rotational motion part 1

8/3/2019 Rotational Motion Part 1

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PHYSICS 1LOCUS

INTRODUCTION

So far we have dealt mostly with the translational motion of single particles or rigid bodies, that is, of bodies whose

parts all have a fixed relationship with each other. No real body is truly rigid, but many bodies, such as molecules,

steel beams, and planets, are rigid enough so that, in many problems, we can ignore the fact that they warp, bend,

or vibrate. A rigid body moves in pure translation if each particle of the body undergoes the same displacement

as every other particle in any given time interval, as discussed in the previous topic.

In this topic we are interested in rotation. For the time being we again restrict ourselves to single particles

and to rigid bodies. Figure 7.1 shows the rotational motion of a rigid body about a fixed axis, in this case the z-axis

of our reference frame. Let P represents a point in the body, arbitrarily selected and at a distance r from the z-axis.

As the body rotates, the point P rotates in a circle of radius r about the z-axis. We then say that : A rigid bodymoves in pure rotation if every point of the body moves in a circle, the centres of which are on a straight

line called the axis of rotation. When a body moves in pure rotation all the points on the body move in parallel

planes and axis of rotation is perpendicular to these planes. If we

draw a perpendicular from any point in the body to the axis, each

such line will sweep through the same angle in any given time

interval as another such line. Thus we can describe the pure rotation

of a rigid body by considering the motion of any one of the particles

(such as P) that make it up. (We must ru le out, however, particles

that are on the axis of rotation. Why?)

r

y

x

P

O

z

fig. 7.1:The general motion of a rigid body is a combination of translation

and rotation however, rather than one of pure rotation. If a body

moves neither in pure translation nor in pure rotation, as shown in

figure 7.2, how should we proceed then ?

A body of arbitrary

shape rotating about the z-axis.



PHYSICS 2LOCUS

Let us move to the frame attached with some arbitrary

point, P, on the body. In this frame the point P must

be at rest. As the body is rigid, all other points on the

body maintain their distances from the point P and

hence they move in a circle or we can say that the

body moves in pure rotation about an axis passingthrough the pointP, as shown in figure 7.3. Therefore,

to take the advantage of this fact, while analyzing the

general motion of a rigid body in some frame x - y - z,

first we should analyze its motion in some frame x'-

y'- z' attached to the some arbitrary point P on the

body and then add the effect of motion of the frame

x'- y'- z' itself with respect to the frame x- y- z.

Generally, the selected point P is the centre of mass

of the body. This particular choice has following two

great advantages:

(1) analysis of a system is much simpler in itsC

frame, as we saw in the chapter 6,

(2) the analysis of the motion of the centre of mass

itself is also much simpler as compared to the

any other point on the body.

Hence, the smartest way to analyze the general motion of a body is to analyze the motion of the body in its C frame

and then add the effect of the motion of the C frame itself.

ANGULAR VELOCITY AND ANGULAR ACCELERATION:

Let us observe the point P in figure 7.1 by looking downward on it from

above, along the z-axis, as shown in figure 7.4 The point P moves in a

plane parallel to the x- y plane in a circle of radius r . As the body is rigid,

we can tell exactly where the entire rotating body is in our reference

frame if we know the location of any single particle (P) of the body inthis frame. Thus, for the kinetics of this case, we need only consider the

two-dimensional motion of a point in a circle.

θr

y

x

P

S

O

fig. 7.4: The Point is moving in acircle of radius in a planeparallel to the - plane.

Pr

x y

In the same figure the angleθ is the angular position of the particle

Pwith respect to the reference position. We arbitrarily choose the positive

sense of rotation in figure 7.4 to be counter clockwise, so that θ increases

for counterclockwise rotation and decreases for clockwise rotation.

In radians θ is defined by the relation

θ θ θ θ θ = s/r , ...(7.1)

in which s is the arc length shown in figure 7.4.



PHYSICS 3LOCUS

At time 1t the angular position of P is 1θ and at a later time 2t its

angular position is 2θ , as shown in figure 7.5. The angular

displacement of P is 2 1θ θ θ ∆ = − during the time interval 2 1.t t t ∆ = −

The average angular speed ω of point P in this time interval is

defined as

2 1

2 1

∆ – = =∆ –

θ θ θ ω

t t t ...(7.2)

P(t )2

P(t )1

y

x Oθ1θ

2

∆θ

fig. 7.5: The Point has displaced

through an angle in time

P

∆θ∆t t t (= – ).2 1

Theinstantaneous angular speed ω ω ω ω ω is defined as the limit approached

by this ratio as ∆t approaches zero:

→∆ 0

∆= lim =

∆t

θ d θ ω

t dt ...(7.3)

For a rigid body all radial lines fixed in it perpendicular to the axis of rotation rotate through the same angle in thesame time, therefore, the angular speed ω about this axis is the same for each point in the body. Thus ω is

characteristic of the body as a whole. Its units are commonly taken to be radians/sec or revolutions/sec (1 rev/sec

= 2π rad/sec).

