Rotational Motion Part 1

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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 body moves 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?) 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 ?

Rotational Motion

z r

P

y x O fig. 7.1: A body of arbitrary shape rotating about the z-axis.

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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 passing through the point P, 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 its C 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 in this frame. Thus, for the kinetics of this case, we need only consider the two-dimensional motion of a point in a circle.

y P r

θ

S

O

x

In the same figure the angle θ is the angular position of the particle P with 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 = s/r,

...(7.1)

fig. 7.4: The Point P is moving in a circle of radius r in a plane parallel to the x-y plane.

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

Rotational Motion

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At time t1 the angular position of P is θ1 and at a later time t2 its angular position is θ 2 , as shown in figure 7.5. The angular displacement of P is ∆θ = θ 2 − θ1 during the time interval ∆t = t2 − t1.

y P(t2) ∆θ

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

ω =

∆θ θ2 – θ1 = ∆t t 2 – t1

∆θ dθ = ∆t → 0 ∆ t dt

x

...(7.2)

The instantaneous angular speed ω is defined as the limit approached by this ratio as ∆t approaches zero: ω = lim

θ θ1 2

P(t1)

...(7.3)

fig. 7.5: The Point P has displaced through an angle ∆θ in time ∆t (= t2 – t 1).

For a rigid body all radial lines fixed in it perpendicular to the axis of rotation rotate through the same angle in the same 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 ω1 and ω2 be the instantaneous angular speeds at the time t1 and t2 respectively; then the average angular acceleration α of the particle P is defined as α =

∆ω ω2 − ω1 = ∆t t 2 + t1

...(7.4)

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

∆ω dω = ∆t dt

...(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, ωi , and initial angular position, θi , are given then integration can be used to find ω(t) and θ(t). We have, dω =α dt

⇒ ⇒

dω = α ⋅ dt ω

t

ωi

0

∫ dω = ∫ α ⋅ dt

[Let at some time t, angular speed is ω.]

t

⇒

ω − ωi = ∫ α ⋅ dt 0

Rotational Motion

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t

⇒

ω = ω i + ∫ α ⋅ dt

⇒

dθ = ωi + ∫ α ⋅ dt dt 0

⇒

t  dθ = ω i ⋅ dt +  ∫ α ⋅ dt  ⋅ dt 0 

...(7.6)

0

t

θ

⇒

t  θ = ω + α ⋅ d dt dt   ⋅ dt i ∫ ∫ ∫∫  θi 0 00 

⇒

t  θ − θi = ωi t + ∫  ∫ α ⋅ dt  ⋅ dt   00 

⇒

t  θ ( t ) = θ i + ω i t + ∫  ∫ α ⋅ dt  ⋅ dt  00

t

t

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

t

t

...(7.7)

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

and

ω( t ) = ω i + αt

[From equation (7.6)]

...(7.8)

1 θ ( t ) = θi + ωi t + αt 2 2

[From equation (7.7)]

...(7.9)

If α is constant and θi = 0, then, we have, θ ( t ) = ωi t +

1 + αt 2 2

...(7.10)

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

ω2 = ωi2 + 2 ⋅ α ⋅ θ

...(7.11)

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

θ  →S ω  →v α  →a

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RELATION BETWEEN ANGULAR VARIABLES AND LINEAR VARIABLES: y

From equation (7.1), we have

a v

s = θ⋅r ω

Therefore, speed of the point P, v=

⇒

ds dt

=

d (θ ⋅ r ) dt

=

dθ ⋅r dt

O

[∵ r is constant ]

v = ωr

α

P r

S

θ

x

fig. 7.6: Linear and angular variables are shown together.

