# Rotational motion

If an object rotates about a fixed axis, then we call that motion, rotational motion. In that sense, circular motion is also a rotational motion. But here, we specifically will focus on rotation in which the rotation axis is within the object. i.e., when considering the earth, its motion around the sun is circular motion, while the rotation about its axis is the rotational motion.

In kinematics, you studied 1 and 2-D motion of objects, where objects move from one place to another. Now, we will call that translational kinematics and the motion, translational motion to distinguish it from rotational kinematics and rotational motion. And we will call, the velocity, $v$, and the acceleration, $a$ as linear velocity and linear acceleration to distinguish it from their rotational counterparts.

### Rotating rigid body

Consider a rigid body. What is a rigid body?. A solid object, where the distance between any two points in the object is constant, is called a Rigid body. Any external force acting on a rigid body does not deform the object so that the distance between any two points remains the same. Although there is no perfect rigid body exists, we will consider all the objects as rigid body when studying its rotational motion.

Assume, the rigid body is rotating about a fixed axis. Let us take the rigid body as a collection of large number of small objects, called point objects. As the rigid body rotates, all the point objects rotate on circles of different radii. For a given point object, the radius is its distance from the rotation axis of the rigid body. Point objects closer to the axis of rotation moves slower than those far away from it. It is like a merry-go-round, a person sitting farther from the center (axis of rotation) moves faster than the one closer to it.

#### Angular displacement

Consider a point object on a rigid body that is at a distance $r$ from the axis of rotation of the rigid body. When the rigid body rotates, the point object move on a circle of radius $r$. If the point object moves a distance, $\Delta l$ on the circle, then its angular displacement is defined as,

$\Delta \theta=\dfrac{\Delta l}{r}$

If you take any points on the rigid body, you get the same value for $\Delta \theta$. So we can say, $\Delta \theta$ is the angular displacement of the rigid body.SI unit of angular displacement is radian (rad.).

When a point object completes one circle, angular displacement of the object is $360^\circ$. Angular displacement for a complete circle is

$\Delta \theta=\dfrac{2\pi r}{r}=2\pi\:rad.$

i.e., $2\pi\:rad.=360^\circ$.

(or)

$1\: rad. =\dfrac{360}{2\pi}\:degrees$.

This is the conversion factor between radian and degrees.

#### Angular velocity

Angular velocity tells us how fast an object is rotating. If an object's angular displacement during a time interval, $\Delta t$ is $\Delta \theta$, then the angular velocity of the object is

$\omega=\dfrac{\Delta {\theta}}{\Delta t}$

Angular velocity is a vector, we call its magnitude angular speed. Unit of angular velocity is $rad./s.$

The velocity, $v$, that you learned in the one and two dimensional motion, now we call linear velocity to distinguish it from the angular velocity. The angular and linear velocity are related as you can see now.

If a point object in a rigid body travels a distance, ${\Delta l}$ on its circular path during the time, ${\Delta t}$, then, the angular displacement

$\Delta \theta=\dfrac{\Delta l}{r}$

Dividing by $\Delta t$,

$\dfrac {\Delta \theta}{\Delta t}=\dfrac{1}{r} \dfrac{\Delta l}{\Delta t}$

Left hand side is the angular velocity of the point object (or the rigid body) and on the right hand side, $\Delta l/\Delta t$ is nothing but the linear velocity, $v$ of the point object. So we have,

$\omega =\dfrac{v}{r}$

(or)

$v=r\omega$

The angular velocity and linear velocity are related by this equation.

If $T$ is the period of rotation of a rigid body, i.e., the time for one complete rotation, then a point object in the rigid body at a distance, $r$ from the rotation axis is in circular motion with velocity,

$v=\dfrac{2\pi r}{T}$.

Substituting $v=r\omega$ and the frequency of rotation $f=1/T$, and canceling $r$, we get

$\omega=2\pi f$

The angular velocity is related to the frequency of rotation.

#### Direction of angular velocity

Angular velocity is a vector like linear velocity. Direction of angular velocity is determined by the right hand rule. When you curl your fingers of the right hand around the rotation axis and point in the direction of rotation, then the thump points in the direction of the angular velocity.

#### Angular acceleration

It is the rate of change of angular velocity. We use the letter $\alpha$ for angular acceleration. Angular acceleration is

$\alpha=\dfrac{\Delta \omega}{\Delta t}$.

SI unit of angular acceleration is $rad/s^2$.