If the angular speed of P is not constant, then the particle has an angular acceleration. Let 1ω and 2ω be the

instantaneous angular speeds at the time 1t and 2t respectively; then the average angular acceleration α of the

particle P is defined as

−

+

2 1

2 1

∆= =∆

ω ω ωα

t t t

...(7.4)

The instantaneous angular acceleration α α α α α is the limit of this ratio as ∆t approaches zero, or

∆= =∆ t

ω d ωα

t d ...(7.5)

Because ω is the same for all points in the rigid body, α must be the same for each point and thus α , like ω , is a

characteristic of the body as a whole. Its units are commonly taken to be radian/sec² or revolution/sec².

If θ (t ) is given, ω (t ) can also be found by differentiating it with respect to the time and by differentiating ω (t ) with

respect to the time, α (t ) can also be found. This is very similar to what we did in chapter 1 (kinematics).

If α (t ) and initial angular speed, ,iω and initial angular position, ,iθ are given then integration can be used to find

ω (t ) and θ (t ). We have,

d

dt

ω α =

⇒ d dt ω α = ⋅

⇒0i

t

d dt

ω

ω

ω α = ⋅∫ ∫ [Let at some time t , angular speed is ω.]

⇒0

t

i dt ω ω α − = ⋅∫



PHYSICS 4LOCUS

⇒ = ⋅∫ 0

t

i dt ω α ...(7.6)

⇒ 0

t

i

d

dt dt

θ

ω α = + ⋅∫

⇒0

t

id dt dt dt θ ω α

= ⋅ + ⋅ ⋅ ∫

⇒0 0 0i

t t t

id dt dt dt

θ

θ

θ ω α

= + ⋅ ⋅

∫ ∫ ∫ ∫ [Let at some time t , angular position is θ.]

⇒ 0 0

t t

i i

t dt dt θ θ ω α

− = + ⋅ ⋅ ∫ ∫

⇒

∫ ∫ 0 0

( )

t t

i i t t dt dt + + ⋅ ⋅θ ω α ...(7.7)

If angular acceleration, α , is constant, then, we have,

( ) ω +i ω t = αt [From equation (7.6)] ...(7.8)

and21

( ) = + + 2i i θ t θ ω t αt [From equation (7.7)] ...(7.9)

If α is constant and 0,iθ = then, we have,

21( ) = + +

2i θ t ω t αt ...(7.10)

Eliminating ‘t ’ from (7.8) and (7.10), we get,

⋅ ⋅2 2= + 2i ω ω α θ ...(7.11)

Here you should note that the rotation of a particle about a fixed axis has a formal correspondence to the translationmotion of a particle (or a rigid body) along a fixed direction. The correspondence is as follows:

S

v

a

θ

ω

α

→

→

→



PHYSICS 5LOCUS

RELATION BETWEEN ANGULAR VARIABLES AND LINEAR VARIABLES:

From equation (7.1), we have

s r θ = ⋅

Therefore, speed of the point P,

y

x Oθ

ωα

r

a

v

S

P

fig. 7.6: Linear and angular variablesare shown together.

dsv

dt =

( )d r

dt

θ ⋅=

d r

dt

θ = ⋅ [ ]is constantr ∵

⇒ =v ωr ...(7.12)

The relation between tangential linear acceleration of the point P and angular acceleration is obtained by taking the

derivative of equation (7.12) with respect to time:

(= rate of change of speed)t

dva

dt =

( )d r

dt

ω =

d r

dt

ω = ⋅ [ ]is constantr ∵

⇒ ⋅=t a α r ...(7.14)

Each point on the body has a radial linear acceleration, the centripetal acceleration which points inward along the

radial line, and has the magnitude

22

= =n

va ω r

r ...(7.14)

A particle moves in a circle of radius 100 m with constant speed of 20 m/s.

(a) What is its angular velocity in radians per second about the centre of the circle?

(b) How many revolutions does it make in 30 seconds?

Solution: From equation (7.12), we have,

v r ω =

⇒ v r ω =

20rad/s

100=

0.2 rad/s=



PHYSICS 6LOCUS

From equation (7.2), we have, average angular velocity,

t

θ ω

∆=

∆

⇒ t θ ω ∆ = ⋅ ∆

t ω = ⋅ ∆ As is constant,ω ω ω =

(0.2 30) radians= ×

6 radians=

Therefore, number of revolutions =2

θ

π

∆

6

2π

=

3revolutions.

π =

A wheel starts from rest and has a constant angular acceleration 2 rad/s².

(a) What is the angular velocity after 5 s?

(b) How many revolutions has it made in 5s?

(c) After 5s, what is the speed and acceleration of a point 0.3 m from the axis of rotation?

Solution: Using equation (7.8), we have

i t ω ω α = +(0 2 5) rad/s= + ×10 rad/s=

Using equation (7.10), we have,

21

2it t θ ω α = +

210 2 5 rad

2

= + × ×

25 rad=Therefore, number of revolutions

2

θ

π =

25revolutions

2π =

If point P be at a distance of 0.3 m away from the axis of rotation, then, at t = 5 s its speed,

v r ω =(10 0.3) m/s= ×3 m/s=

At the same moment the radial acceleration of the point P is 2 230 m/sr a r ω = = and the tangential acceleration of

the point P is 20.6 m/s .t a r α = =



PHYSICS 7LOCUS

A solid body rotates about a stationary axis so that its angular velocity depends on the rotation angle θ as

0 0,k ω ω θ ω = − andk are positive constants. At the moment t = 0, the angle θ = 0. Find the time dependence of

(a) the rotation angle(b) the angular velocity.