...(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: at =

⇒

dv (= rate of change of speed) dt

=

d (ω r ) dt

=

dω ⋅r dt

at = α ⋅ r

[∵ r is constant ] ...(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 v2 = ω2 r an = 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 = ωr ⇒

ω=v r =

20 rad/s 100

= 0.2 rad/s Rotational Motion

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

ω = ⇒

∆θ ∆t

∆θ = ω ⋅ ∆t

 As ω is constant, ω = ω 

= ω ⋅ ∆t = (0.2 × 30) radians = 6 radians

Therefore, number of revolutions =

∆θ 2π

=

6 2π

=

3 revolutions. 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 ω = ωi + αt = (0 + 2 × 5) rad/s = 10 rad/s Using equation (7.10), we have, 1 θ = ωit + αt 2 2 1   =  0 + × 2 × 52  rad   2 = 25 rad θ Therefore, number of revolutions = 2π 25 revolutions = 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 = ωr = (10 × 0.3) m/s = 3 m/s At the same moment the radial acceleration of the point P is ar = ω 2 r = 30 m/s 2 and the tangential acceleration of the point P is at = α r = 0.6 m/s2 .

Rotational Motion

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A solid body rotates about a stationary axis so that its angular velocity depends on the rotation angle θ as ω = ω 0 − kθ , ω 0 and k 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, ⇒

ω = ω 0 − kθ dθ = ω 0 − kθ dt θ

⇒

dθ ∫ ω 0 − kθ = ∫ dt 0 0

⇒

−

⇒

ln(ω 0 − kθ ) − ln ω 0 = − kt

⇒

ln

⇒

ω 0 − kθ = e − kt ω0

⇒

kθ = ω 0 − ω 0e − kt

⇒

θ=

ω0 (1 – e – kt ) k

ω=

dθ dt

Now,

t

[∵

At t = 0, θ = 0]

1 θ t ln(ω 0 − kθ ) 0 = t 0 k

ω 0 − kθ = − kt ω0

(

)

ω  d  0 1 − e − kt  k  =  dt

ω 0  de − kt  = 0 −  k  dt  = ω 0 e − kt

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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 angular quantities, as shown in figure 7.6 (a) and (b) ! ω

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

fig. 7.6(b): In this case the direction of the angular velocity of the disc is pointing perpendicularly downwards from the plane of rotation of the disc.

! ω

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). ! α ! ω α

! ω ω

fig. 7.7(a): Angular acceleration has the same direction as that of angular velocity. Angular velocity is increasing.

α

ω

! α fig. 7.7(b): Angular acceleration has the opposite direction as that of angular velocity. Angular velocity is decreasing.

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. Rotational Motion

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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.

π 2

2 Axis

Axis 1

π 2

Axis 1

2 Axis

(a): Original orientation of book. It is then rotated by π/2 radians (rad) about Axis 1.

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

Axis 1

2 Axis

(c): Orientation after a subsequent rotation of π /2 rad after Axis 2.

(d): Original orientation of book.

π 2

t on r F

(e): Orientation after a rotation of π/2 rad about Axis 2.

(f): Orientation after subsequent rotation of π/2 rad about Axis 1.

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, finite rotations cannot be represented as vectors

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But, as the two angular displacements are made smaller, the difference between two sums disappear rapidly. If the angular displacements are made infinitesimal, the order of addition no longer affects the result. Hence infinitesimal " " " "  dθ  "  dω  angular displacements are vectors. Hence, dθ , ω  = and α  = are vector quantities.  dt   dt  " " " " " " Figure 7.9 shows the vectors r , v , at , ar , ω and α for a rotating particle P. The angular quantities are along the axis of rotation, point along the direction given by the right hand rule. We can prove that z ω or α v or aT

and where and

" " " v = ω× r " " " a = at + ar " " " at = α × r " " " ar = ω × v .

P

...(7.15 a)

aR

r y

...(7.15 b) ...(7.15 c). O

x

fig. 7.9

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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 θ = 3.0 − 2.0t 3 , 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. Find a formula to convert radians per second into revolutions per minute.

3. 4.

5.

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 ? 1 A turntable rotating at 33 rev/min is shut off. It brakes with constant angular acceleration and comes to rest 3 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 and ac of 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? ω

(a)

Rotational Motion

ω

t

(b)

ω

t

(c)

ω

t

(d)

ω

t

(e)

t

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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 accelerate translationally 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. F O

NO ROTATION, i.e., α = 0 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. α=0

O

F

BODY STARTS ROTATING, i.e., it has some angular acceleration fig. 7.10(b) α=0

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).

