greatest. g is greatest at the poles, since R is least. It has intermediate values at a point between the equator and poles.

Mechanics

Re

Rp

Figure 1.3

Effect of Rotational Motion of the Earth on Acceleration due to Gravity We know that the earth has rotational motion of period of 24 hours about its own axis. Let us assume that the earth to be a sphere of radius R and mass M. Let the axis of rotation be ZOZ′ as shown in Figure 1.4, and its angular velocity be ω. Let us consider the case of a particle of mass m at P so that OP makes an angle λ with OE. Here λ is the latitude of the particle. w

Z N C

Fc

P R FG λ

W

O

F A

E

S Z’

Figure 1.4

It is obvious from the above figure that the particle is moving in a circle of radius CP = R cos λ with angular velocity ω. The force of the attraction on the particle towards the centre is given by FG =

where

GM m R2

M = Mass of the earth, and R = Radius of the earth.

The centrifugal force on the body due to rotational motion of the earth acts along radius of the circular path and in outward direction. The magnitude of the centrifugal force on the body is given by FC = Mass of the body × Radius of circular part × ω2 13

= m × R cos λ × ω 2

Physics

= m R ω 2 cos λ

The resultant of FC and FG has a magnitude and is given by FG2 + FC2 + 2 FG FC cos (180o − λ )

F=

⎧⎪⎛ G M m ⎞ 2 ⎫⎪ GM m 2 2 2 o = ⎨⎜ + ω λ + ω λ ( m R cos ) 2 . . m R cos cos (180 − λ ) ⎬ 2 ⎟ R2 ⎩⎪⎝ R ⎠ ⎭⎪ ⎧⎪⎛ G M m ⎞ 2 ⎫ 2G M m 2 2 4 2 2 2 ⎪ = ⎨⎜ + ω λ − ω m R cos . m R cos λ ⎬ ⎟ 2 R2 ⎩⎪⎝ R ⎠ ⎭⎪

=

GM m R2

1+

R 4 ω4 cos 2 λ G2 M 2

−

2 R 3 ω2 cos 2 λ GM

∴ The second term under square root, i.e.

R3 ω4 cos 2 λ G2 M 2

is negligible and hence

can be neglected. Therefore, 1

GM m ⎛ 2 R 3 ω2 cos 2 λ ⎞ 2 F= 1 − ⎜ ⎟⎟ GM R 2 ⎜⎝ ⎠ =

GM m R

2

− R ω2 m cos 2 λ

[Apply Binomial Expansion)

The resultant force F is not directed towards the centre. If the body is released from rest, it moves in the direction of F. If acceleration in this direction is g′ then mg ′ =

GM m R

2

− R ω2 m cos 2 λ

mg ′ = mg − R ω2 m cos 2 λ

or

g ′ = g − R ω2 cos 2 λ

or

. . . (1.11)

Thus, acceleration due to gravity decreases due to rotational motion of the earth. Let us consider the following two cases : Case 1

The value of ‘g’ at pole; here λ = 90o g pole = g − R ω2 cos 2 90o = g Hence, there is no effect of rotational motion of earth on the value of g at poles. Case 2

The value of g at equator : Here, 14

λ = 0o , gequator = g − R ω2 cos 2 λ

= g − R ω2 cos 2 0o

Mechanics

= g − R ω2

Hence, the maximum effect of rotational motion of earth on the value of g is at equator. From the values of g at poles and equator, we may conclude that the value of g at a place increases with the increase in the latitude of place. Example 1.8

If the earth were a perfect sphere of radius 6.37 × 106 m, rotating about its axis with a period of 1 day, how much would the acceleration due to gravity (g) differ from the poles to the equator? Solution

Acceleration due to gravity at the latitude is given by g ′ = g − R ω2 cos 2 λ

where g is the value of acceleration due to gravity in the absence of rotational motion of the earth. We know that, at the pole λ = 90o and at the equator, λ = 0o. g pole = g − R ω2 cos 2 90o =g and

gequator = g − R ω2 cos 2 0o = g pole − R ω2

So,

g pole − g equator = R ω2

Here

R = 6.37 × 106 m

and

ω=

2π 8.64 × 104

rad s −1 2

∴

g pole − g equator

⎛ ⎞ 2π = 6.37 × 10 × ⎜⎜ = 3.37 × 10− 2 ms − 2 4 ⎟ ⎟ ⎝ 8.64 × 10 ⎠ 6

SAQ 2 (a)

Calculate the mass and mean density of earth from the following data : Gravitational constant (G) = 6.67 ×10– 11 N m2 kg– 2 Radius of earth (R) = 6.67 ×106 m Acceleration due to gravity = 9.8 ms– 2.

(b)

If the earth be one-half its present distance from the sun, how many days will be in one year?

(c)

At what height above the earth will the acceleration due to gravity be one-fourth of that on the surface of the earth?

(d)

If the diameter of the earth becomes two times its present value and its mass remains unchanged; then how would the weight of an object on the surface of the earth be affected?

So far, you have learnt to apply the universal law of gravitational to objects on the Earth. It also applies to heavenly objects. Let us consider its application to the motion of 15

Physics

planets. You may know about Kepler’s laws of planetary motion. These laws can be devised from Newton’s law of universal gravitation, we now discuss Kepler’s Laws.

1.4 KEPLER’S LAWS In the early 1600s, Johannes Kepler proposed three laws of planetary motion. Kepler was able to summarize the carefully collected data of his mentor – Tycho Brahe – with three statements, which described the motion of planets in a sun-centered solar system. Kepler’s efforts to explain the underlying reasons for such motions are no longer accepted; nonetheless, the actual laws themselves are still considered an accurate description of the motion of any planet and any satellite. Kepler’s three laws of planetary motion can be described as follows : •

The paths of the planets about the sun are elliptical in shape, with the center of the sun being located at one focus (The Law of Ellipses).

•

An imaginary line drawn from the center of the sun to the center of the planet will sweep out equal areas in equal intervals of time (The Law of Equal Areas).

•

The ratio of the squares of the periods of any two planets is equal to the ratio of the cubes of their average distances from the sun (The Law of Harmonics).

Let us explain these laws. Kepler’s First Law

•

The planets orbit the sun in a path described as an ellipse mean (Figure 1.5).

•

An ellipse is a special curve in which the sum of the distances from every point on the curve to two other points is a constant. The two other points (represented here by the tack locations) are known as the foci of the ellipse.

•

The sun is located at one of the foci of the ellipse.

•

The closer these points are, the more closely that the ellipse resembles the shape of a circle. In fact, a circle is the special case of an ellipse in which the two foci are at the same location. The Law of Equal Areas

An imaginary line drawn from the sun to any planet sweeps out equal areas in equal interval of time.

Figure 1.5

Kepler’s Second Law

16

•

The speed at which planet orbits the sun is constantly changing.

•

A line drawn from the centre of the sun to the centre of the planet would sweep out the same area in equal interval of time. For instance, if that line were drawn from the earth to the sun, then the area swept out by the line every 30 days would be the same. This also means that a planet moves

fastest when it is closet to the sun and slowest when it is farthest from the sun.

Mechanics

Kepler’s Third Law

Sometimes referred to as the law of harmonics : Period(s)

Average Dist. (m)

T2/R3 (s2/m3)

Earth

3.156 × 107 s

1.4957 × 1011

2.977 × 10− 19

Mars

5.93 × 107 s

2.278 × 1011

2.975 × 10− 19

Planet

You may like to solve a few questions based on Kepler’s laws. The law of gravitation has many useful applications in space. The satellites orbiting the earth are an example, you may like to learn about them. Orbital Velocity

Consider a satellite of mass m orbiting the Earth (mass M). If the satellite moves in a circular orbit, then the net centripetal force acting upon the satellite is Fnet =

mv2 r

This net centripetal force is provided by gravitational force Fnet = Fgrav = GM m r2

=

or

v2 =

GM r

or

v =

GM r

GM m r2

mv2 r

where G = 6.67 × 10 − 11 N m 2 /kg 2 , M = the mass of the central body about which the satellite orbits, and r = the radius of orbit for the satellite. At a height h from the surface of the Earth r = R + h where R is the radius of the Earth. Hence,

v =

Now, we also have

g =

GM R + h

. . . (1.12)

GM R2

Therefore, the orbital velocity of a satellite in terms of g is : v =

g R2 R + h

. . . (1.13)

Eqs. (1.12) and (1.13) are used to find the orbital velocity for a satellite orbiting at a height h above the surface of earth, when the satellite is orbiting very close to the surface of the earth. In such case h ≈ 0, then from Eq. (1.12), we get v =

GM R

. . . (1.14)

Also from Eq. (1.13), we get 17

v =

Physics

gR

. . . (1.15)

In substituting g = 9.8 ms– 2, and R = 6.4 × 106 ms– 1 in Eq. (1.15), we get v = 7.92 km/s to revolve in a orbit just near the surface of earth. Time Period of Satellite

Suppose the satellite takes time T to complete one orbit of radius (R + h) around the Earth, i.e. T is its time period. Then if (R + h) is the radius of the orbit in which the satellite revolves with an orbital velocity v, then

v=

2π ( R + h) T

where 2π ( R + h) being distance travelled in unit time. This gives

T =

2 π ( R + h) v

Substituting the value of v in the above equation, we get T =

2 π ( R + h) GM ( R + h)

4 π 2 ( R + h ) 2 4π 2 ( R + h) 4 π 2 ( R + h )3 = ( R + h) 2 = GM GM ⎛ GM ⎞ ⎜ ⎟ ⎝ R + h⎠

or,

T2 =

or

T = 2π

or

T 2 ∝ ( R + h )3

( R + h )3 GM

If r = R + h, where r is the orbital radius, then T 2 ∝ r3

Therefore the square of the time period is directly proportional to the cube of the orbital radius.