Solution:We have, 0 k ω ω θ = −

⇒ 0

d k

dt

θ ω θ = −

⇒00 0

t d

dt k

θ θ

ω θ =

−∫ ∫ [ ]At 0, 0t θ = =∵

⇒ 0 00

1ln( )

t k t

k

θ

ω θ − − =

⇒ 0 0ln( ) lnk kt ω θ ω − − = −

⇒0

0

lnk

kt ω θ

ω

−= −

⇒0

0

kt k e

ω θ

ω

−−=

⇒0 0

kt k eθ ω ω

−= −

⇒ – 0= (1 – )kt ω

θ ek

Now,d

dt

θ ω =

( )0 1 kt d e

k

dt

ω − − =

0

0

0kt

kt

de

k dt

e

ω

ω

−

−

= −

=



PHYSICS 8LOCUS

ROTATIONAL QUANTITIES AS VECTORS:

Angular motion has direction associated with it and is inherently a vector process. But a point on a rotating wheel is

continuously changing its direction and it is inconvenient to track that direction. The only fixed, unique direction for

a rotating wheel is the axis of rotation so, it is logical to choose axial direction as the direction of the angular velocity.

Left with two choices about direction, it is customary to use the right hand rule to specify the direction of angularquantities, as shown in figure 7.6 (a) and (b)

fig. 7.6(a): Right hand rule: turn the fingersof your right hand along the sense of rotation,the direction in which the thumb points is thedirection of the angular velocity. In this casethe direction of the angular velocity of the discis pointing perpendicularly upwards from theplane of rotation of the disc.

ω !

!

ω

fig. 7.6(b): In this case the direction of theangular velocity of the disc is pointingperpendicularly downwards from the planeof rotation of the disc.

Here you should note that nothing moves in the direction of angular velocity, i.e., nothing moves along the axis of

rotation. The direction of angular velocity represents the rotational motion taking place in the plane perpendicular to

the axis. In a similar way we can associate a direction with angular acceleration also. If the angular velocity is

increasing then the direction of the angular acceleration coincides with that of the angular velocity as shown in figure

7.7 (a) and if the angular velocity is decreasing then the direction of the angular acceleration is opposite to that of

angular velocity, as shown in figure 7.7(b).

fig. 7.7(b): Angular acceleration has

the opposite direction as that of angularvelocity. Angular velocity is decreasing.

ω !

ωα

α !

fig. 7.7(a): Angular acceleration has

the same direction as that of angularvelocity. Angular velocity is increasing.

ω !

ωα

α !

Now, we can say that the angular velocity and the angular acceleration are vector quantities. As their directions

coincide with the axis of rotation, these quantities are also called axial vectors.



PHYSICS 9LOCUS

Angular velocity and angular acceleration follow the triangle law of vector addition and their additions are commutative

also but the addition of angular displacements are noncommutative, hence angular displacement is not a vector

quantity. Noncommutative nature of the angular displacement is shown in figure 7.8.

π2

A x i s 2

A x i s 1

(a): Original orientation of book. It is then rotated

by /2 radians (rad) about Axis 1.π(b): Orientation after a rotation /2 rad about

Axis 1.

π

π2

A x i s 2

A x i s 1

(c): Orientation after a subsequent rotation of

/2 rad after Axis 2.π(d): Original orientation of book.

π2

A x i s 2

A x i s 1

(e): Orientation after a rotation of /2 rad

about Axis 2.

π

F r o n

t

π

2

(f): Orientation after subsequent rotation of /2 rad

about Axis 1.

π

fig. 7.8: The orientation in the sixth figure is not the same as in the third. Obviously the commutative law of addition is not satisfied by these rotations. Despite the fact that they have a magnitude and a direction, finiterotations cannot be represented as vectors


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PHYSICS 11LOCUS

1. The axis of rotation of a purely rotating body

(a) must pass through the centre of mass

(b) may pass through the centre of mass(c) must pass through a particle of the body

2. A wheel rotating about a fixed axis has an angular position given by 33.0 2.0 ,t θ = − where θ is measured in

radians and t in seconds. What is the angular acceleration of the wheel at t = 2.0 s?

(a) –1.0 rad/s² (b) –24 rad/s²

(c) –2.0 rad/s² (d) –4.0 rad/s²

(e) –3.0 rad/s².

3. Find a formula to convert radians per second into revolutions per minute.

4. A wheel starts from rest and has a constant angular acceleration of 2 rad/s2.

(a) What is its angular velocity after 5 s? (b) Through what angle has the wheel turned after 5 s?

(c) How many revolutions has it made in 5 s?

(d) After 5s, what is the speed and acceleration of a point 0.3 m from the axis of rotation ?

5. A turntable rotating at1

333

rev/min is shut off. It brakes with constant angular acceleration and comes to rest

in 2 min.