O

F BODY STARTS ROTATING, but with angular acceleration smaller than that in case (b). 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. F⊥ = F sinθ F O

α F BODY DOES NOT ROTATE, i.e., α = 0 fig. 7.10(d)

O

θ r

F| | = F cosθ

α ∝ F sinθ; 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, F⊥ (= F sin 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

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Γ = r ⋅ F⊥

...(7.16 a)

= r ⋅ Fsinθ

...(7.16 b)

Equation 7.16 (b) can be rearranged to obtain, Γ = ( r sin θ ) ⋅ F

= 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.

F

α O

θ

r⊥ =

rs inθ

θ

Hence, a force can not produce a torque about a point if its line of action passes through that point (i.e., r⊥ = 0 ).

fig. 7.11: Γ= r⊥.F = rF sin θ r⊥ is called as moment arm.

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 F Γ direction. Using right hand rule it can be more appropriately said α that torque is on the axis of rotation and is ponting out of the plane θ r O # 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 , α Γ hence, in the vector form, torque can be defined as, ! ! ! " " " fig. 7.12: Γ= r ×F ...(7.17) 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 ML2T −2 . 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:

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! Γ fig.7.13(a)

fig.7.13(b) T

! r

! r

! Γ

T T Γ

Γ

T Top view of above case.

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. O

m, l fig. 7.14(a)

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. R

!

O

O

l/2

!

r

Γ

!

F

mg fig. 7.14(b): About O, torque of reaction force from the hinge, R, must be zero only weight of the rod can providea torque in this case

O

! ! !

fig. 7.14(c): Γ(= r ×F ) is perpendicular to the plane of the page of and is pointing inwards.

l/2

mg fig. 7.14(d): Perpendicularly inward torque implies clockwise rotation in the plane of rotation. It can be proved using right hand rule too.

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 Rotational Motion

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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 is Γ , then,

Γ = weight of the rod × = mg × = mg

l 2

perpendicular distance of O from the line of action of weight

[from figure 7.14 (b)]

l 2

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.

F

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 DISC; Radius = R

O

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 fig. 7.15(a)

F

F

O

(i)

O O

(ii)

F

(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.

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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 to rotate 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. " Fir Fi Consider the situation shown in figure 7.16. A force Fi is acting on the ith particle of mass mi of a wheel which is pivoted about an mi axis through its centre and perpendicular to the plane of wheel. Fit " The component Fir parallel to the radius vector of the element, ri , ri has no effect on the rotation of the body. The tangential component " Fit , which is perpendicular to ri , does affect the rotation of the " body. The torque produced by the force Fi about the centre of the wheel is given by Γ i = Fit ri Using Newton’s second law, the tangential acceleration of the particle is given by

fig. 7.16

[∵ ait = riα ]

Fit = mi ait = mi riα

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,

Γ i = mi ri2α If we now sum over all the particles in the body, we obtain

ΣΓi = Σm i ri2α

(

)

= Σmi ri2 α

[∵α is same for all elements] ...(7.18)

The quantity ΣΓ i is the resultant torque acting on the body, which we denote as Γ net . The sum Σmi ri2 is the property of the wheel and is defined as the moment of inertia I. Therefore,

I = Σmi ri2

...(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 ML2 and is usually expressed in kg -m 2 . For a particle of mass m at a distance r from the axis of rotation, moment of inertia is

I = mr 2

...(7.19 b)

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

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summation in I = Σmi ri2 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 I = ∫ r 2 × dm ...(7.19 c)

r

where the integral is taken over the whole body. The procedure by which the summation Σ of a discrete distribution is replaced by

dm

axis fig. 7.17: Moment of inertia of the element is dI = dm•r ². Moment of inertia of the whole body = sum of moments of inertia of all elements = ∫dI.