#### Centripetal and tangential acceleration of a rotating point object

A point object in a rigid body is in circular motion as the body rotates. Therefore, there is a centripetal acceleration for the point object,

$a_c=\dfrac{v^2}{r}$

Substituting $v=r\omega$, we get

$a_c=r\omega^2$

If the circular motion is not uniform, then there will be an additional acceleration for the point object, called tangential acceleration.

$a_{tan}=\dfrac{\Delta v}{\Delta t}$

Substituting $v=r\omega$,

$a_{tan}=r\alpha$

Tangential acceleration is always tangent to the circle and it is actually the linear velocity $a$ in the translational kinematics.

#### Rotational kinematics

Rotational kinematics is about studying the motion of rotating objects without considering the cause of the motion. Similar to 1-D kinematics, we can obtain a set of four equations to describe rotational motion.

To obtain the rotational kinematic equations, we consider a point object in a rotating rigid body that is at a distance, $r$ from the axis of rotation. Let us take, during a time interval $t$, the rigid body makes an angular displacement, $\theta$ and the point moves a linear distance $\Delta x$ on the circle.

So, we have

$\Delta x=r \theta$; and also we have, $v=r\omega$ and $a=r\alpha$ as $a$ in translation motion is $a_{tan}$.

If we substitute these in the 1-D kinematic equation, we get

$\omega=\omega_0+\alpha t$

$\theta=\dfrac{1}{2}(\omega_o +\omega)t$

$\theta=\omega_0 t+\dfrac{1}{2}\alpha t^2$

$\omega ^2= \omega_0^2+2\alpha \theta$

Since the 1-D kinematic equations are valid for constant (linear) acceleration, rotational kinematic equations are valid when the angular acceleration $\alpha$ is constant.

#### Torque

You learned that to accelerate an object you need to apply a force. i.e, in a translational motion, force change the state of an object. Now, if an object is in rest and you want to rotate the object, it also requires a force. But the angular acceleration of a rotating object depends on three factors, (i) magnitude of the force, (iii) the direction of the force and (iii) the point of application of the force. A physical quantity that takes into account all these aspects is called, torque. So, in rotational motion torque plays the role of force. Torque is also called moment or moment of force.A net torque, rotate an object in rest or change its angular velocity if the object is already in rotation.

Magnitude of torque $(\tau)$ on an object by a force $\vec F$ is defined as

$\tau=rF\sin \theta$

where $r$ is the distance between the axis of rotation (also called pivot) and the point where the force is applied; and $\theta$ is the angle between the force and a line joining the pivot and the point of application of the force.
Torque can also be written as

$\tau=r_\perp F$

where $r_\perp$ is called lever arm or moment arm, is the perpendicular distance between the line of force and the axis of rotation.

#### Direction of torque

Torque is a vector. Its direction is determined by the right hand rule: stretch your fingers in the direction from the pivot to the point where the force is being applied, then swirl your fingers in the direction of the force, now the thumb points to the direction of the torque.

#### Sign convention for torque

When you apply a torque on an object it tends to rotate the object. A torque can rotate an object either in the clockwise or in the counter clockwise direction. If there are more than one torque acting, then the torques that tend to rotate the object in the clockwise direction is added together and so the counter clockwise direction is added together. We need a sign convention for torques. It is common to choose the following sign convention, if the force tends to rotate the object in the counter clockwise, torque is positive, and the torque is negative, if the force tends to rotate the object in the clockwise direction.

#### Torque on a rotating point object

Again, consider a point object in a rigid body. To rotate the object, you need to apply a force on the object tangent to the circle. Calculating the torque on the object,

$\tau=rF$.

where $r$ is the radius of the circle.

Since the force ccelerate the object tangentially, we have

$F=ma_{tan}=mr\alpha$

Substituting this in the torque equation,

$\tau=mr^2\alpha$.

#### Torque on a rotating rigid body

We have the torque acting on a rotating point object. By using that we can find the torque on a rotating rigid body. What we need to do is, assume the rigid body is made up of large number of point objects. Assume there are $n$ number of point objects. As the rigid body rotates, all point objects rotate with same angular acceleration, but rotate in different radius. Let $m_1, m_2, m_3, ..., m_n$ are the masses of the point objects and $r_1, r_2, r_3, ..., r_n$, as the radius of their circular path. If you look at the torques on all the point objects, they all point to the same direction, so we can simply add them to get the magnitude of the total torque on the rigid body. The torque on the rigid body is therefore,
$$\tau=m_1r_1^2\alpha+m_2r_2^2\alpha+...m_nr_n^2\alpha=(\sum_{i} m_ir_i^2)\alpha=I\alpha$$
where $$I = \sum_i m_ir_i^2=m_1r_1^2+m_2r_2^2+...m_nr_n^2$$

$I$ is called the moment of inertia of the rigid body about the rotating axis.