SAQ 3 (a)

A satellite is moving in a circular orbit round the earth with a speed of Re g . Find the altitude of the satellite above the earth. 2 Re = Radius of the earth.

(b)

An earth satellite revolves in a circular orbit at a height of 300 km above the earth’s surface.

(c)

18

(i)

What is the speed of the satellite, assuming the earth’s radius to be 6380 km and g to be 9.80 ms– 2?

(ii)

What is the period τ?

(iii)

What is the radial acceleration of the satellite?

Consider an earth satellite so positioned that it appears stationary to an observer on the earth and serves the purpose of a fixed relay station for inter-continental transmission of television and other communication. What would be the height at which the satellite should be positioned and what would be the direction of its motion? Acceleration due to gravity = 9.8 ms– 2, radius of the earth = 6.38 × 106 m.

So far you have learnt about the universal law of gravitational and its application to planetary motion and orbiting satellites. Let us now discuss the motion of extended bodies which cannot be regarded as point masses so far we have assumed them to be point masses.

Mechanics

1.5 MOTION OF EXTENDED BODIES Roll a ball on the floor. What kind of motion does it execute? It moves from one point to another – it translates. It also rotates on its axis while moving. Thus, a body under the action of forces may have two types of motion : •

Translational motion

•

Rotational motion

Translational Motion

•

It is caused by the application of single or a number of forces equivalent to a single force.

•

To produce a given acceleration the greater is mass of the body the greater is the force required.

•

The property of a body to remain in the state of rest until it is acted by unbalanced force is called inertia.

•

The mass of a body is a measure of its inertia (state of rest).

Rotational Motion

•

Rotational motion is caused and altered not by a single force but by a couple or a number of forces reducible to a couple.

•

A couple produces angular acceleration just as a force produces linear acceleration.

•

An analogous quantity in rotational motion, which is a measure of the inertia of a rigid, rotating body, is called its moment of inertia or simply rotational inertia about a given axis.

•

The inertia in rotational motion not only depends on the mass of the rigid body but also on the distribution of the mass in the body about the axis of rotation.

This brings us to concept of moment of inertia. Moment of inertia of a body which is rotating about an axis passing through a point O and perpendicular to the plane of the page.

Let m1 be the mass of a particle in the body at a distance r1 from the axis of rotation; Let m 2 be the mass of another particle at a distance r2 from the axis of rotation; Let mn be the mass of the nth particle at a distance rn from the axis of rotation. ω

Figure 1.6

19

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We now take the product of the mass of each group with the square of its distance from the axis. Thus, we get the quantities, m1 r12 , m2 r22 , . . . , mn rn2 . Then the moment of inertia of the body, about the given axis is given by n

I = m1 r12 + m2 r22 + . . . + mn rn2 = ∑ mi ri2 i =1

The moment of inertia of a rigid body about a given axis is, therefore, defined as the sum of the products of the mass of each and every particle of the body and the square of its distance from the given axis.

Considering the body as composed of discrete point masses, if we regard it as made up of matter which is continuously distributed, then the process of summation should be replaced by the process of integration. We imagine the whole body to be divided into infinitesimal elements, each of mass dm. Let r be the distance of one such element from the axis of rotation. Then, its M.I. about the given axis will be dI = r 2 dm and the M.I. of the whole body about the given axis will be

I=

∫ dI = ∫ r

2

dm

The limits of integration depend upon the shape and size of the body and the position of the axis.

SAQ 4 Calculate the moment of inertia of a cylinder of length 150 cm, radius 5 cm and density 8 g/cm3 about the axis of the cylinder.

1.6 ANGULAR VELOCITY AND ANGULAR MOMENTUM The angle described by a rotating body per unit time is called its angular velocity (ω) or it is defined as the time rate of change of angular displacement at any instant.

ω=

θ t

Angular velocity ω has same direction as the direction of θ. The direction of θ is determined by the axis of rotation. By right hand rule, if the fingers are curled around the axis in the direction of rotation, then the outstretched thumbs gives the direction of θ. [Note : 1 revolution = 2π radians; 1 rev/min = 2π/60 rad/sec.] If θ is in radians and t in seconds, the unit of angular velocity is the rad/sec. Let us consider a particle moving with the uniform speed v in a circle of radius r, as shown in Figure 1.7 below. This particle travels the distance s = vt in time t. The angle through which it moves in that time t is

θ= 20

s vt = , r r

So that its angular velocity is

Mechanics

v θ = =ω r t

r V

θ

V

Figure 1.7

Relationship between v and ω

The linear speed of a particle performing circular motion is the product of its angular velocity and the radius of its path. The expression relating to v, r and ω is as follows. v=r×ω

Here, ω is measured in radians. •

The axis of rotation of a rigid body turning in place is that line of the particles which does not move.

•

All other particles of the body move in circles about the axis.

•

Since v = r ω, the farther particle is from the axis, the greater the linear speed, although all the particles of the body have the same angular velocity.

Angular Acceleration

If magnitude v of the instantaneous linear velocity of a particle moving in a circle is changing continuously, i.e. the motion of the particle is non-uniform circular motion, then, its angular velocity ω will also change continuously. If the angular velocity of a body changes from the original value ω0 to the new angular velocity ω in a time interval t, the angular acceleration α of the body is

α=

ω − ω0 t

•

The unit of angular acceleration is rad/sec2.

•

The direction of α is same as the direction of ω.

α

v

ω

0

r

Figure 1.8

SAQ 5

21

Physics

(a)

A particle of mass 1 kg starts from rest and moves in a horizontal circle of radius 0.5 cm with a uniform angular acceleration of 2 rad/s2. Find its angular speed, linear speed, and the magnitude of the centripetal force acting on it, at the end of one minute.

(b)

Determine the angular speeds of the second hand, the minute hand, and the hour hand of a clock.

Angular Momentum

The final concept we develop for rotational motion is that of angular momentum. First we develop the concept for a single particle, and then generalize for a system of particles. Angular Momentum for a Single Particle Consider a single particle of mass m travelling with a velocity v a radius r from an axis, as shown in Figure 1.9. O

v

r

θ P

Figure 1.9 : A Single Particle Moving with Respect to an Axis, O

The angular momentum of the single particle, then, is defined as : L = r . m . v . sin θ Notice that this equation is equivalent to L = r . p . sin θ where p is the linear momentum of the particle. A particle does not need to move in a circular path to possess angular momentum. However, when calculating angular momentum, only the component of the velocity moving tangentially to the axis of rotation is considered (explaining the presence of sin θ in the equation). Another important aspect of this equation is that the angular momentum is measured relative to the origin chosen. This choice is arbitrary, and our origin can be chosen to correspond to the most convenient calculation. Angular momentum about a point is defined as the moment of its linear momentum about that point.

Because angular momentum is the cross product of position and linear momentum, the angular momentum formula is expressed in vector notation r r r as : L = r × p . This equation provides the direction of the angular momentum vector: it always points perpendicular to the plane of motion of the particle and can be determined by the right handed screw rule. Angular Momentum and Net Torque

22

It is possible to derive a statement relating angular momentum and net torque. Unfortunately, the derivation requires quite a bit of calculus, so we will simply ⎡ dp ⎤ revert to the linear analogue. Recall that ∑ F = ⎢ ⎥ . In a similar way, ⎣ dt ⎦

∑τ =

Mechanics

dL dt

A net torque changes a particle’s angular momentum in the same way that a net force changes a particle’s linear momentum. In circumstances of rotational motion, however, we usually deal with rigid bodies. In such cases the definition of the angular momentum of a single particle is of little use. Thus we extend our definitions to systems of particles. Angular Momentum of Systems of Particles

Consider a rigid body rotating about an axis. Each particle in the body moves in a circular path, implying that the angle between the velocity of the particle and the radius of the particle is 90°. If there are n particles, we find the total angular momentum of the body by summing the individual angular momentum : L = 11 + 12 + 13 + . . . + 1n

Now we express each l (angular momentum) in terms of the particle’s mass, radius and velocity : L = r1 m1 v1 + r2 m2 v2 + . . . + rn mn vn

We now substitute ω for v using the equation v = ω r : L = r1 m1 r1 ω1 + r2 m2 r2 ω2 + . . . + rn mn rn ωn

However, in a rigid body, each particle moves with the same angular velocity. Thus Since

n

L = ω ∑ mi ri2 i =1

n

I = ∑ mi ri2 , L = I ω i =1

where I is known as moment of inertia about the axis of rotation. Here we have a concise equation for the angular momentum of a rigid body. From this equation for a rigid body we can also generate a statement relating external torque and total angular momentum :

τext = ∑

dL dt

Just as an external force changes the total linear momentum of a system of particles, an external torque changes the angular momentum of a rigid body. To illustrate this very simple concept, we examine a very simple situation. Consider a bicycle wheel. By pedaling the bike we exert a net external torque on the wheel, causing its angular velocity to increase, and thus its angular momentum to follow suit. Conservation of Angular Momentum

The total vector angular momentum of a body or a system remains constant, if the external torque acting on the body or the system is zero. This is known as the principle of conservation of angular momentum. It can be proved as follows. We have earlier proved that, for a rigid body, rotating about a given axis, 23

Physics

τ = Iα = I

dω dt

Now, for a rigid body, rotating about a given axis, I remains constant.

