# Rotational Dynamics Notes | Class 12

Rigid body

A body is said to be rigid if its molecular separations are constant even when it is in translational or rotational motion.

Moment of force:

Moment of force is defined as the product of force acting on an object and its perpendicular distance from the axis of rotation. It is a type of static force. In physics, moment of force is a measure of its tendency to cause a body to rotate about a specific point or axis.

From figure,

Moment of force = F × OA

[Where the axis of rotation passes through O.]

Or,        Moment of force = F × OB sin$\theta$       [$\because$ sin$\theta$ = $\frac{OA}{OB}$ $\therefore$ OA = OB sin$\theta$]

$\therefore$ Moment of force = F r sin$\theta$

Torque ($\tau$):

Torque is the turning effect of force. It is a movement type force. Torque is the measure of force that causes an object to rotate about an axis. Just like how force causes an object to accelerate in linear kinematics, torque is what causes an object to acquire angular acceleration. Mathematically, it can be defined as the product of force and its perpendicular distance from the axis of rotation. It is denoted $\tau$  by and given by

$\tau$ = F r sin$\theta$

In vector notation, above result can be expressed as

$\overrightarrow{\tau }$ = $\overrightarrow{r}$ × $\overrightarrow{F}$

Special Cases:

1. When $\theta$ = 0o,

Torque, $\tau$ = F r sin0o = 0 (Minimum)

1. When $\theta$ = 90o,

Torque, $\tau$ = F r sin90o = F.r (Maximum)

Parallel Forces:

Two forces are said to be parallel if their lines of action are parallel to each other.

Couple:

Two equal and unlike parallel forces acting on an object at different points is known as couple.

Moment of Inertia (I)

In linear motion mass of a body is the measure of inertia (i.e. the measure of difficulty to change of state of rest or uniform motion.) Similarly, in rotational motion the moment of inertia of a body is the measure of rotational inertia (i.e. the measure of difficulty to change the state of rest or uniform rotational motion.)

Mathematically, the moment of inertia of a rigid body about a particular axis is defined as the sum of products of all particles constituting the body and square of their perpendicular distance from the axis of rotation. It is denoted by ‘I’ and given by I = $\Sigma$mr2. Its unit is Kgm2 and the dimensional formula is [ML2T0].

Moment of inertia depends upon mass, radius and distribution of mass as well.

The radius of gyration of a rigid body rotation about a particular axis is defined as the perpendicular distance of its centre of mass from the axis of rotation.

Let us consider a rigid body consisting of a large number of particles of mass m1, m2, m3, …… mn at a distance of r1, r2, r3, ……….. rn  from the axis of rotation. From the definition of moment of inertia, we can write

I = m1r12 + m2r22 + m3r32 + …… + mnrn2

For m1 = m2 = m3 = …………. mn = m.

I = mr12 + mr22 + mr32 + …… mrn2 ………. (i)

Let C.M. be the centre of mass of the rigid body which is at a distance of R from the axis of rotation. This means ‘R’ is the radius of gyration of the given body. Let ‘M’ be the whole mass of the given body. Again, the moment of inertia can be written as

I = MR2 ………. (ii)

From (i) and (ii)

MR2 = m (r12+r22+r32)

Or, nmR2 = m(r12+r22+r32+….. +rn2)

$\therefore$ R = $\sqrt{\frac{{{r}_{1}}^{2}+{{r}_{2}}^{2}+{{r}_{3}}^{2}+…….+{{r}_{n}}^{2}}{n}}$,

This is the required expression for the radius of gyration.

Moment of inertia of a uniform rod with axis of rotation passing through its centre and perpendicular to its length.

Let us consider a uniform rod of length ‘l‘ and mass ‘m’ let the axis of rotation of the rod YOY’ passes through its centre and is perpendicular to its length. Let ‘dm’ be the small mass of a small section ‘dx’ that is at a distance of ‘x’ from the axis of rotation.

Since mass of ‘l‘ length of rod = m

Mass of unit length of rod = m/l

and, mass of dx length of rod = $\frac{m}{l}$ dx = dm (say)

Now, the moment of inertia of elementary mass ‘dm’ is given as

dI = dm x2

Or, dI = $\frac{m}{l}$.dx.x2

Or, dI = $\frac{m}{l}$.x2dx ………. (i)

Total moment of inertia of the whole rod can be obtained by integrating (i) from x = –l/2 to x = l/2, i.e.