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:

m

m

r

I = mr 2

Rotation about the axis axis

(b)

axis

fig. 7.18

Two small balls, each of mass m, with perpendicular bisector of the line joining the two balls as the axis of rotation:

m

r

r

m

m

m Rotation about the axis axis

axis fig. 7.19 (a)

(c)

I =Σ mi ri2 = mr 2 + mr 2 = 2mr 2

fig. 7.19 (a)

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: m

m

l

l

m

m l l

l

r

m axis fig. 7.20 (a) Rotational Motion

= mr 2 + mr 2 + mr 2

r

r

l Rotation about the m about

axis fig. 7.20 (b)

I = Σ I i = Σmi ri2 l  = 3mr 2 = 3m  sec3 θ 2  = 3m

l2 4 • 4 3

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An uniform rod of mass m and length l with the line passing through its centre and perpendicular to its length as the axis of rotation:

m, l

m, l

x

axis

dx dm

Top view axis

axis fig. 7.21(a)

fig. 7.21(b)

dI = dm.x2 I = dI

fig. 7.21(c)

Using equation 7.19, moment of inertia of the element is

dI = dm ⋅ x 2 m  =  ⋅ dx ⋅ x 2  l 

Therefore, moment of inertia of the rod is

I = ∫ dI +l 2

=

+l 2

m 2 m 2 ∫ l x ⋅ dx = l ∫ x ⋅ dx −l 2 −l 2 +l 2

+l 2 m l3  m =   = l 3  −l 2 l  3  − l 2 3l

(e)

=

m  l 3  l 3   m 2l 3 ml 2 =  − − = 3l  8  8   3l 8 12

=

ml 2 12

An uniform rod of length l and mass m with the line passing through one of its ends and perpendicular to its length as the axis of rotation:

m, l

axis

m, l

dm x

top view axis fig.7.22(a)

Rotational Motion

fig.7.22(b)

dx dI = dm.x

axis fig.7.22(c)

2

I = dI

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

m m l3 ml 2 = ∫ dI = ∫ dm ⋅ x = ∫ x 2 ⋅ dx = = 3 l l 30 0 l

2

(f)

An uniform ring of mass m and radius r with the line passing through its centre and perpendicular to its plane as the axis of rotation: dm

m, r, ring m, r, ring axis axis (a)

axis dI= dm•r 2 (c)

top view (b) fig 7.23

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

= ∫ dI

= ∫ dm ⋅ r 2

= r 2 ∫ dm

= mr 2

∵ all elements are equidistant   from the axis  ∵ dm = m   ∫ 

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

r

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 is mr 2 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 mr 2 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 still mr 2 only.

m

I = mr2

axis

m fig. 7.24(b)

r

I = mr 2 ring

axis

r

m

I = mr2 fig. 7.24(c)

hollow cylinder axis

Rotational Motion

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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). m,a

m,a

ma2 I = 12

ma2 I= 3

stretch stretch

ma I = 12

ma2 I= 3

2

b

b a

a 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 mass m with the line passing through its centre and perpendicular to its plane as the axis of rotation:

dx m, r, ring

dm x

dI = dm•x2 I = dI

axis

axis Top view (b)

(a)

(c)

fig. 7.26

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

dI = dm ⋅ x 2  m  =  2 ⋅ 2π x ⋅ dx ⋅ x 2  πr  =

 Using I = mr 2 for an uniform ring    mass per unit area   mass of the × =  element  area of the element  

2m 3 ⋅ x ⋅ dx r2

The moment of inertia of the disc,

I = sum of moments of inertia of all elements = ∫ dI Rotational Motion

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r

2m = 2 ∫ x 3 ⋅ dx r 0

(h)

=

2m r 4 ⋅ r2 4

=

mr 2 2

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: dm R1

R1

r dr

R2 axis top view (b)

(a)

R2 (c)

fig. 7.27

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 each ring 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 mass dm, is shown in figure 6.27(c), r being its radius and it has a thickness dr, then its moment of inertia,

dI = dm ⋅ r 2   M = 2 × 2π r ⋅ dr  ⋅ r 2 z 2  π R2 − π R1  =

2M r 3 ⋅ dr 2 − R1

R22

Therefore, moment of inertia of the given body is

I = ∫ dI 2M = 2 R2 − R12 =

R2

∫

R1

2M R24 − R14 r ⋅ dr = 2 × 4 R2 − R12 3

M ( R12 + R22 ) 2

NOTE: It could also be calculated using the method I actual = I remaining + I removed , Rotational Motion

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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, dm = mass per unit area × area of the ring