τ=I

d ω d ( I ω) dL = = dt dt dt

If the external torque acting on the body, τ = 0

τ=

dL =0 dt

Hence L is constant. Now, L is constant implies that I ω = constant. ω increases when I decreases and vice versa. If I changes to I1 then, in the absence of an external torque, ω will changes to ω1 such that ω > ω1 and I < I1. This principle is made use of by ballet dancers, on the distribution of mass about the axis of rotation; a large variation in its value can be brought about by extending or pulling in the limbs. Thus, ballet dancers and ice skaters can decrease or increase the angular speed of a spin by stretching out or pulling in their hands. Axis of Rotation

ω

Figure 1.10

1.7 MOMENT OF INERTIA AND RADIUS OF GYRATION In Newtonian mechanics the mass of a body is taken to be a constant quantity. The moment of inertia of a rigid body is, however, not a unique property of the body, but is found to depend on the position of the axis. •

This is, because, when the position of the axis changes, the distribution of mass about the axis also changes.

•

Moment of inertia of any rigid body about a given axis can be expressed as the products of the mass of the body and the square of a certain distance from the axis, i.e. we can write I = mk 2 . The quantity k is called the radius of gyration of the body about the given axis.

•

For a solid, homogeneous, sphere rotating about a diameter

I=

24

2 M r2 5

k2 =

i.e.

2r 2 5

•

If we imagine that the whole mass of the body is concentrated at a distance k from the axis, then the M.I. of this imaginary point mass about the given axis will be the same as that of the actual body about the same axis.

•

Thus, if we imagine that the whole mass of solid, homogeneous, sphere is 2 r from a diameter then the moment of inertia concentrated at a distance 5 of this imaginary point mass about the diameter will be 2 I = mk 2 = m r 2 which is the same as the moment of inertia of the actual 5 sphere about the same diameter. We, therefore, define the radius of gyration as follows .

Mechanics

Radius of gyration of a rigid body about a given axis as the distance from the axis at which the whole mass of the body must be supposed to be concentrated so that this imaginary point mass has the same moment of inertia as the actual body, about the given axis. Its SI unit is meter.

•

The value of k gives us an idea about how the mass of the body is distributed about the axis of rotation.

•

If the value of k is small the mass must be concentrated close to the axis and if it is large the mass must be spread out over a large region around the axis.

SAQ 6 (a)

A homogenous solid sphere has a mass of 20 kg. Its radius of gyration about a diameter is 10 cm. Find its (i) M.I, (ii) angular momentum and (iii) kinetic energy, when it is rotating about a diameter at 120 r.p.m.

(b)

A constant torque of magnitude 5000 Nm acting on a body increases its angular velocity from 2 rad/s to 20 rad/s in 9 seconds. Calculate the M.I. about the axis of rotation.

Theorem of Parallel Axis

Principle of Parallel Axes According to this principle, the M .I. of a rigid body about any axis is equal to the sum of its M.I. about a parallel axis through its centre of mass and the product of the mass of the body and the square of the distance between the two axes.

•

Consider a rigid body of mass m rotating about an axis passing through a point O and perpendicular to the plane of the page. Let I0 be the M.I. of the body about this axis.

•

Take a parallel axis passing through the point C the centre of mass of the body. Let Ic be the M.I. of the body about this axis, then if OC = h by the principle of parallel axes I = I c + m h 2 .

25

•

Physics

Take an infinitesimal element of the body of mass dm situated at a point B, in the plane of the page. From point B drop perpendiculars BA on to OC produced and join BC.

ω B

O

A

C

Figure 1.11

•

As the triangle → OBA is a right angled triangle, we have OB 2 = OA2 + BA2 = (OC + CA) 2 + BA2

= OC 2 + 2OC . CA + CA2 + BA2

Also, from triangle CBA, CB 2 = BA2 + CA2 OB 2 = OC 2 + 2OC . CA + CB 2

or,

OB 2 dm = OC 2 dm + 2OC . CA dm + CB 2 dm

or,

∫ OB

or,

I 0 = mh2 + 2OC

or,

I 0 = I c + mh2 + 2h

2

∫

dm = OC 2 dm +

∫ 2OC . CA dm + ∫ CB

2

dm

∫ CA . dm + Ic ∫ CA . dm

As the sum of turning moments of the weight of the particles about the centre of the mass is zero, i.e. ∫ CA . dm = 0 . Therefore,

I0 = Ic + m h2

From above relation it is clear that the moment of inertia of a body about an axis through the centre of mass is least and less than about any other parallel axis. Theorem of Perpendicular Axis According to this principle, the M.I. of a rigid plane lamina about an axis perpendicular to its plane is equal to the sum of its moments of inertia about any two mutually perpendicular axis in its plane and meeting in the point where the perpendicular axis cuts the lamina.

26

•

A lamina is a thin plate (a body whose thickness is negligible compared to its surface area).

•

Let OX and OY be two mutually perpendicular axis in the plan of a rigid lamina and let OZ be an axis passing through the point O and perpendicular to the plane of the lamina.

•

By principle of perpendicular axis, Iz = Ix + Iy.

•

Take an infinitesimal element of the lamina of mass dm situated at a point C whose coordinates are x and y respectively from the X and Y-axis. If the distance of C from O is r, we have OA = x and OB = y .

Mechanics

IZ D

A

x

B

y

O r

Iy

C dm

Ix

Figure 1.12

•

Now moment of inertia of the lamina about Z-axis,

and

Iz =

∫ OC

Ix =

∫ CA

Iy =

∫ CB

∫r

2

. dm =

2

dm =

∫y

2

dm =

∫x

2

2

2

. dm

dm dm

In triangle OCB, OC 2 = CB 2 + OB 2 = x 2 + y 2 = r 2

r 2 dm = x 2 dm + y 2 dm

∫r

2

dm = ∫ x 2 dm + ∫ y 2 dm

Iz = Therefore,

∫r

2

dm =

∫x

2

dm +

∫y

2

dm = I y + I x

Iz = I y + Ix .

1.8 WORK, POWER AND ENERGY In this section, you will be introduced to the work, power and energy. The terms ‘work’ and ‘energy’ are quite familiar to us and we use them in our day-to-day life situation. However, our usage of these terms may be often vague and may not correspond to the precise scientific definitions. For instance, the word ‘work’ may refer to any kind of activity both mental and physical. You are supposed to be ‘working’ hard before your term end examination. But, according to the scientific definition, work is done by a force acting on a object when the point of application of the force moves through some distance in which force has a component along the line of motion. Then the concept of potential energy (associated with the position) and kinetic energy (associated with the motion) of an object are illustrated.

1.8.1 Work In this Sub-section, the motions of objects will be analyzed in perspective of work and energy. The effect that work has upon the energy of an object (or system of objects) will be investigated; the resulting velocity and/or height of the object can then be predicted from energy information. It is important to first have a clear understanding of a few basic terms such as •

Work

•

Mechanical energy

•

Potential energy

•

Kinetic energy

•

Power

Work is defined as a force acting upon an object to cause a displacement. 27

Physics

In order for a force to qualify as having done work on an object, there must be a displacement and the force must cause the displacement. There are several good examples of work which can be observed in everyday life – a weightlifter lifting a barbell above her head and a shot-putter launching the shot, etc. In each case described here there is a force exerted upon an object to cause that object to be displaced. Mathematically, work can be expressed by the following equation W = F ⋅ d ⋅ cos θ where F = force, d = displacement, and the angle (theta) is defined as the angle between the force and the displacement vector. The angle is not just any ‘ole angle’, but rather a very specific angle. The angle measure is defined as the angle between the force and the displacement. To gather an idea of its meaning, consider the following three scenarios. Scenario A

A force acts rightward upon an object as it is displaced rightward as shown in Figure 1.13. In such an instance, the force vector and the displacement vector are in the same direction. Thus, the angle between F and d is 0 degrees. Scenario

Direction of Force and Displacement

θ Degrees

A

d

0

F B

d

180

F C

d

90 F

Figure 1.13

Scenario B

A force acts leftward upon an object which is displaced rightward as shown in Figure 1.13. In such an instance, the force vector and the displacement vector are in the opposite direction. Thus, the angle between F and d is 180 degrees. Scenario C

A force acts upward upon an object as it is displaced rightward as shown in Figure 1.13. In such an instance, the force vector and the displacement vector are at right angles to each other. Thus, the angle between F and d is 90o. A vertical force can never cause a horizontal displacement; thus, a vertical force does not do work on a horizontally displaced object. When a force is exerted on an object at an angle to the horizontal, only a part of the force contributes to (or causes) a horizontal displacement. In the case of work (and also energy), the standard metric unit is the Joule (abbreviated “J”). One Joule is equivalent to one Newton of force causing a displacement of one metre. In other words, The Joule is the unit of work. 1 Joule = 1 Newton × 1 metre 1 J = 1 Nm Calculating the amount of work done by forces : 28

Mechanics

F = 100 N F = 150 N 30o

F = 100 N

15 kg 15 kg 15 kg A 100 N force is applied to move a 15 kg object a horizontal distance of 5 m at constant speed A

An upward force 150 N is applied to lift a 15 kg object to a height of 5 m at constant speed

A 100 N force is applied at o angle of 30 to the horizontal to move a 15 kg object at a constant speed for a horizontal distance of 5 m B

C

Figure 1.14

Case A W = F ⋅ d ⋅ cos θ = 100 ⋅ 5 ⋅ cos 0 = 500 J In this case force and displacement are in the same direction, the angle is 0 degree. Case B W = F ⋅ d ⋅ cos θ = 100 ⋅ 5 ⋅ cos 30o = 433 J

In this case force is 30o with the horizontal. Thus the angle between F and d is 30 degree. Case C W = F ⋅ d ⋅ cos θ = 150 ⋅ 5 ⋅ cos 0 = 750 J In this case the applied force is 150 N since 15 kg weight is lifted up at constant speed. Since F and d points to the same direction, the angle is 0o. Work Done by Variable Force

There are many physical situations in which force acting on a body is not constant, e.g. •

Gravitational force

•

Electrostatic force

•

Force exerted on a body by a spring.