$\int\limits_{-l/2}^{l/2}{{}}$dI    =   $\int\limits_{-l/2}^{l/2}{{}}$$\frac{m}{l}.x2.dx Or, I = \frac{m}{l}$$\int\limits_{-l/2}^{l/2}{{}}$x2dx

Or, I = $\frac{m}{l}$$\left[ \frac{{{x}^{3}}}{3} \right]$$\underset{-l/2}{\overset{l/2}{\mathop{\underset{{}}{\overset{{}}{\mathop{{}}}}\,}}}\,$

Or, I = $\frac{m}{3l}$$\left[ {{\left( \frac{l}{2} \right)}^{3}}-{{\left( \frac{-l}{2} \right)}^{3}} \right] Or, I = \frac{m}{3l}$$\left[ \frac{{{l}^{3}}+{{l}^{3}}}{8} \right]$

Or, I = $\frac{m}{3l}$. $\frac{2{{l}^{3}}}{8}$

$\therefore$ I = $\frac{\mathbf{m}{{l}^{\mathbf{2}}}}{\mathbf{12}}$

This is the required expression for moment of inertia of uniform rod with axis of rotation passing through its centre and perpendicular to the length.

Note

The moment of inertia of a uniform rod with its axis passing through one of its ends and perpendicular to its length is given by

I = $\frac{\mathbf{1}}{\mathbf{3}}\mathbf{m}{{l}^{\mathbf{2}}}$      [ Hint: Take limit: 0 to l ]

Moment of inertia of a uniform rod with axis of rotation passing through one of its ends and perpendicular to its length.

Let us consider a uniform rod of length ‘l‘ and mass ‘m’ let the axis of rotation of the rod YOY’ passes through one of its ends and is perpendicular to its length. Let ‘dm’ be the small mass of a small section ‘dx’ which is at a distance of ‘x’ from the axis of rotation.

Since mass of ‘l‘ length of rod = m

Mass of unit (1) length of rod = $\frac{m}{l}$

and, mass of dx length of rod = $\frac{m}{l}$ dx = dm (say)

Now, the moment of inertia of elementary mass ‘dm’ is given as

dI = dm x2

Or, dI = $\frac{m}{l}$.dx.x2

Or, dI = $\frac{m}{l}$.x2dx ………. (i)

Total moment of inertia of the whole rod can be obtained by integrating (i) from x = 0 to x = l, i.e.

$\int\limits_{0}^{l}{{}}$dI  = $\int\limits_{0}^{l}{{}}$$\frac{m}{l}.x2.dx Or, I = \frac{m}{l}$$\int\limits_{0}^{l}{{}}$x2dx

Or, I = $\frac{m}{l}$$\left[ \frac{{{x}^{3}}}{3} \right]$$\underset{0}{\overset{l}{\mathop{\underset{{}}{\overset{{}}{\mathop{{}}}}\,}}}\,$

Or, I = $\frac{m}{l}$$\left[ {{\left( \frac{{{l}^{3}}}{3} \right)}^{{}}}-\left( \frac{{{0}^{3}}}{3} \right) \right] Or, I = \frac{m}{l}$$\left[ \frac{{{l}^{3}}}{3} \right]$

Or, I = $\frac{m{{l}^{2}}}{3}$

$\therefore$ I = $\frac{\mathbf{1}}{\mathbf{3}}\mathbf{m}{{l}^{\mathbf{2}}}$

This is the required expression for moment of inertia of uniform rod with axis of rotation passing through one of its ends and perpendicular to the length.

Rotational kinetic Energy of a Rigid Body:

Let us consider a rigid body consisting of  a large number of particles of mass m1, m2, m3, …….. mn at a distance of r1, r2, r3, …… rn respectively from the axis of rotation. Let the body rotate with constant angular velocity ‘$\omega$’ about the given axis as shown in figure above.

Kinetic energy of particle- m1 = $\frac{1}{2}$m1v12

K.E. of particle- m1 = $\frac{1}{2}$m1 $\omega$12 r12 [$\because$ v = $\omega$r]

Similarly, K.E. of particle- m2 = $\frac{1}{2}$m2 $\omega$22 r22

K.E. of particle- mn = $\frac{1}{2}$mn $\omega$n 2 rn2

Now, total kinetic energy of rotating body can be written as

Rotational kinetic energy = $\frac{1}{2}$m1 $\omega$12 r12 + $\frac{1}{2}$m2 $\omega$22 r22 + $\frac{1}{2}$m3 $\omega$32 r32  + …..  $\frac{1}{2}$mn $\omega$n2 rn2

For a given rigid body, the angular velocities of all particles are the same. i.e.