(a) Find the angular acceleration. (b) What is the average angular velocity of the turntable ?

(c) How many revolutions does it make before stopping ?

6. A particle moves in a circle of radius 100 m with constant speed of 20 m/s.

(a) What is its angular velocity in radians per second about the center of the circle?

(b) How many revolutions does it make in 30s ?7. A disk of radius 10 cm rotates about its axis from rest with constant angular acceleration of 10 rad/s2.

At t = 5s what are

(a) the angular velocity of the disk and

(b) the tangential and centripetal acceleration at anda

cof a point on the edge of the disk ?

8. A horizontal disk with a radius of 10 cm rotates about a vertical axis through its center. The disk starts from

rest at t = 0 and has a constant angular acceleration of 2.1 rad/s². At what value of t will the radial and

tangential components of the line acceleration of a point on the rim fo the disk be equal in magnitude?

(a) 0.55 s (b) 0.63 s

(c) 0.69 s (d) 0.59 s

(e) 0.47 s.

9. The ball shown in the figure will loop-the-loop if it starts from a point

high enough on the incline. When the ball is at point A, the centripetal

force on it is best represented by which of the following vectors?

(a) 1 (b) 2

(c) 3 (d) 4

(d) 4

(e) 5

10. The graphs below show angular velocity as a function of time. In which one is the magnitude of the angular

acceleration constantly decreasing?

t

ω

t

ω

t

ω

t

ω

t

ω

(a) (b) (c) (d) (e)



PHYSICS 12LOCUS

ROTATIONAL EFFECT OF A FORCE:

Before introducing any new physical quantity, let us analyze the following case:

An uniform rod of mass m and length l is placed on a smooth horizontal surface with one of its ends pivoted at point

O about which the rod can rotate freely. Now, if we apply a force F at the other end of the rod then what would

happen? In our daily life we come across various similar cases. We know that the rod would not acceleratetranslationally along F because it is pivoted at O but the rod may rotate. Hence, alongwith producing a translational

acceleration a force can produce rotational (or angular) acceleration as well. The effect of force causing angular

acceleration in a body is defined as torque of that force.

In figure 7.10 (a), F is applied along the length of the rod. In this case F does not produce rotation hence torque of

F is zero.

NO ROTATION, i.e., = 0α

F

O

fig. 7.10(a)

In figure 7.10 (b), F is applied perpendicular to the length of the rod. In this case F produces rotation in the rod,

hence, torque of F is nonzero.

BODY STARTS ROTATING,i.e., it has some angular acceleration

F

O

α = 0

fig. 7.10(b)

In Figure 7.10 (c), F is applied perpendicular to the length

of the rod but it is applied at the mid-point of the rod not at

the free end. F produces angular acceleration in this case

too but angular acceleration produced in this case is exactly

half of that produced in the case (b). Hence, torque of F

in this case is half of that in case (b).

BODY STARTS ROTATING,but with angular accelerationsmaller than that in case (b).

F O

α = 0

fig. 7.10(c)

In figure 7.10 (d), F is applied at the pivoted end of the rod. In this case no rotation is produced, hence, torque of

F is zero.

BODY DOES NOT ROTATE,

i.e., = 0α

FO

fig. 7.10(d)

F

O

α

r

θ

F F ⊥ = sinθ

F F | | = cosθ

α F sin ;θ α r ∝ ∝fig. 7.10(e)

Summary of all four cases is given in figure 7.10(e).F is applied at a distance r from the pivot and at an angle θ with

the length of the rod. It is obvious that only perpendicular component of the force, ( sin )F F θ ⊥ = , causes rotation

and rotation caused (i.e., angular acceleration α ) is directly proportional to the magnitude of .F ⊥ Again, it is also

learnt that α is also directly proportional to r . Hence, torque of F is directly proportional to r and .F ⊥ Therefore,

torque of F , denoted by ,Γ can be defined as



PHYSICS 13LOCUS

⊥⋅Γ = r F ...(7.16 a)

⋅= sinr F θ ...(7.16 b)

Equation 7.16 (b) can be rearranged to obtain,

( sin )r F θ Γ = ⋅

⊥ ⋅= r F ...(7.16 c)

Hence, torque of a force can be calculated using either of two ways:

Multiply perpendicular component of the force with the distance of

point of application of force from O or multiply force with the

perpendicular distance of O from the line of action of force.

Hence, a force can not produce a torque about a point if its

line of action passes through that point (i.e., r ⊥ = 0 ).

O

F α

θ

θ

fig. 7.11: = = sinΓ θ

r rF ⊥

⊥

.F

r is called as moment arm.

s i n

r

r ⊥ =

θ

Torque is a rotational analogue of force. If you apply a net force on a body, body gets an acceleration along the

direction of the net force and the magnitude of acceleration is directly proportional to the magnitude of the net force.

Similarly, when we apply a net torque on a body, body gets an angular acceleration along the direction of the net

torque and the magnitude of the angular acceleration is directly proportional to the magnitude of the net torque.