The force may change in (i) magnitude (ii) direction and (iii) magnitude as well as direction. Let us now consider the case in which magnitude of the forces changes. Suppose the body moves along the x-axis. Force acting on the body is also along the x-axis. The body moves from x = a to x = b. Wab = ∑ Δ W = ∑ F ⋅ Δ x y

F (x)

x a

Δx

Figure 1.15

b

29

Physics

The exact value of Δx can be calculated by taking limit as Δx tends to zero. b

b

Wab = lim ∑ F Δx = Δx → 0 a

∫

F dx

a

Total work done is given by above expression.

SAQ 7 (a)

Jai carries a 200 N suitcase up three flights of stairs (a height of 10.0 m) and then pushes it with a horizontal force of 50 N at a constant speed of 0.5 m/s for a horizontal distance of 35 meters. How much work is done on his suitcase during this entire motion?

(b)

Jai with a mass of 80 kg runs up three flights of stairs in 12.0 sec. Jai has gone a vertical distance of 8 m. Determine the amount of work done by Jai to elevate his body to this height. Assume that her speed is constant.

1.8.2 Power The quantity work has to do with a force causing a displacement. Work has nothing to do with the amount of time that this force acts to cause the displacement. Sometimes, the work is done very quickly and other times the work is done rather slowly. For example, a rock climber takes an abnormally long time to elevate her body up a few meters along the side of a cliff. On the other hand, a trail hiker might elevate her body a few meters in a short amount of time. The two people might do the same amount of work, yet the hiker does the work in considerably less time than the rock climber. The quantity which has to do with the rate at which a certain amount of work is done is known as the power. The hiker has a greater power rating than the rock climber. Power is the rate at which work is done. It is the work/time ratio. Mathematically, it is computed using the following equation.

Power =

Work done Time

In an amount of work ΔW is done in a very short interval of time Δt, then the instantaneous power delivered is

P=

ΔW Δt

If an amount of work W is done in a total time t, then the average power is

Pav =

W t

Obviously, if P does not vary with time, then P = Pav and the total work done W = P t. The standard metric unit of power is the Watt. As is implied by the equation for power, a unit of power is equivalent to a unit of work divided by a unit of time. Thus, a Watt is equivalent to a Joule/second. For historical reasons, the horsepower is occasionally used to describe the power delivered by a machine. One horsepower is equivalent to 746 Watts.

1.8.3 Energy Work is a simple notion, but it brings us to be complicated and many-sided concepts of energy. We so far understood that – something has energy means that it is capable of 30

doing work on something else. On other hand, when we do work on something, we have added to it an amount of energy equal to be work done.

Mechanics

What properties can a body have that may converted into work? Or what form of energy can take? •

Potential energy

•

Kinetic energy

Potential Energy

Potential energy is the energy stored in a body or a system by virtue of its position in a field of force or by its configuration. An object can store energy as the result of its position. For example, the heavy ram of a pile driver is storing energy when it is held at an elevated position. This stored energy of position is referred to as potential energy. Potential energy is the stored energy possessed by an object by virtue of its position in a field of force or by its configuration. The above example illustrates gravitational potential energy. •

Gravitational potential energy is the energy stored in an object as the result of its vertical position (i.e. height). The energy is stored as the result of the gravitational attraction of the Earth for the object.

•

The gravitational potential energy of the heavy ram of a pile driver is dependent on two variables – the mass of the ram and the height to which it is raised.

•

There is a direct relation between gravitational potential energy and the mass of an object; more massive objects have greater gravitational potential energy.

•

There is also a direct relation between gravitational potential energy and the height of an object; the higher that an object is elevated, the greater the gravitational potential energy. These relationships are expressed by the following equation : PEgrav = m g h

Another form of potential energy which we will discuss now is elastic potential energy. Elastic potential energy is the energy stored in elastic materials as the result of their stretching or compressing. Elastic potential energy can be stored in rubber bands, springs, an arrow drawn into a bow, etc.

The heavy ram of a pile driver possess stored energy – potential energy

Figure 1.16 : Example of Potential Energy

Springs are a special instance of a device which can store elastic potential energy due to either compression or stretching. •

A force is required to compress a spring; the more compression there is, the more force which is required to compress it further. 31

Physics

•

For certain springs, the amount of force is directly proportional to the amount of stretch or compression (x); the constant of proportionality is known as the spring constant (k).

Fspring = k . x •

If a spring is not stretched or compressed, then there is no elastic potential energy stored in it. The spring is said to be at its equilibrium position. The equilibrium position is the position that the spring naturally assumes when there is no force applied to it.

•

In terms of potential energy, the equilibrium position could be called the zero-potential energy position.

•

There is a special equation for springs which relates the amount of elastic potential energy in terms of the amount of stretch (or compression) and the spring constant. The equation is

PEspring =

1 2 kx 2

•

To summarize, potential energy is the energy which an object has stored due to its position relative to some zero position.

•

An object possesses gravitational potential energy if it is positioned at a height above (or below) the zero height position.

•

An object possesses elastic potential energy if it is at a position on an elastic medium other than the equilibrium position.

Kinetic Energy

Kinetic energy is the energy which is possessed by a body by virtue of its motion. An object which has motion – whether it be vertical or horizontal motion – has kinetic energy. There are many forms of kinetic energy : •

Vibrational (the energy due to vibrational motion).

•

Rotational (the energy due to rotational motion).

•

Translational (the energy due to motion from one location to another).

In this section we will focus upon translational kinetic energy and rotational kinetic energy. Translational Kinetic Energy The amount of translational kinetic energy which an object has depends upon two variables: the mass (m) of the object and the speed (v) of the object. The following equation is used to represent the kinetic energy (KE) of an object.

KE = where

1 m v2 2

m = mass of object, and v = speed of object.

This equation reveals that the kinetic energy of an object is directly proportional to the square of its speed. Kinetic energy is a scalar quantity; it does not have a direction. It is described by magnitude alone. The standard metric unit of measurement for kinetic energy is the Joule. 32

1 Joule = 1 kg

m2

Mechanics

s2

Rotational Kinetic Energy Consider a rigid body of mass M rotating about an axis passing through a point O and perpendicular to the plane of the page. The body possesses kinetic energy due to its motion and this energy is called kinetic energy of rotation. ω m1 P O

Figure 1.17

Take an infinitesimal element of mass m1 situated at a point P in the plane of the page and at a distance r1 from the axis. Then, at any given instant if ω is the angular speed of the body, the linear speed of the element will be v = r ω. The kinetic energy of the element will be

1 1 m1 . v12 = m1 . r12 ω12 2 2 We can similarly find the kinetic energies of all the infinitesimal elements of which the body is made. Then the total kinetic energy of the body will be

E=

1 1 1 m1 . r12 ω2 + m2 . r22 ω2 + . . . + mn . rn2 ω2 2 2 2 n 1 1 1 1 mi . ri2 ω2 = ω2 ∑ mi . ri2 = ω2 I = I ω2 2 2 2 i =1 2 i =1 n

E= ∑

n

M.I. of the body about the given axis, I = ∑ mi ri2 . i =1

If we compare the above expression with the expression for the kinetic energy of a body in translational motion, then it can be seen that I is analogous to the mass M in translational motion. In case if the centre of mass of the body also moves, the particle will possess in addition to the kinetic energy of rotation, the kinetic energy of translational motion. Total kinetic energy of the body = KE of rotation + KE due to translatory motion

=

1 1 I ω2 + M v 2 2 2

Hence when a body is rolling on a horizontal plane, the total energy of the body will be

=

1 1 I ω2 + M v 2 2 2

33

Physics

=

1 1 M K 2 ω2 + M v 2 (in terms of K (radius of gyration) 2 2

=

1 M ( K 2 ω2 + v 2 ) 2

=

1 M ω2 ( K 2 + r 2 ) 2

=

1 M ω2 r 2 2

=

1 M v2 2

⎛ K2 ⎞ ⎜⎜ 2 + 1⎟⎟ ⎝ r ⎠

⎛ K2 ⎞ ⎜⎜ 2 + 1⎟⎟ ⎝ r ⎠

SAQ 8 (a)

A solid cylinder rolls down an inclined plane. Its mass is 2 kg and radius 0.1 m. If the height of the inclined plane is 4 m, what is its rotational kinetic energy when it reaches the foot of the plane? Assume the surfaces are smooth.