The moment of inertia of a rigid body depends on the mass, shape and the axis of rotation. With the use of calculus, you can find the moment of inertia of rigid bodies of different shapes. Following table, gives you the moment of inertia of objects of different shapes and axis of rotation.

Object Axis of rotation Moment of inertia
Uniform sphere Through center $\dfrac{2}{5}MR^2$
Solid cylinder Through center $\dfrac{1}{2}MR^2$
Hollow cylinder Through center $\dfrac{1}{2}M\big(R_1^2+R_2^2\big)$
Long uniform rod (or cylinder) Through center $\dfrac{1}{12}Ml^2$
Long uniform rod (or cylinder) Through one end $\dfrac{1}{3}Ml^2$
Thin hoop Through center $MR^2$
Thin hoop Through central diameter $\dfrac{1}{2}MR^2+\dfrac{1}{12}Mw^2$
Rectangular plate Through center $\dfrac{1}{12}M\big(l^2+w^2\big)$

#### Rotational kinetic energy

An object in rotation has kinetic energy, like an object in translational motion. To find the kinetic energy of a rotating object, we use the same strategy that we used to find the torque. Assume the rigid body is made up of large number of point objects. As the body rotates each point object rotates with same angular velocity but with different linear velocities because of different radius. Total kinetic energy of the body is the sum of the kinetic energies of the point objects.

$KE_{rot}=\dfrac{1}{2}m_1v_1^2+\dfrac{1}{2}m_2v_2^2+\dfrac{1}{2}m_3v_3^2+...+\dfrac{1}{2}m_nv_n^2$.

where $\omega$ is the angular velocity of the rigid body.

Substituting $v=r\omega$, we have

$KE_{rot}=\dfrac{1}{2}m_1r_1^2\omega^2+\dfrac{1}{2}m_2r_2^2\omega^2+\dfrac{1}{2}m_3r_3^2\omega^2+...+\dfrac{1}{2}m_nr_n^2\omega^2$.

(or)

$KE_{rot}=\dfrac{1}{2}(\sum_{i} m_ir_i^2)\omega^2$.

$KE_{rot}=\dfrac{1}{2}I\omega^2$

where $I$ is the moment of inertia of the object about the axis of rotation.

#### KE of a rolling object

If an object undergoes both rotational and translational motion on a surface, then we call that rolling motion. If you ride a bike, the wheels of the bike are in rolling motion as they rotate about their axle and also translate from one place to another. A rolling object has both translational and rotational kinetic energy. Further, if you roll an object, it will rotate about its center of mass. Therefore, total kinetic energy of a rolling object is

$KE=\dfrac{1}{2}mv^2+\dfrac{1}{2}I_{cm}\omega^2$

where $I_{cm}$ is the moment of inertia of the object about its center of mass.

#### Work in rotation

Torque does work on a rotating object in the same way a force does work on an object that makes linear displacement.

If a torque $\tau$ applied on an object rotate the object by an angle $\theta$, then the work done on the object by the torque is

$W=\tau \theta$

#### Power in rotation

Power is the rate of work by force. In rotation, we use torque instead of force, that is rate of work by torque.

Power generated by torque is

$P=\tau \omega$

#### Angular momentum

Angular momentum in rotational motion is analogous to the linear momentum in translational motion. Angular momentum $(L)$ of a rotating object is defined as,

$L=I\omega$

Unit of angular momentum is $kg.m^2/s$. Since the angular momentum is proportional to the angular velocity, its direction is same as the angular velocity.

#### Torque and angular momentum

Torque and angular acceleration are related by

$\tau=I\alpha$.

$=I\dfrac{\Delta \omega}{\Delta t}$.

$=\dfrac{\Delta (I\omega)}{\Delta t}$.

(or)
$\tau=\dfrac{\Delta L}{\Delta t}$

#### Conservation of angular momentum

When there is no net torque on an object, we have

$\tau=\dfrac{\Delta L}{\Delta t}=0$.

$\Delta L=0$

$L_1=L_2$

That is the angular momentum is constant.Therefore, when there is no net torque acting on a rotating object, the angular momentum of the object is conserved.