Figure 7.10 (e) is redrawn in figure 7.12. When F "

is applied, as shown in figure, the rod rotates in anticlockwise

direction in the given plane. As the torque and angular acceleration

should have same directions, the torque is also in the anticlockwise

direction. Using right hand rule it can be more appropriately saidthat torque is on the axis of rotation and is ponting out of the plane

of rotation, denoted by # symbol (A direction perpendicular to

the plane of paper and pointing inwards is denoted by ⊗ symbol).

Here direction of torque coincides with the direction of ,r F ×""

hence, in the vector form, torque can be defined as,

Γ = ×" ""

r F ...(7.17)

θO

Γ α

r

F

α Γ

fig. 7.12: =Γ r ×F !

! !

!

Hence, torque of a force,"

F, with respect to some point O is defined as the cross product of the position

vector of the point of application of "

F with respect to O and the force ⋅"

F

Unit of torque is N-m and it has the dimensions 2 2 ML T

− . You may not that it has the same dimensions as that of

work. Torque is also called the moment of force.

Some more illustrations are shown in figure 7.13 to decide the direction of torque:



PHYSICS 14LOCUS

Γ

T

Γ

T

Top view of above case.

!

!

r

T

Γ

T

Γ !

fig.7.13(a)

r

!

fig.7.13(b)

Top view of above case.

fig.7.13: Torque on a ring about its centre due to a force in the plane of the ring and acting tangential

A thin uniform rod of mass m and length l is supported by a fixed support about which it can rotate freely, as shown

in figure 7.14 (a). Find the torque on the rod about point O when the rod is horizontal.

fig. 7.14(a)

O m l ,

Solution:There are two forces acting on the rod:

(1) the weight of the rod, mg, in the vertically downward direction;

(2) the reaction force from the hinge, R. About the point O, the reaction force from the hinge does not produce

any torque because its line of action passes through that point. Only weight of the rod produces torque

about O.

O l /2

mg

R

About , torque of reaction force from the hinge, ,must be zero only weight of the

rod can providea torque in this case

O R

fig. 7.14(b): fig. 7.14(d): Perpendicularly inwardtorque implies clockwise rotation inthe plane of rotation. It can be proved

using right hand rule too.

O

mg

l /2

!

!

!

Γ

r O

F

fig. 7.14(c): Γ (= × ) is

perpendicular to the plane of thepage of and is pointing inwards.

r F !

! !

The torque of the weight of the rod about O is pointing perpendicularly inside the page, as suggested in figure 7.14

(c). Perpendicularly inside direction of torque means it tends to produce rotation in the clockwise sense in the given

plane, as shown in figure 7.14(d). When you will turn the fingers of your right hand along the sense of rotation, thumb



PHYSICS 15LOCUS

would point along the direction of the torque. But, at this juncture, I must tell you that in simple cases you need not

to first find the direction of torque by finding the direction of r F ×""

and then decide the sense of rotation. You can

know it directly by just observing the direction of force and its point of application only. For an example, in figure

7.14 (d) it is quite obvious that mg would try to rotate the rod about the point O in the clockwise direction only.

Now let us find the magnitude of the torque of the weight of the rod. If it isΓ

, then,

perpendicular distance of weight of the rod

from the line of action of weight

O

Γ = ×

2

lmg= × [ ]from figure 7.14 (b)

2

lmg=

From the previous discussion it is obvious that the net torque on the rod about O is equal to the torque due to its

weight only.

Three tangential forces, each of magnitude F , are

applied on a disc of radius R. Find the net torque

on the disc due to these three forces about the

centre of the disc O.

F

F

F

ODISC;Radius = R

fig. 7.15(a)

Solution: A close look of the situation would let you know that

each force tends to rotate the disc in the same direction (clockwise

direction), as shown in figure 7.15(b).

F

O

(i)

F

O

(ii)F

O

(iii)

fig. 7.15(b)

If you use r F ×""

to decide the direction of the torque of the each force, then, also you would get the same result.

Each of the three forces provide torque perpendicularly inward, i.e., they try to rotate the disc in the clockwise

direction.

Magnitude of the torque due to each force is F . R (force × perpendicular distance of O from the line of action of

force), hence, net torque about O due to all the forces is 3FR.



PHYSICS 16LOCUS

ROTATIONAL INERTIA:

If you want a body to remain static then not only the net force on it should be zero but the net torque on it (about any

stationary point) should also be zero. Hence, a necessary condition for a body not to rotate is that the resultant

torque about any point be zero. Although this condition is necessary, it is not a sufficient condition for a body not

to rotate. If a body is set rotating about an axis and there is no external torque acting, the body will continue torotate with constant angular velocity. (This is the same as the case of linear motion. A particle can be at rest and

remain at rest only if the resultant force is zero, but this is not a sufficient condition for being at rest. If the resultant

force is zero, the particle would be moving with constant velocity). Hence, a body maintains its rotational state

unless an external unbalanced torque is applied on it.

The physical quantity measuring the tendency of a body to maintain its rotational state is called as rotational inertia

or moment of inertia of the body.