(b)

A flywheel of mass 2 kg has a radius of 0.2 m. It is making 240 rpm. What is the torque necessary to bring it to rest in 20 s? If the torque is due to a force applied tangentially on the rim of the flywheel, what is the magnitude of the force?

Conservation of Energy

Kinetic Energy – Work Energy Principle Consider a body of mass m, being acted upon by a resultant accelerating force F be moving with the velocity v then by Newton’s second law of motion, we have

F =m

dv dt

dv is the acceleration of the particle. The work done by the force in dt displacing the body. where

W = =

∫ F . dr dv

∫ m dt dr

=m

∫

dr dv = m dt

v

1

∫ v . dv = 2 m v

2

0

Since the energy of the particle is its capacity to do work, we have Kinetic Energy, (Kinetic energy) T = We know that

1 m v2 2

F =m

dv dt

Multiplying both sides by ‘v’ in equation, we have 34

F . v = mv d dt

F .v =

Mechanics

dv dt ⎛1 2⎞ ⎜ mv ⎟ 2 ⎝ ⎠

We know that power is the rate of change of work.

Therefore

P=

dW = F .v dt

P=

dW d ⎛1 2⎞ = F .v = ⎜ mv ⎟ dt dt ⎝ 2 ⎠

Let the initial velocity of the particle be u and final velocity at any instant t be v then by integrating above equation, we have t

d dt

∫

W =

0

v

=

∫ u

=

⎛1 2⎞ ⎜ m v ⎟ dt 2 ⎝ ⎠

⎛1 ⎞ d ⎜ m v2 ⎟ = 2 ⎝ ⎠

v

⎡1 2⎤ ⎢2 mv ⎥ ⎣ ⎦u

1 1 mv 2 − mu 2 2 2

= Final KE – Initial KE. W = change in the kinetic energy. From this we conclude that the work done by the net force on a particle equals the change in the particles kinetic energy : Wtot = K2 – K1 = Δ K

(work-energy theorem)

This is known as work energy principle or work energy theorem. When Wtot is positive; K2 is greater than K1, the kinetic energy increases, and the particle is going faster at the end of the displacement than at the beginning. When Wtot is negative, the kinetic energy decreases and the speed is less after the displacement. When Wtot = 0, the initial and final kinetic energies K1 and K2 are the same and the speed is unchanged. We stress that the work-energy theorem by itself tells us only about changes in speed, not velocity, since the kinetic energy carries no information about the direction of motion. Conservative Forces

The potential energy is by virtue of its position, a body is capable to do work by virtue of its position, configuration or state of strain. The measure of this capability of the body is known as potential energy. It is usually denoted by U or V. Let the position of the body at any instant be represented by a vector r and its standard position, a position at which the potential energy becomes zero, i.e. it loses its capacity to do any work by r0. The potential energy at this instant V is given by r0

V =

∫

r

F . dr = −

r

∫

F . dr

r0

If the force F and displacement dr are in the same direction, we have r

V =−

∫ r0

F dr 35

Physics

Differentiating above relation, we have

dV dV = − F or F = − dr dr Hence the potential energy, V, may also be defined as a function of position whose negative gradient gives the force. From this we can also define a conservative force as equal to the negative gradient of potential energy V. The Law of Conservation of Energy : The Energy Function

Let a particle be displaced from position A to position B under the influence of a conservative force. The amount of work done on particle by the definition of kinetic energy will be given by B

WAB =

∫

F . dr = TB − TA

A

= Change in the kinetic energy and according to the definition of potential energy it will be given by A

WAB =

∫

F . dr = VA − VB

B

From above relation of work in terms of kinetic energy and potential energy, we have W AB = TB − TA = V A − VB V A + TA = VB + TB V A + TA = VB + TB = E (a constant)

The above relation shows that the sum of the kinetic and potential energies of a body under conservative force is a constant and this law is known as the law of conservation of mechanical energy. If the body is acted by non-conservative force, this law does not hold good. This can be seen from the following example. Let us slide a body on a rough fixed surface by applying force. We have to do work, part of which is used for providing the motion to the body and part of it is used for overcoming the friction, i.e. the work is done against the force of friction, which is a non-conservative force. Now when we slide the body back to its original position the friction force direction is changed and instead of recovering the work done on it during initial displacement we have to do work against the friction again. Thus, the total mechanical energy is not conserved and decreases by an amount equal to the work done against the non-conservative force.

1.9 GYROSCOPE A Gyroscope is any rotating body that exhibits two fundamental properties: inertia and precession. The term gyroscope is commonly applied to spherical, wheel-shaped, or diskshaped bodies that are universally mounted to be free to rotate in any direction; they are often used to indicate movements in space. Gyroscopes can be very perplexing objects because they move in peculiar ways and even seem to defy gravity. These special properties make gyroscopes extremely important in everything from your bicycle to the advanced navigation system on the space shuttle. The effect of all this is that, once you spin a gyroscope, its axle wants to keep pointing in the same direction. This is often called gyroscopic stability. With gyroscopic stability, a spinning object resists changes to its axis of rotation because an applied force moves along with the object itself. 36

Precession of Gyroscope

If you have ever played with toy gyroscopes, you know that they can perform all sorts of interesting tricks. They can balance on string or a finger; they can resist motion about the spin axis in very odd ways; but the most interesting effect is called precession. This is the gravity-defying part of a gyroscope.

Mechanics

Precession occurs when the center of gravity of a spinning object is not in a straight line. Precession can be pictured when we see a toy top or a gyroscope “wobble.” The most important source of precession we encounter everyday is with the planet Earth. The Earth demonstrates precession, or wobble, as it spins on its axis. Example : Gyroscopic bicycle wheel is able to hang in the air like this due to precession effect.

Figure 1.18 : Gyroscopic Bicycle

Consider a disc which is rotating about an axis passing through its centre and at right angles to its plane. Angular momentum of the rotating disc is along its axis of rotation and its direction is given by the right hand thumb rule. If now a torque is applied to the rotating disc r r • If torque τ direction is same as that of angular momentum L -angular momentum increases. r r • If τ is applied in the opposite direction to that L -angular momentum decreases. r However in both cases, the change ΔL is along the direction of torque acting. r • If torque τ direction is not the same as that of angular momentum r L -magnitude and direction of angular momentum changes. r • Torque τ acts in a direction perpendicular to that of angular r momentum L -magnitude remains unchanged but the direction of angular momentum changes. The Gyroscope is Rotating

ω

F

F

L=Iω L

τ

Spin Axis

+dL

L

−dL

τ

F

F

Figure 1.19 : The Gyroscopie is Rotating

Precessional Motion

Suppose a disc is rotating about an axis passing through its centre and at right angles to its plane. If the disc is now acted upon by a constant torque at right angles to the axis of rotation of the disc, the latter changes its plane of rotation by turning about a third mutually perpendicular axis. This turning motion is called precession and torque causing it is called precessional torque.

37

Physics

Frequency of Precessional Motion of a Simple Gyroscope

•

A simple gyroscope consists of a disc which can freely spin about a axle which is pivoted at a distance r from the centre of mass of the disc.

•

Let I be the moment of inertia of the disc about its axis of rotation.

•

If the disc rotates with an angular speed ω, the angular momentum r r associated with it is given by the equation L = I ω and is directed along the axle.

•

The torque about the point O due to the weight of the disc is given by Mg R and the direction of this torque is perpendicular to the axle. r This torque perpendicular to the direction of angular momentum L causes angular momentum to change in the direction of torque (perpendicular to the axle).

•

•

Hence, the axle moves in the direction of torque and the disc rotating about the axle turns about the pivot. The change in the angular momentum in time r r Δt in the direction of torque τ acting is given by ΔL = τ Δt .

τ

ΔL L

R Mg

L + ΔL

Δθ L

ΔL

Figure 1.20

•

If torque applied is larger, the change in angular momentum is also larger.

•

If Δθ be small, ΔL = L Δθ and the rate at which the axle rotates about its vertical axis, i.e. rate of precession is given by

Δθ 1 ΔL 1 τ Δt τ Mg R = = = = Δt L Δt L Δt L Iω •

•

Δθ Mg R = , the left hand side is the angular velocity of precession Δt Iω motion and is seen to be inversely proportional to the angular momentum. The frequency of precession is given by

Δθ = 2πν Δt ν= •

If k be the radius of gyration of the disc, L = I ω = Mk 2 ω ∴

38

1 Mg R 2π I ω

ν=

1 Mg R 1 gR = 2 2π Mk ω 2π k 2 ω

This expression gives the frequency of precessional motion. •

Mechanics

Hence the time period is given by ∴

T =

2π k 2 ω gR

Precession of a Spinning Top

Consider the spinning of top about an axis passing through point O. The weight Mg of the top acts through its C.G vertically downwards. If R be the distance of the C.G from O, the torque τ acting on the top is given by τ = Mg R sin α Also

ΔL = L sin α × Δθ

∴

Δθ =

ΔL L sin α

Hence the rate of precession is given by

Δθ 1 ΔL τ Mg R sin α Mg R Mg R = = = = = L sin α L Iω Δt L sin α Δt L sin α As the angular velocity decreases due to friction at the point O and air resistance, the rate of precession increases. Uses of Gyroscopes

The effect of all this is that, once you spin a gyroscope, its axle wants to keep pointing in the same direction. If you mount the gyroscope in a set of gimbals so that it can continue pointing in the same direction, it will. This is the basis of the gyrocompass. If you mount two gyroscopes with their axles at right angles to one another on a platform, and place the platform inside a set of gimbals, the platform will remain completely rigid as the gimbals rotate in any way they please. This is this basis of Inertial Navigation Systems (INS). In an INS, sensors on the gimbals’ axles detect when the platform rotates. The INS uses those signals to understand the vehicle’s rotations relative to the platform. If you add to the platform a set of three sensitive accelerometers, you can tell exactly where the vehicle is heading and how its motion is changing in all three directions. With this information, an airplane’s autopilot can keep the plane on course, and a rocket’s guidance system can insert the rocket into a desired orbit!