=

M × ( 2π r ⋅ Rdθ ) 4π R 2

dm

R

dθ

M = × ( 2π R cos θ ⋅ Rdθ ) 4π R 2 =

r = Rcosθ

r

θ

Reference line

M cos θ ⋅ dθ 2

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

dI = dm ⋅ r 2

fig. 7.28

M  2 =  cos θ ⋅ dθ  ( R cos θ )  2  =

MR 2 cos3 θ ⋅ dθ 2

Therefore, the moment of inertia of the given hollow sphere is I = sum of moments of inertia of all elemental rings

= ∫ dI MR 2 = 2

+π 2

∫

−π 2

MR 2 4 = × 2 3 =

Rotational Motion

cos3 θ ⋅ dθ

[Solve the integral on your own.]

2 MR 2 3

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Icm

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 moment of inertia about an axis through the centre of mass of a body to that about a second parallel axis. Let I cm 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

I = I cm + mh 2

I

z y

x

O (C.M.)

...(7.20) d

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 with the 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

(

square of perpendicular dI cm = dm ⋅ distance from the z-axis

fig. 7.29(a) Icm z

I (x,y,z)

y

)

x

O (C.M.)

= dm ⋅ ( x 2 + y 2 )

Now, consider another axis parallel to the z-axis and meeting the x-y plane at the point (a, b) as shown in figure 7.29 (b). The moment of inertia of the same element about this axis is

d fig. 7.29(b)

2

dI = dm ⋅  ( x − a) + ( y − b)  2

y

 Using change of refrence   from method 

I

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

d

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

z

I = ∫ dI

Icm

b x

a

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

= ∫ dm( x 2 + y 2 ) + ∫ dm(a 2 + b2 ) − 2a ∫ dm ⋅ x − 2b ⋅ ∫ dm ⋅ y = ∫ dI cm +

(∫ dm) ⋅ d

2

= I cm + m ⋅ d 2 − 0 − 0

⇒ Rotational Motion

− 2a ∫ dm ⋅ x − 2b∫ dm ⋅ y

∵ dm( x 2 + y 2 ) = dI cm    2 2 2  and a + b = d 

∵ dI cm = I cm ; dm = m; ∫  ∫   ∫ dm ⋅ y = m ⋅ ycm = 0.

∫ dm ⋅ x = m ⋅ xcm = 0;  

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

I = I cm + mh 2

(a) A thin uniform rod:

=

ml 2 l + m   2 12

=

ml 2 ml 2 + 12 4

=

ml 2 3

l ml 2 Icm= 12

I

2

(b) A uniform ring:

I = I cm + mh 2 = mr 2 + mr 2

r

= 2mr 2

Icm

I

(c) a uniform disc:

I = I cm + mh 2 =

mr 2 + mr 2 2

=

3 2 mr 2

r

Icm

I

fig.7.30: Applications of parallel axis theorem

Rotational Motion

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25

z

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 lies in 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.

y

x r

Iz = I x + I y

...(7.21)

y x

O

Therefore, we have,

dm

fig. 7.31

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

dI x = dm ⋅ x 2 The moment of inertia of the element about the y-axis is dI y = dm ⋅ y 2 The moment of inertia of the element about the z-axis is dI z = dm ⋅ r 2 = dm ⋅ ( x 2 + y 2 ) = dm ⋅ x 2 + dm ⋅ y 2 = dI x + dI y

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

I z = ∫ dI z = ∫ dI x + ∫ dI y = Ix + I y

⇒

Rotational Motion

Iz = I x + I y

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Applications of perpendicular axis theorem: Some applications of perpendicular axis theorem are shown in figure 7.32: I0 I

I0 = I + I

I

I0

⇒

I = I0 /2 = mr 2 2

I I Top view fig. 7.32(a): Finding moment of inertia of a ring about its diameter I I0

I0 = I + I

I

⇒

I0

I = I0 2 = mr 2 2 2

I I

=

mr 2 4

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

I0

ma2 I1= 12

ma2 I1= 12

I0 = I1+ I2 2

ma I2= 12

I0

2

Top view

ma I2= 12

2

ma 2 mb + = 12 12 2 a2+ b =m 12

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

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

Rotational Motion

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