Consider the situation shown in figure 7.16. A force iF "

is acting on

the ith particle of mass im of a wheel which is pivoted about an

axis through its centre and perpendicular to the plane of wheel.The component ir F parallel to the radius vector of the element, ,ir

"

has no effect on the rotation of the body. The tangential component

,it F which is perpendicular to ,ir "

does affect the rotation of the

body. The torque produced by the force iF "

about the centre of the

wheel is given by

i it iF r Γ =

F i

r i

mi

F ir

F it

fig. 7.16Using Newton’s second law, the tangential acceleration of the particle is given by

it i it i iF m a m r α = = [ ]it ia r α =∵

where α is the angular acceleration of the particle which is same as that of the whole body.

Combining the two expressions above, we get,

2i i im r α Γ =

If we now sum over all the particles in the body, we obtain

2ΣΓ = Σi i i m r α

( )2= Σ i i m r α [ is same for all elements]α ∵ ...(7.18)

The quantity iΣΓ is the resultant torque acting on the body, which we denote as .net Γ The sum 2i im r Σ is the

property of the wheel and is defined as the moment of inertia I.

Therefore, 2= Σ i i I m r ...(7.19 a)

The moment of inertia depends upon the distribution of mass relative to the axis of rotation. For the same

body moment of inertial can be deferent for different axes. Moment of inertia has the dimensions 2 ML and is

usually expressed in 2-kg m .

For a particle of mass m at a distance r from the axis of rotation, moment of inertia is

2= I mr ...(7.19 b)

For a body that is not composed of discrete point masses but is instead a continuous distribution of mass, the



PHYSICS 17LOCUS

summation in 2i i I m r = Σ becomes an integration. We imagine the

body to be subdivided into infinitesimal elements, each of mass

dm, as shown in figure 7.17. Let r be the distance from such an

element to the axis of rotation. Then the moment of inertia is

obtained from

∫ 2= × I r dm ...(7.19 c)

dmr

axis

fig. 7.17: Moment of inertia of the elementis = ². Moment of inertia of thewhole body = sum of moments of inertia of

all elements =

dI dm r

dI.

•

∫

where the integral is taken over the whole body. The procedure by

which the summation Σ of a discrete distribution is replaced by

the integral for∫ a continuous distribution is the same as that

discussed for the centre of mass.

For bodies of irregular shape the integrals may be difficult to evaluate. For bodies of simple geometrical shape the

integrals are relatively easy when the axis of symmetry is chosen as the axis of rotation.

CALCULATION OF MOMENT OF INERTIA:(a) A small ball of mass m at a distance r from the axis:

axis

Rotationabout theaxis

fig. 7.18

I mr = 2mr

axis

m

(b) Two small balls, each of mass m, with perpendicular bisector of the line joining the two balls as the axis of

rotation:

2

2 2

22

i i I m r

mr mr

mr

=Σ

= +

=

mr r m

axis

fig. 7.19 (a)

axis

mm

Rotationabout theaxis

fig. 7.19 (a)

(c) Three balls, each of mass m, at the vertices of an equilateral triangle of side length l with the line passing

perpendicularly through the centroid of the triangle as the axis of rotation:

axis

l

l

l

m

m

m

fig. 7.20 (a)

axis

r

Rotationabout theabout

r

r

l

l

l

m

m

m

fig. 7.20 (b)

2

2 2 2

2

2

2

3 3 sec3

24

34 3

i i i I I m r

mr mr mr

lmr m

lm

ml

θ

=Σ = Σ

= + +

= =

= •

=



PHYSICS 18LOCUS

(d) An uniform rod of mass mand length l with the line passing through its centre and perpendicular to its length

as the axis of rotation:

m, l

axis

m, l axis

Top view

x dx

dm

dI dm.x =2

I = dI axis

fig. 7.21(a) fig. 7.21(b) fig. 7.21(c)

Using equation 7.19, moment of inertia of the element is

2dI dm x= ⋅

2mdx x

l

= ⋅ ⋅

Therefore, moment of inertia of the rod is

I dI = ∫ 2 2

2 2

2 2

l l

l l

m m x dx x dx

l l

+ +

− −

= ⋅ = ⋅∫ ∫

23

23

22

3 3

ll

ll

m l ml

l l

++

−−

= =

3 3 3 22

3 8 8 3 8 12

m l l m l ml

l l

= − − = =

=

2

12

ml

(e) An uniform rod of length l and massm with the line passing through one of its ends and perpendicular to its

length as the axis of rotation:

axis

fig.7.22(a)

m, l axis m, l

top view

axis

dm

dx x

dI dm.x =2

I = dI fig.7.22(b) fig.7.22(c)



PHYSICS 19LOCUS

sum of moments of inertia of all elements about the same axis I =

32 2

0 03

llm m l

dI dm x x dxl l

= = ⋅ = ⋅ = =∫ ∫ ∫ 2

3

ml

(f) An uniform ring of mass mand radius r with the line passing through its centre and perpendicular to its plane


m r , , ring

m r , , ring dm

axis axis

axis(a)

top view(b)

fig 7.23

dI dm r =(c)

•2

sum of moments of inertia of all the elements of the ring about the chosen axis I =

2 2

dI dm r r dm= = ⋅ =∫ ∫ ∫ all elements are equidistantfrom the axis

∵

2= mr dm m = ∫ ∵

NOTE: The above result could also be established in the following way:

I mr =

axis

r m

ring

2

I mr =

axis

r m 2

fig. 7.24(c)

I mr =

axis

m

hollowcylinder

2

r

fig. 7.24(b)

fig. 7.24(a)

Consider the point mass m at a distance r from the axis of

rotation, as shown in figure 7.24 (a). In this case the moment

of inertia of the mass m about the chosen axis is2

mr Now,

if we redistribute this mass and make a ring of radius r

about the same axis, as shown in figure 7.24 (b), the entire

mass still remains at the same distance from the axis and

hence the moment of inertia is 2mr only. Now, if we stretch

the ring along the length of the axis and make a hollow

cylinder of radius r , as shown in figure 7.24 (c), the entire

mass is still at the same distance from the chosen axis and

hence the moment of inertia is still2

mr only.