SAQ 9 (a)

What is a Gyroscope?

(b)

What is Precession?

1.10 CENTROIDS, CENTRE OF GRAVITY AND CENTRE OF MASS It is often required to define a body such that the length of wire, area of a rectangular plate; the volume, the mass or the gravitational forces acting on a body may be assumed to be concentrated at that point. Such points are often called as central points. Some commonly used central points are :

39

Physics

•

Centroid of the length of a curve, line.

•

Centroid of the volume of a body.

•

Mass centre of the mass of a body.

•

Centre of gravity of the gravitational forces acting on a body.

Centre of Gravity

Centre of gravity of a body is a point through which the resultant of the distributed gravity forces act irrespective of the orientation of the body. y

ΔM1

x1

Xc

c

ΔM2

yc

ΔM3

y1

o

x

Figure 1.21

Consider a body of mass M. Let this body be composed of ‘n’ number of masses ΔΜ1, ΔΜ2, ΔΜ3, . . . , ΔΜn, distributed within the body such that

Μ = ΔΜ1 + ΔΜ2 + . . . + ΔΜn The distance of these masses with respect to the axes be, (x1, y1), (x2, y2), . . . , (xn, yn). Let the centre of gravity of the whole mass M lie at a distance (xc , yc) with respect to the reference axes. Let us assume that the gravitational field is uniform and parallel. Gravitational force acting on the mass ΔΜ1 = ΔΜ1 g. Similarly, we can find the gravitational forces acting the masses ΔΜ2, ΔΜ3, . . . ΔΜn. Τo find the resultant of parallel forces ΔΜ1 g, ΔΜ2 g, ΔΜ3 g, . . . , ΔΜn g, we apply the principle of moments. The moment of the resultant of all forces about the y-axis = the sum of the moments of all the forces about the y-axis : Mg xc = ( ΔM 1 g ) x1 + ( ΔM 2 g ) x2 + . . . + ( ΔM n g ) xn

or,

xc =

(ΔM1 ) x1 + (ΔM 2 ) x2 + .. . + (ΔM n ) xn M n

xc =

∑ (ΔM i ) xi

i =1 n

∑ (ΔM i )

i =1 n

Similarly,

yc =

∑ (ΔM i ) yi

i =1 n

∑ (ΔM i )

i =1

Centre of Mass

40

It is the point where the entire mass of the body may be assumed to be concentrated.

Mechanics

The centre of mass and centre of gravity of a body are different only when the gravitational field is not uniform and parallel. The earth exerts a gravitational force on each of the particles forming a body. These forces can be replaced by a single equivalent force equal to the weight of the body and applied at the center of gravity for the body. The centroid of an area is analogous to the center of gravity of a body. The concept of the first moment of an area is used to locate the centroid. Concept of Centroid

One Dimensional Body Consider a body in the shape of a curved homogeneous wire of uniform cross-section and of a length L.

Figure 1.22

Divide the length of the wire into small elements of lengths ΔL1, ΔL2, ΔL3, . . . , ΔLn. Let the uniform area of cross section = A, density of the wire is ρ. The mass M of the wire of length L = A ρ L The mass of an element of length ΔL1 = ΔM1 ΔM1 = A ρ ΔL1 Similarly, the masses ΔM2, ΔM3, . . . , ΔMn of other elements can be determined. Let the distance of the centers of these lengths with respect to the axes be (x1, y1), (x2, y2), . . . , (xn, yn). Applying the principle of moments,

xc = xc =

∑ (ΔM i ) xi ∑ (ΔM i )

( A ρ Δ L1 ) x1 + ( A ρ Δ L2 ) x2 + ( A ρ Δ L3 ) x3 + . . . + ( A ρ Δ Ln ) xn ( A ρ Δ L1 ) + ( A ρ Δ L2 ) + ( A ρ Δ L3 ) + . . . + ( A ρ Δ Ln ) n xc =

A ρ (Δ L1 x1 + Δ L2 x2 + Δ L3 x3 + . . . + Δ Ln xn ) A ρ (Δ L1 + Δ L2 + Δ L3 + . . . + Δ Ln )

41

n

Physics

xc =

∑ ( ΔLi ) xi

i =1 n

∑ (ΔLi )

i =1 n

Similarly, yc =

∑ (ΔLi ) yi

i =1 n

∑ (ΔLi )

i =1

Because the density, ρ, and the area of the cross section, A, are constant over the entire length of the wire, the coordinates of CG of the wire become the co-ordinates of the centroid of the wire : generally called the centroid of a line segment. Centroid of Two-dimensional Body

We can derive the centroid of an area as we have derived for the line segment.

xc =

∑ (ΔAi ) xi ∑ (ΔAi )

yc =

∑ (ΔAi ) yi ∑ (ΔAi )

xc, yc are the centroids of the plate, generally called as the coordinates of the centroid of an area. Again, as the density and thickness of the plate are constant over the entire area so, the coordinates of the CG become the coordinates of the centroid of the area.

Figure 1.23

Generally the term centroid is used for the centre of the gravity of a geometrical figure and the term centre of gravity is used when referring to actual physical bodies. Determination of Centroid and Centre of Gravity

By considering a plane figure to be made up of a number of small elements of length or area, we cannot generate the true shape of the figure. To generate the shape of a figure, we have to make the size of these elements very small and their number very large.

42

Mathematically, we replace the terms ΔL and ΔA in the early expressions of centroid and CG by the integral term. Since ΔL and ΔA is considered infinitesimally small, expressions of centroid and CG are rewritten as

∫ x dL xc = ∫ dL yc =

∫ x dA xc = ∫ dA

∫ y dL ∫ dL

yc =

∫ y dA ∫ dA

∫ x dm xc = ∫ dm yc =

Mechanics

∫ y dm ∫ dm

where dL, dA and dm denote the length, area and mass respectively of a differential element chosen and the (xc, yc) the coordinates of its centroid. The integral ∫ x dA is known as the first moment of area w.r.t. the y-axis. Similarly, the integral

∫ y dA

denotes the first moment of area w.r.t. the x-axis.

Figure 1.24

xA=

∫

x dA = Qy

= first moment with respect to y.

yA=

∫

y dA = Qx

= first moment with respect to x.

xL=

∫

yL =

∫

x dL y dL

Figure 1.25

Example 1.9

Compute the centroid of a triangular lamina. Solution

Consider the right triangle depicted below. Suppose that the right triangle is constructed from a material with a uniform density of ρ units of mass per unit of area.

43

y

Physics

h f (x)

x

o dx

x b

Figure 1.26

We are going to compute the moment of the triangle about the y-axis. •

To simplify the calculation, we cut the triangle up into a set of thin vertical strips.

•

We will compute the moment of each of the strips about the y-axis and then integrate to add up all of the moments.

The moment of a strip centered at x is given by Moment = Mass × Distance = m ( x) x The mass of the strip, m (x), is given by the density ρ times the area of the strip. The area is the product of the little bit of thickness dx and the height of the strip, f (x). m ( x) λ = ρ A ( x) = ρ f ( x) dx A simple argument shows that

f ( x) =

h x b

h ⎛ h ⎞ Hence, Moment of the small strip w.r.t. y-axis = ⎜ ρ x dx ⎟ x = ρ x 2 dx b ⎝ b ⎠ b

and Total moment of all such strips w.r.t. y-axis =

∫ 0

h h x3 ρ x 2 dx = ρ b b 3

b

0

To compute the distance from the reference line to the center of mass, we use the formula Total moment = (Total mass) d c

or

dc =

Total moment Total mass

The mass is easy to compute.

Total mass = ρ × Area = ρ ×

Thus

44

1 2 b hρ 2 dc = 3 = b 1 ρ hb 3 2

1 hb 2

2 b , the triangle will 3 balance exactly on the line. If you look back up at the picture of the triangle above, you will see that this answer appears to be perfectly reasonable.