$\omega$1 = $\omega$2 = $\omega$3 = …………… $\omega$n = $\omega$ (say)

Then above equation becomes,

Rotational K.E = $\frac{1}{2}$m1 $\omega$2 r12 + $\frac{1}{2}$m2 $\omega$2 r22 + $\frac{1}{2}$m3 $\omega$2 r32…. $\frac{1}{2}$mn $\omega$2 rn2

Or, rotational K.E. = $\frac{1}{2}$ (m1 r12+m2 r22+ …. +mn rn2) × $\omega$2

Or, rotational K.E. = $\frac{1}{2}$ ($\Sigma$mr2) $\omega$2

Or, rotational K.E. = $\frac{1}{2}$  I × $\omega$2           [$\because$I = $\Sigma$mr2]

$\therefore$ Rotational K.E. = $\frac{\mathbf{1}}{\mathbf{2}}$I $\omega$2

This is the required expression for rotational kinetic energy of a rotating rigid body.

Relation between torque ($\tau$) and momentum of inertia.

The torque of a rigid body rotating about a particular axis is defined as the product to force applied and the perpendicular distance of the object from the axis of rotation. It is denoted by ‘$\tau$’ and given as $\tau$ = F× r

It is the turning effect of force on an object it can also be defined as the rate of change of angular momentum, i.e. $\tau$  = $\frac{dL}{dt}$, Where L is angular momentum.

Let us consider a rigid body consisting of a large number of particles of masses m1, m2, m3… mn at a distance of r1, r2, r3, …… rn respectively from the axis of rotation. Let the body rotate with constant angular velocity ‘$\omega$’ about the given axis as shown in figure above.

Acceleration of particle- m1 = α1 r1

[$\because$ V = $\omega$r ⇒ $\frac{dv}{dt}$ =  $\frac{d\,(\omega \,r)}{dt}$ ⇒ a = $\alpha$.r]

Then, force on particle m1 = m1 $\alpha$r1                [$\because$ F = ma]

Or, torque on particle-m1 = (m1 $\alpha$r1) × r1        [$\because$ Torque, $\tau$ = F ×r]

Torque on particle-m1 = m1 $\alpha$r12

Similarly,

Torque on particle-m2 = m2 $\alpha$r22

Torque on particle-m3 = m3 $\alpha$r32

Torque on particle-mn = mn $\alpha$rn2

Now the total torque on the rigid body can be expressed as

$\tau$ = m1 $\alpha$1 r12 + m2 $\alpha$2 r22 + m3 $\alpha$3 r32 + ……. mn $\alpha$n rn2

For a given rigid body, the angular velocity of all particles are the same and hence the angular acceleration are also the same. i.e.

$\alpha$ 1 = $\alpha$ 2 = $\alpha$ 3 = …. $\alpha$ n = $\alpha$ (say)

Then above equation becomes,

$\tau$  = (m1 r12 + m2 r22 + m3 r32 + ……. mn rn2) $\alpha$

$\tau$ = $\Sigma$mr2 × $\alpha$

Since, moment of inertia, I = $\Sigma$mr2

$\therefore$    Torque, $\tau$  = I $\alpha$

This is the required expression for torque in terms of moment of inertia.

Angular momentum (L) and its relation with moment of inertia:

The angular momentum of a rigid body rotating about a particular axis is defined as the product of linear momentum and perpendicular distance from the axis of rotation. It is denoted by ‘L’ and given by

$\overrightarrow{L}$= $\overrightarrow{r}$ ×  $\overrightarrow{P}$

Taking magnitude only, L = P × r

Let us consider a rigid body consisting of a large number of particles of masses m1, m2, m3 …. mn at a distance of r1, r2, r3, ….. rn respectively from the axis of rotation. Let the body rotate with angular velocity ‘$\omega$’ about the given axis as shown in figure above.