PHYSICS 20LOCUS

Same approach could be use to find the moments of inertia of uniform rectangular plates using the results for uniform

rods, as shown in figure 7.25(a) and 7.25(b).

stretch

m a,

a

b

I =ma12

2

I =ma12

2

stretch

m a,

a

b

I =ma3

2

I =ma3

2

axis axis

fig. 7.25 (b)fig. 7.25 (a)

We can say that when the mass of a system is redistributed in such a way that each element shifts parallel to the axis

of rotation (so that its distance from the axis remains the same), the moment of inertia of the system remains

unaffected by the redistribution.

(g) An uniform disc of radius r and massm with the line passing through its centre and perpendicular to its plane


(c)

dI dm x = •

I dI =

2

m r , , ring

axis

(a)

axis

(b)

fig. 7.26

dx x

dm

Top view

The top view of the disc is shown in figure 7.26 (b). To calculate the moment of inertia of the disc about the

chosen axis, it is divided into a large number of elements. Each element being a ring with the same centre as

that of the disc. Such an element is shown in figure 7.26 (c). Mass of the element is dm, radius is x and

thickness is dx. The moment of inertia of the element is

2dI dm x= ⋅

2Using for an uniform ring I mr =

2

22

m x dx x

r π

π

= ⋅ ⋅ ⋅ mass per unit area

mass of the=

elementarea of the element

×

3

2

2m x dx

r = ⋅ ⋅

The moment of inertia of the disc,

sum of moments of inertia of all elements I =

dI = ∫



PHYSICS 21LOCUS

3

2

0

2r

m x dx

r = ⋅∫

4

2

2

4

m r

r

= ⋅

2

=2

mr

(h) A disc of uniform mass density having mass M and radius R with a concentric circular hole of radius r with

the line passing through the centre of the disc and perpendicular to the plane of the disc as the axis of

rotation:

(c)

dr

axis

(a)

top view

fig. 7.27

R1

R2

R1

R2

r

dm

(b)

To find the moment of inertia of the given body about its axis of symmetry we follow the same approach as

we did in the last case. Divide the body in several elemental rings, calculate the moment of inertia of eachring and then add the moments of inertia of all elemental rings to get the moment of inertia of the whole body.

Such an element, having massdm, is shown in figure 6.27(c), r being its radius and it has a thickness dr , then

its moment of inertia,

2dI dm r = ⋅

2

2 22 1

2 M

r dr r z R R

π π π

= × ⋅ ⋅ −

32 22 1

2 M r dr

R R= ⋅−

Therefore, moment of inertia of the given body is

I dI = ∫ 2

1

4 43 2 1

2 2 2 22 1 2 1

2 2

4

R

R

M M R Rr dr

R R R R

−= ⋅ = ×

− −∫

( )2 2

1 2

2 M R R=

NOTE: It could also be calculated using the method

,actual remaining removed I I I = +



PHYSICS 22LOCUS

which is exactly what we have done in centre of mass.

(i) A hollow sphere of mass M and radius R with the line coinciding with its diameter as the axis of rotation:

To find the moment of inertia of the given hollow sphere, let us divide it into several elemental rings having

the same axis as that of the hollow sphere, as shown in figure 2.28. Let the elemental ring shown in figure is

at an angular position θ from the reference line. If dm be the mass of the ring, then,

mass per unit area area of the ringdm = ×

( )2

24

M r Rd

Rπ θ

π = × ⋅

( )2

2 cos4

M R Rd

Rπ θ θ

π = × ⋅

fig. 7.28

dm

θ

R

r

Reference

line

axis

r R= cosθ

dθ

cos2

M d θ θ = ⋅

The moment of inertia of the elemental ring, about the chosen axis is

2dI dm r = ⋅

( )2

cos cos2

M d Rθ θ θ

= ⋅

2

3cos2

MRd θ θ = ⋅

Therefore, the moment of inertia of the given hollow sphere is

sum of moments of inertia of all elemental rings I =

dI = ∫ 22

3

2

cos2

MRd

π

π

θ θ

+

−

= ⋅∫

24

2 3

MR= × [Solve the integral on your own.]

22=

3 MR



PHYSICS 23LOCUS

PARALLEL AXIS THEOREM:

We can often simplify the calculation of moments of inertia for

various bodies by using general theorems relating the moment of

inertia about one axis of the body to that about another axis.