This says that if we lay a line parallel to the y-axis at x =

Mechanics

Example 1.10

Determine the centroid of an area of a semi-circle as shown in Figure 1.27. y D A a

θ

C

x F

G x

B

E x

dx

Figure 1.27

Solution

You will note that it is not possible to divide a semi-circle into a few suitable standard areas for each of which magnitude as well as location of its centroid is known. Hence in such a situation, we should consider a very large number of very thin strips of width dx and assume that its centroid is at the centre of the strip. Consider the semi-circle with centre C and its straight boundary ECD along y-axis. The axis CX is therefore the axis of symmetry of the area. This means that portion of quadrant CFE is the mirror image of quadrant CFD. Let us consider of an area as a strip AB at a distance x from C with width dx tending to zero. If a is the radius of semi-circle, then

AB = 2a sin θ

Elemental area

dA = 2a sin θ dx

where

x = a cos θ

or

dx = − a sin θ dθ A=

π 2

∫

dA

0

=

π 2

∫

2a 2 sin 2 θ d θ

0

Since

cos 2θ = 1 − 2 sin 2 θ

2sin 2 θ = 1 − cos 2θ

∴

A = a2

π 2

∫

(1 − cos 2θ) d θ

0

sin 2θ ⎞ ⎛ = a2 ⎜ θ − ⎟ 2 ⎠ ⎝

π 2

0

45

Physics

Since value of θ = ∴

π , 2 A=

π a2 2

is already a known value of a semi-circle, since area of full circle is π a 2 . Considering the first moment of elemental areas about y-axis, we have A x = ∑ Ai xi =

π 2

∫

2a 2 sin 2 θ d θ × a cos θ

0

π

⎡ sin 3 θ ⎤ 2 = 2a 3 ⎢ ⎥ ⎣⎢ 3 ⎦⎥ 0 π a2 2a 3 x= 2 3

x=

∴

4a = 0.424 a 3π

Example 1.11 Determine the centroid of the shaded area shown in Figure 1.28. B

A

a1 = 100 mm C

200 mm

H

D

G E a2 = 60 mm 100 mm

100 mm

100 mm

y B

A

A1 (100, 100) A3

H

A2 C

C1 42.4 mm

x

D

Figure 1.28

Solution

The shaded area A can be considered as algebraic sum of three areas A1, A2 and A3 as shown in Figure 1.28. Now

A = A1 + A2 − A3

where

A1 = 200 × 200 = 40,000 mm 2 with centroid at (100, 100)

A2 =

and 46

π × 1002 = 15,700 mm 2 with centroid at (242.4, 100) 2

⎛ − π × 602 ⎞ 2 A3 = ⎜ ⎟⎟ = − 5655 mm with centroid at (100, 25.44) ⎜ 2 ⎝ ⎠

Sl. No.

Area Ai (mm2)

xi (mm)

Ai xi (mm3)

yi (mm)

Ai yi (mm3)

1

A1 = 40,000

100

400 × 104

100

400 × 104

2

A2 = 15,700

242.4

380 × 104

100

157 × 104

3

A3 = − 5655

100

− 56 × 104

25.44

− 14 × 104

∴

Mechanics

∑ Ai = 50,045 mm 2

∑ Ai xi = 724 × 104 mm3 ∑ Ai yi = 543 × 104 mm 3

∴

x=

y=

724 × 104 5.0045 × 104

543 × 104 5.0045 × 104

= 144.67 mm

= 108.50 mm

Example 1.12

Determine the CG of the thin uniform wire bent into shape ABC as shown in Figure 1.29. y 4m C

3m B

G (x, y) x 5m y

A

x

Figure 1.29

Solution

Let G be the location of CG of the bar as shown in Figure 1.29. L = L1 + L2 = 5 + 5 = 10 m L x = L1 x1 + L2 x2 = 5 × 0 + 5 × 2 = 10

x =1m Taking moment about x-axis, L y = 5 × 2.5 + 5 × 6.5 or,

y = 4.5 m

Example 1.13

Determine the CG of a wire of uniform cross-section bent into shape of a semi-circle as shown in Figure 1.30. 47

Physics

Take radius of wire = a. y B a δθ

δθ θ

x C

G x

A

Figure 1.30

Solution Refer Figure 1.30 showing length of wire π a, where a = radius of semi-circle with centre at 0.

For element (δL) = a dθ +

L=

π 2

∫π

−

Lx =

+

∫

∫π

δL =

−

2

π 2

π 2

a dθ = a θ

2

−

δL × x = a 2

π 2

∫π

−

= a 2 sin θ ∴

x=

π 2 −

= π a (i.e. the perimeter of semi-circle)

π 2

π 2

cos θ d θ

2

= 2a 2

2a 2 2a = = 0.637 a πa π

SAQ 10 (a)

Determine the centroid of the area as shown in Figure 1.31 where D is the contact point of the circle to the edge GF. y

2 cm

E

F 4 cm

a = 4 cm D 12 cm

G

B 2 cm

O

A 10 cm

48

Figure 1.31

x

(b)

Determine the position of centroid of a quadrant OAB of a circle, where arc AB subtends an angle of 90o at the centre O and radius OA = a.

(c)

Determine the centroid of the area shown in Figure 1.32, where portion BCE is a quadrant of circle of radius = 10 cm.

Mechanics

y

D

E

10 cm

A

B

C

10 cm

x

10 cm

Figure 1.32

(d)

Determine the centroid of length in form of a channel ABCD as shown in Figure 1.33. 10 cm D

C

20 cm

x

M

A

B

Figure 1.33

1.11 SUMMARY Let us summarise what we have learnt in this unit. Motion is a common phenomenon but by mechanics point of view, it is an extremely important phenomenon. In this unit, we have understood the cause of motion and found that a body at rest may be set into motion, on being acted upon by external forces. We have also discussed Newton’s law of gravitation. A particle may move along a circular path with uniform or variable velocity. We have also discussed angular velocity, as well as, angular momentum. The concept of radius of gyration has also been introduced while discussing moment of inertia. In this unit, we also learnt that whenever a constant force F acts on an object while it experiences a displacement, ‘d’, we say that the force does work W on the object. The amount of work done is a scalar quantity, and is calculated from W = F . d. The concepts of work, power, and energy have been explained in detail in this unit along with different forms of energy. The concept of gyroscope has also been introduced. Centre of gravity is a point where the entire mass or weight is assumed to be concentrated. Centre of mass, for a body with a system of particles is the point where the entire mass of the system is assumed to be concentrated. Generally, a given plane area can be divided into number of smaller areas Ai for which the centroid Gi (xi, yi) are known. The location of the centroid for a composite area can be computed with the help of following expressions :

A = ∑ Ai , x =

∑ Ai xi , A

y=

∑ Ai xi A 49

Physics

1.12 KEY WORDS Centre of Mass

: The point where the entire mass of a body may be assumed to be concentrated.

Centre of Gravity

: The point at which the entire weight acts irrespective of the orientation of the body.

Centroid

: For geometrical figures like line, areas and volumes ‘CG’ is known as centroid and depends only on the geometry of the body.

Parallel-axis Theorem

: Moment of inertia of an area about an axis is equal to the sum of (a) moment of inertia about an axis passing though the centroid parallel to the given axis and (b) the product of area and square of the distance between the two parallel axis.

Perpendicular-axis Theorem

: Moment of inertia of a plane lamina about an axis perpendicular to the lamina and passing through its centroid is equal to the sum of the moment of inertia of the lamina about two mutually perpendicular axes passing through the centroid and in the plane of the lamina

Izz = Ixx + Iyy

50

Newton’s First Law

: Everybody continues in its state of rest or of uniform motion in a straight line unless it is compelled to change that state by forces acting on it.

Newton’s Second Law

: The rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction in which the force acts.

Newton’s Third Law

: To every action there is always an equal and opposite reaction.

Power

: Power is defined as the rate at which work is being done.

Linear Momentum

: Linear momentum of a particle at any instant is the product of its mass (m) and its velocity (v) at that instant.

Work

: The work done by a constant force F acting on a particle is the product of the component of force in the direction of displacement and the magnitude of displacement.

Radius of Gyration

: It is defined as the distance from the axis of rotation at which, if whole mass of the body were supposed to be concentrated, the moment of inertia would be same as with the actual distribution of the mass of body into small particles. It is denoted by K. (I = MK2).

Energy

: The energy of a body is defined as its capacity of doing work. It is a scalar quantity.

Potential Energy

: It is the energy possessed by a body by virtue of its position or configuration (shape or size).

Keplar’s Law of Planetary Motion

: (i)

Every planet moves in a elliptical orbit with the sun at one of the focal points.

(ii)

The radius vector drawn from the sun to any planet sweeps out equal areas in equal time intervals.

(iii)

The square of the orbital period of any planet is proportional to the cube of the semi-major axis of the elliptical orbit.

Kinetic Energy

: The energy possessed by a body by virtue of its motion is called kinetic energy.

Angular Velocity

: The rate of change of angular displacement with time is known as angular velocity.

ω=

Mechanics

dθ dt

: It is inherent property of a body which offer resistance to change in its state of rest or of uniform motion.

Inertia

1.13 ANSWERS TO SAQs SAQ 1

(a)

Cannon mass (M) = 20000 kg, shell mass (m) = 100 kg. Its velocity is 800 ms– 1. Let recoil velocity be V then MV + mv = 0, by principle of conservation of momentum or

V =−

100 × 800 = − 4.0 ms −1 20000

(Direction of common velocity opposite to that of the shell.) Final velocity of cannon (v) = 0, initial velocity u = 4.0 ms– 1. Let a = Acceleration of cannon, and t = Time taken to stop cannon in its recoil distance of s (s = 0.4 m). v 2 = u 2 + 2 as

Then or

a=

1 × {0 − 4.02 } = − 20 ms − 2 (retardation) 2 × 0.4

The retarding force P required = M × a = 20000 × 20 = 4 × 105 N Also,

v = u + at

⇒

0 = 4.0 − 20 t

⇒

t = 0.20 sec.

Let M1 = 7 kg, and M2 = 12 kg, for mass M1; Figure 1.34(a) is free body diagram for mass M1.