From the definition, angular momentum of particle m1 is

L1 = P1 r1 = m1 v1 × r1

Or,  L1 = m1 × $\omega$1r1 × r1

Or,  L1 = m1 $\omega$1 r12

Similarly,

angular momentum of particle- m2, L2 = m2 r22 $\omega$2

angular momentum of particle- m3, L3 = m3 r32 $\omega$3

angular momentum of particle- mn, Ln = mn rn2 $\omega$n

Now, total angular momentum of the rigid body is

L = m1 r12 $\omega$1+ m2 r22 $\omega$2 + m3 r32 $\omega$3 + ….. mn rn2 $\omega$n

For a given rigid body, the angular velocity of all particles are the same. i.e.

$\omega$1= $\omega$2 = $\omega$3 …….  $\omega$n = $\omega$ (say)

L = (m1 r12 + m2 r22 + m3 r32 + …. mn rn2) $\omega$

= $\Sigma$mr2 × $\omega$

Since, moment of inertia, I = $\Sigma$mr2

$\therefore$ L = I$\omega$

This is the required expression for angular momentum of a rigid body.

Principle of conservation of Angular Momentum:

Principle of conservation of Angular Momentum states “In the absence of external force, the total angular momentum of an isolated rotating object remains constant.” i.e.

L = I$\omega$ = constant

Let us consider a rigid body of moment of inertia (I) rotates about a particular axis with angular velocity ‘$\omega$’. From the definition of torque.

Torque ($\tau$) = $\frac{dL}{dt}$………… (i)

Also, $\tau$ = F × r …………. (ii)

In the absence of external force, i.e. F = 0

$\tau$ = 0 × r = 0

From (i)

$\frac{dL}{dt}$ = 0       [ $\because $$\tau = 0] Or, dL = 0.dt Integrating above equation, we get, \int\limits_{}^{}{{}}dL = \int\limits_{}^{}{{}}0.dt Or, L = constant Or, L = I\omega = constant Or, I1\omega 1 = I2\omega 2 \therefore I1\omega 1 = I2\omega 2 = L = Constant This is conservation of angular momentum. Note 1. Since I\omega = constant I × \frac{2\pi }{T}= Constant [\because$$\omega$ = $\frac{2\pi }{T}$]

$\therefore$ I  ∝ T

2. Since I$\omega$ = constant

I × 2$\pi$f = constant         [$\because$$\omega$ = 2$\pi$f]

$\therefore$ f $\frac{\mathbf{1}}{\mathbf{I}}$

Short Questions:

1. If the ice on the polar caps of the earth melts, how will it affect the duration of the day? Explain.
2. Can you distinguish a raw egg and a hard boiled egg by spinning each one on the table? Explain.
3. If earth shrinks, how will the duration of a day be affected?
4. The cap of a bottle can be easily opened with the help of two fingers rather than with one finger. Why?
5. A ballet dancer stretches her hands when she wants to come to rest. Why?
6. A fan with blades takes a longer time to come to rest than without blades. Why?
7. A ballet dancer can increase her arms to reduce her motion. Explain
8. A dancer girl is rotating over a turntable with her arms outstretched. If she lowers her arms, how does this affect her motion?
9. Explain why spokes are fitted in the cycle wheel.
10. What is the counterpart of the mass and force in rotational motion?
11. If meteorites strike the earth, the earth will slow down slightly. Why?

Couple, Moment of Couple and Expression for work done by Couple.

Couple:

Two equal and opposite forces (unlike force) along a body at two different points which rotate or trend to rotate the object is called couple.

Moment of couple:

Moment of couple can be defined as the product of one of its force and perpendicular distance between them.

Moment of couple is denoted by ‘$\tau$’ and given by

$\tau$ = F×2r

Work done by couple:

Let us consider a couple (two equal and opposite forces) acting tangentially on a wheel at two opposite points A and B, which rotates the wheel through angle $\theta$.

Work done by the couple is

W = F × $\overset\frown{AA’}$ + F × $\overset\frown{BB’}$ ……… (i)

Since, $\theta$ = $\frac{l}{r}$= $\frac{\overset\frown{AA’}}{r}$= $\frac{\overset\frown{BB’}}{r}$

$\therefore$  $\overset\frown{AA’}$= $\theta$r           and          $\overset\frown{BB’}$ = $\theta$r

$\therefore$ W = F × $\theta$r + F × $\theta$r

$\Rightarrow$ W = 2Fr $\theta$ ……….. (ii)

Here, total torque can be written as

$\tau$ = F × r + F × r = 2Fr ……. (iii)

From (ii) and (iii)

W = $\tau$ $\theta$……….. (iv),

This is the required expression for work done by the couple.