Steiner’s theorem or parallel axis theorem, relates the momentof inertia about an axis through the centre of mass of a body to that

about a second parallel axis. Let cm I be the moment of inertia

about an axis through the centre of mass of a body and I be that

about a parallel axis a distance h away. The parallel axis theorem

states that

2cm I I mh= + ...(7.20)

I I

z

y

x

(C.M.)O

cm

d

fig. 7.29(a)

where m is the mass of the body,

Proof: Consider the situation shown in figure 7.29 (a). A body of

mass M is shown with its centre of mass coinciding withthe origin of the reference frame. The moment of inertia of

an element of mass dm at the point ( x, y, z), as shown in

figure 7.29(b), about the axis coinciding with the z-axis

and hence through the centre of mass of the body is

I I

z

y

x

(C.M.)O

cm

d

fig. 7.29(b)

(x,y,z)

( )square of perpendiculardistance from the z-axiscmdI dm= ⋅

2 2( )dm x y= ⋅ +

Now, consider another axis parallel to the z-axis and meeting the x- yplane at the point (a,b) as shown in figure 7.29 (b). The moment

of inertia of the same element about this axis is

2 2( ) ( )dI dm x a y b = ⋅ − + −

Using change of refrencefrom method

y

I

d b

a z x

I cm

fig. 7.29(c): Top view of - plane. x y

2 2 2 2( ) ( ) 2 2dm x y dm a b a dm x b dm y= + + + − ⋅ ⋅ − ⋅ ⋅

Therefore, the moment of inertia of the whole body about this axis is

I dI = ∫ 2 2 2 2

( ) ( ) 2 2dm x y dm a b a dm x b dm y= + + + − ⋅ − ⋅ ⋅∫ ∫ ∫ ∫

( ) 2 2 2cmdI dm d a dm x b dm y= + ⋅ − ⋅ − ⋅∫ ∫ ∫ ∫ 2 2

2 2 2

( )

and

cmdm x y dI

a b d

+ =

+ =

∵

2 0 0cm I m d = + ⋅ − −

; ; 0;

0.

cm cm cm

cm

dI I dm m dm x m x

dm y m y

= = ⋅ = ⋅ = ⋅ = ⋅ =

∫ ∫ ∫ ∫

∵

⇒ 2= +cm I I md



PHYSICS 24LOCUS

Applications of parallel axis theorem: Some applications of parallel axis theorem are shown in figure 7.30.

2cm I I mh= +

I =ml 12

2

I cm

(a) A thin uniform rod:

l

22

12 2

ml lm= +

2 2

12 4

ml ml= +

2

3

ml=

2cm I I mh= +

I I

(b) A uniform ring:

cm

r

2 2mr mr = +

22mr =

2cm I I mh= +

I I

(c) a uniform disc:

cm

r

22

2

mr mr = +

23

2mr =

fig.7.30:Applications of parallel axis theorem



PHYSICS 25LOCUS

PERPENDICULAR AXIS THEOREM:

The plane figure theorem or perpendicular axis theorem relates

the moments of inertia about two perpendicular axes in a plane

figure to the moment of inertia about a third axis perpendicular to

the figure. If x, y, z, are perpendicular axes for a figure which liesin the x- y plane, the moment of inertia about the z-axis equals the

sum of the moments of inertia about the x and y axes.

Therefore, we have,

x

dm

x

y

O

r

y z

fig. 7.31= + z x y I I I ...(7.21)

Before applying this theorem you must make sure that the body is in x-y plane only and the third axis ( z-axis in this

case) must pass perpendicularly through the intersection points of x and y axes.

Proof: The moment of inertia of the chosen element in figure 7.31, about the x-axis is

2 xdI dm x= ⋅

The moment of inertia of the element about the y-axis is

2 ydI dm y= ⋅

The moment of inertia of the element about the z-axis is

2 zdI dm r = ⋅

2 2( )dm x y= ⋅ +

2 2dm x dm y= ⋅ + ⋅

x ydI dI = +

Therefore, the moment of inertia of the whole planar body about the z-axis is

z z I dI = ∫

x ydI dI = +∫ ∫ x y I I = +

⇒ + z x y I I I


http://slidepdf.com/reader/full/rotational-motion-part-1 26/26

PHYSICS 26LOCUS

Applications of perpendicular axis theorem: Some applications of perpendicular axis theorem are shown in

figure 7.32:

I I 0

I

I

I 0

I

0

0

2

/2

2

I I I

I I

mr

= +

⇒ =

=

fig. 7.32(a): Finding moment of inertia of a ring about its diameter

Top view

I

I 0

I

fig.7.32(b): Finding moment of inertia of a disc about its diameter.

I

I 0

I 0

0

2

2

2

2 2

4

I I I

I I

mr

mr

= +

⇒ =

=

=

Top view

1 I I I = 2+

2

12

ma+=

2

12

mb

m=

2

12

a b+2

I

I 0 I = ma12

2

I =ma12

2

1

2 I =ma12

2

2

I =

ma

12

2

1

0

fig.7.32(c): Finding moment of inertia of a rectangular plate about an axis passing through its axis andperpendicular to its plane.

0

Top view

NOTE: For a square plate, put a = b.

rotational motion part 1

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