(b)

M1 a = T − M1 g

⇒

7 a = T − 7 × 9.8

. . . (1)

T T M1 a

51

M1 B

B

M2 a B

M1 g B

B

B

Physics

(a)

(b) Figure 1.34

For mass M2; Figure 1.34(b) is free body diagram for mass M2. M2 a = M2 g − T

12 a = 12 × 9.8 − T

. . . (2)

Adding Eq. (1) and (2), we get 7a + 12a = 12 × 9.8 − 7 × 9.8 ⇒

a=

5 × 9.8 = 2.58 ms − 2 19

From Eq. (1), we have T = 7 a + 7 × 9.8 = 7 × 2.58 + 7 × 9.8 = 86.66 N SAQ 2

(a)

The gravitational attraction of the earth of mass M and radius R on a body of mass, m on the surface of earth is given by mg =

GM m R2

where g is acceleration due to gravity

M =

g R 2 9.8 × (6.37 × 106 ) 2 = = 6 × 1024 kg 11 − G 6.67 × 10

If ρ be the density of earth, then

M = or (b)

ρ=

4 π R3 ρ 3

3M 4π R

3

=

3 × 6 × 1024 6 3

4π × (6.37 × 10 )

= 5.5 × 103 kg/m3

Let the radius of the earth’s orbit round the sun be R. When it is reduced to half, the new distance becomes R/2. As the earth is moving round the sun, centripetal force is provided by the gravitational attraction of the sun on the earth. Hence

mv 2 G M m = R R2 where the symbols have their usual meaning,

v2 =

52

GM R

. . . (1)

If T be the time period of the earth to complete one revolution round the sun, then

v=

2πR T

. . . (2)

Mechanics

From Eq. (1) and (2), we get 4 π2 R 3 GM

T2 =

. . . (3)

Let T1 be the time period of revolution of earth round the sun, when the distance between the two is R/2. Hence

4 π2 R 3 8G M

T12 =

. . . (4)

Dividing Eq. (4) by Eq. (3), we have

T12 =

(c)

T2 T , T1 = 8 2 2 365

∴

T1 =

We know

g′ = g

⇒ here

g′ =

⇒

g R2 =g 4 ( R + h) 2

⇒

R 1 = R+h 2

⇒

R=h

= 129 days

2 2

(Q 1 year = 365 days)

R2

( R + h) 2

g 4

At a height, which is equal to the radius of the earth. (d)

We know, W = mg =

GM m

, when the diameter and hence radius R of the R2 earth becomes two times, the weight of the object becomes one fourth of its previous value.

SAQ 3

(a)

mv 2 , where (Re + h) = orbital radius of ( Re + h) the satellite, Re = the radius of the earth, and m = mass of satellite.

We know centripetal force =

Now,

mv 2 GM m = ( Re + h) ( Re + h) 2

⇒

v2 =

⇒

Re g g Re2 = 2 ( Re + h)

⇒

Re + h = 2 Re

⇒

h = Re

GM ( Re + h) [Q G M = g Re2 ]

53

Physics

(b)

(i)

(ii)

We know

mv 2 GM m = ( Re + h) ( Re + h) 2

⇒

v2 =

GM Re + h

⇒

v2 =

g Re2 Re + h

⇒

v=

g Re2 Re + h

⇒

=

9.8 × (6.38 × 106 ) 2 6.68 × 10

Time period = =

(iii)

= 7727.62 ms −1 (Q Re + h = 6680 km)

2π ( Re + h) v 2 × 3.14 × 6.680 × 106 = 5431.38 sec or 1.51 hrs 7727.62

Radial acceleration =

v2 ( Re + h)

=

(c)

6

7727.622 6.68 × 10

6

= 8.94 ms − 2

⎛ T2 ⎞ We know, R 3 = ⎜ 2 ⎟ g Re2 ⎜ 4π ⎟ ⎝ ⎠

where R = Orbital radius of the satellite, Re = Radius of the earth, and T = Time period. In this case, T = 1 day = 86400 seconds and substituting the value of Re and g, we get, 1

⎡⎛ 1 ⎞ ⎤3 R = ⎢⎜ 2 ⎟ (86400)2 × 9.8 × (6.38 × 106 ) 2 ⎥ ⎣⎝ 4π ⎠ ⎦ = 42, 250 × 103 m = 42, 250 km

Therefore, the height of the satellite above the surface of the earth h = R − Re = 35,870 km

Note : If a satellite is placed in orbit at this height above the equator it will appear stationary. SAQ 4

For a solid cylinder, moment of inertia =

1 MR 2 . 2

where M = Mass of the cylinder. Now,

M = Density × Volume = (π r 2 l ) × ρ

54

= ( π × 0.052 × 1.50) × 8000 = 94.25 kg

MI = =

1 M R2 2

Mechanics

1 × 94.25 × 0.052 = 0.1178 kg m 2 2

SAQ 5

(a)

m = 1 kg, r = 0.5 cm = 0.005 m, ω0 = 0, α = 2 rad/s2, t = 60 seconds. ω = ω0 + α t

We know

= 0 + 2 × 60 = 120 rad/sec. v=ωr

= 120 × 0.005 = 0.6 ms −1

Centripetal force = (b)

m v 2 1 × 0.62 = = 72 N r 0.005

Angular speed for second hand =

2π π = rad/sec. 60 30

Angular speed for minute hand =

2π π rad/sec. = 60 × 60 1800

Angular speed for hour hand =

2π π rad/sec. = 12 × 60 × 60 21600

SAQ 6

(a)

m = 20 kg, radius of gyration K = 10 cm = 0.10 m We know moment of inertia = m K2 = 20 × 0.12 = 0.20 kg m 2

Now,

ω=

2π N 2π × 120 ⇒ω= = 4π 60 60

Angular momentum = I ω = 0.20 × 4π = 2.5132 kg m 2 s − 1

1 I ω2 2

Kinetic energy =

1 × 0.2 × (4π) 2 = 15.79 J 2

= (b)

We know τ = I α where

τ = Torque, I = Moment of inertia, and α = Angular acceleration.

α= Now,

ω − ω0 20 − 2 = = 2 rad s − 2 9 t

τ=I α 5000 = I × 2 ⇒ I = 2500 kg m 2 55

Physics

SAQ 7

(a)

Net work done = 3 × (200 × 10 + 50 × 35) = 11250 J

(b)

Work done = m g h = 80 × 9.8 × 8 = 6272 J.

SAQ 8

We know τ = I α

ω0 =

2π N 2π × 240 ⇒ ω0 = = 8π 60 60

Now,

ω = ω0 − α t

⇒

0 = 8π − α × 20

⇒

α=

8π rad/s 2 20

I =

m r2 1 = × 2 × 0.22 = 0.04 kg m 2 2 2

τ=I α

τ = 0.04 ×

8π = 0.05 Nm 20

We know, τ = F r , where r = radius of the flywheel.

F=

⇒

0.05 = 0.25 N 0.2

SAQ 9

Please refer to the preceding section. SAQ 10

(a)

Referring to Figure 1.35, total area is divided into three parts, A1, A2 and A3. y

2 cm 8 cm

4 cm

a = 4 cm

A3

8 cm

A2

A1

2 cm x

Figure 1.35

Area Mark

Area (cm )

(cm)

yi (cm)

20

5

1

2

A1

56

xi

Ai xi 3

Ai yi

(cm )

(cm3)

100

20

A2

20

1

7

20

140

A3

50.26

6

8

301.59

402.12

Total

90.26

∴

(b)

x=

421.59 = 4.67 cm 90.26

y=

562.12 = 6.23 cm 90.26

Mechanics

We have already determined position of centroid of semi-circular area A, which is as under

4a 3π

x=

Let x1 be the distance of centroid of upper quarter of circle of area A1 shown shaded in the Figure 1.36. A1 =

π a2 A = 4 2

y

y

X1

B a a

C

A1 x

O x

O

A2 x1 a X2

Figure 1.36

Area of lower quarter A2 =

π a2 = A1 4

By observation, x2 = x1 By considering moments about OB, A x = A1 x1 + A2 x2

=

A A x1 + x1 = x1 A 2 2

x1 = x =

∴

4a 3π

Similarly we can prove

y1 = y = (c)

4a 3π

Dividing the area into two parts (i)

Square area A1 = 10 × 10 cm 2 , and 57

Physics

(ii)

Quadrant of circle of A2 =

π a2 = 25 π = 78.54 cm 2 4

x2 = 10 + Area Mark A1 A2 Total

∴ (d)

Area (cm2) 100 78.54 178.54

4 × 10 = 14.24 cm 3π

xi

yi

Ai xi

Ai yi

5 14.24

5 4.24

500 1118.4 1618.4

500 333.3 833.3

x = 9.06 cm;

and

y = 4.67 cm

If C is the centroid of the channel-shaped length, it is noted that this channel-shaped length is symmetrical about axis CX which is parallel to flanges AB and CD. It is also to be noted that this length is not symmetrical about other centroidal axis ey which is parallel to the web BC as seen in Figure 1.37. y

C

D 10 cm x 10 cm

B

A 10 cm

Figure 1.37

Let x be the perpendicular distance of C from BC. Total length L = L1 + L2 + L3 = 40 cm

Considering moment about BC L × x = ∑ Li xi

40 x = 2 (10 × 5) = 100 ∴

58

x = 2.5 cm