Power:

Now, differentiation equation (iv) with respect to time.

$\frac{dW}{dt}$ = $\frac{d\,(\tau \,\theta )}{dt}$

Or, P = $\tau$.$\frac{d\theta }{dt}$= $\tau$ $\omega$

$\therefore$ P = $\tau$ $\omega$, which is the required expression for power of couple.

Total kinetic Energy of a Rolling Body:

Let us consider a spherical rigid body of mass m, moment of inertia I and radius ‘R’ is rolling with linear velocity ‘v’ and angular velocity ‘$\omega$’ on a horizontal track as shown in figure above. Let ‘K’ be the radius of gyration of the given rigid body. The total kinetic energy of the rolling body in the sum of translational K.E. and rotational K.E., i.e.

K.Etotal = K.Etans+ K.E.rot

K.Etotal = $\frac{\mathbf{1}}{\mathbf{2}}$mv2 + $\frac{\mathbf{1}}{\mathbf{2}}$I$\omega$2

In terms of radius of gyration, moment of inertia can be expressed as

I = mK2

$\therefore$ K.E.toal . $\frac{1}{2}$mV2 + $\frac{1}{2}$mK2 $\omega$2

Since, v = $\omega$R $\Rightarrow$ $\omega$ = $\frac{v}{R}$

Then, K.E.total = $\frac{1}{2}$mv2 + $\frac{1}{2}$mK2 $\frac{{{v}^{2}}}{{{R}^{2}}}$

Or,     K.E.total = $\frac{\mathbf{1}}{\mathbf{2}}$mv2 $\left( \mathbf{1}+\frac{{{\mathbf{K}}^{\mathbf{2}}}}{{{\mathbf{R}}^{\mathbf{2}}}} \right)$

This is the required expression for total kinetic energy of a rolling body.

Acceleration of a Rolling body along an Inclined plane:

Let us consider a spherical body of mass ‘m’ and radius ‘R’ that starts to roll down along an inclined plane. Let ‘h’ be the height and ‘s’ be the length of the inclined plane which makes an angle $\theta$ with the horizontal. If the object rolls down from rest its initial velocity, (u) is equal to zero. When the object rolls down, it loses its potential energy and gains K.E.

From the conservation of mechanical energy gain in K.E. = loss in P.E.

Or, $\frac{1}{2}$mv2 $\left( 1+\frac{{{K}^{2}}}{{{R}^{2}}} \right)$= mgh

Or, v2 = $\frac{2gh}{1+\frac{{{k}^{2}}}{{{R}^{2}}}}$

Or, v = $\sqrt{\frac{2g\,s\,\sin \theta }{1+\frac{{{K}^{2}}}{{{R}^{2}}}}}$………. (i)     [$\because$Sin$\theta$ = h/s → h = s sin$\theta$]

We know,

v2 = u2 + 2as

Or, $\frac{2g\,s\,\sin \theta }{1+\frac{{{K}^{2}}}{{{R}^{2}}}}$= 0 + 2as

Or, a = $\frac{g\,\,\sin \theta }{1+\frac{{{K}^{2}}}{{{R}^{2}}}}$……… (ii)

This is the required expression for acceleration of a rolling body is an inclined plane.

Time to reach the ground (t):

We have, v = u + at

From equations (i) and (ii),

Or,  $\sqrt{\frac{2g\,s\,\sin \theta }{1+\frac{{{K}^{2}}}{{{R}^{2}}}}}$= 0 + $\frac{g\,\,\sin \theta }{1+\frac{{{K}^{2}}}{{{R}^{2}}}}$.t

Squaring both sides

$\frac{2g\,s\,\sin \theta }{1+\frac{{{K}^{2}}}{{{R}^{2}}}}$= $\frac{{{g}^{2}}si{{n}^{2}}\theta }{{{\left( 1+\frac{{{k}^{2}}}{{{R}^{2}}} \right)}^{2}}}$.t2

Or, $\sqrt{\frac{2s(1+{{k}^{2}}/{{R}^{2}})}{gh/s}}$= ts

$\therefore$   t = $\sqrt{\frac{2{{s}^{2}}(1+{{k}^{2}}/{{R}^{2}})}{gh}}$

This is the required expression for time to reach the ground level for a body rolling down an inclined plane.