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 motion, translational or linear motion to distinguish it from the rotational motion.

Rotating rigid body

In the figure below, a rotating rigid body is shown. To study the rotational motion of the rigid body, we consider it as a collection of large number of small objects, called point objects. When the rigid body rotates, all the point objects rotate about an invisible line, called the axis of rotation of the body. So, the point objects are in circular motion on different circles. For a point object, the radius of the circle is the distance of it from the axis of rotation. In the figure, the circular motion of two point objects are shown. Point objects closer to the axis of rotation moves slower than those far away from it. For example, in a merry-go-round, a person sitting farther from the center (through which the axis of rotation passes) moves faster than the one closer to it.

Angular displacement

Definition of angular displacement

Consider a point object in 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, $\Delta \ell$ on the circle, then its angular displacement, $\Delta \theta$ is defined as,

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

If you take any point on the rigid body, all of them have the same angular displacement. So, $\Delta \theta$ is the angular displacement of the rigid body. Angular displacement is a vector as it has a direction.

The SI unit of angular displacement is radian (rad.). The common unit of angular displacement or angle is degree ($^\circ$). We can find a relation between the radian and degree as follows.

When a point object makes one complete rotation, its angular displacement is $\Delta \theta= 360^\circ$. And, the distance traveled by the point object in one complete rotation is equal to the circumference of the circle. That is, $\Delta \ell = 2\pi r$. Substituting these in $\Delta \theta=\Delta \ell/r$, we get

$360^\circ=\dfrac{2\pi r}{r}\:rad.$

Canceling out the $r$'s,

$360^\circ=2\pi\:\:rad.$

$\boxed{180^\circ=\pi\:\:rad.}$

Angular velocity

If $\Delta \theta$ is the angular displacement of an object in a time interval, $\Delta t$, then the angular velocity of the object is

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

We use the greek letter omega ($\omega$) for angular velocity.

Unit of angular velocity is $rad./s.$

In the figure below you can see the rotating direction of a disc and the direction of its angular velocity.

Angular velocity and linear velocity

Let us again consider a point of object of a right body. Assume in a small time interval, $\Delta t$, the point object travels a distance, $\Delta \ell$ on its circular path. So, the angular displacement of the point object is

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

Dividing by $\Delta t$,

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

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

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

$\boxed{v=r\omega}$

Note that the angular velocity of all the points of a rigid body is the same, but the linear velocity is different at different points as they are at different distance, $r$ from the axis of rotation.

Angular speed, period and frequency of rotation of a rigid body

The time for one full rotation of a rigid body is called the period, $T$ of rotation. In one full rotation, a point object of a rigid body travels a distance equal to the circumference of the circle. Therefore, the linear speed of the point object can be written as

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

Substituting, $v=r\omega$, we get,

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

Canceling the $r$'s,

$\omega=\dfrac{2\pi}{T}$

The reciprocal of period is the frequency of rotation, that is, $f=1/T$. Substituting this in the equation, we get,

$\boxed{\omega=2\pi f}$

So, from the frequency of rotation of a rigid body, you can find its angular speed.

Angular acceleration

When there is a change in angular velocity of an object, there is an angular acceleration. We use the letter, alpha ($\alpha$) for angular acceleration. The angular acceleration of a rigid body is

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

where $\Delta \omega$ is the change in angular velocity of the rigid body in a time interval, $\Delta t$.

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

Centripetal and tangential acceleration of a point object

  $a_c=\dfrac{v^2}{r}$

Substituting, $v=r\omega$, and simplifying, we get

$\boxed{a_c=r\omega^2}$

If the angular speed of the rigid body is not constant, then the speed (the linear speed) of the point objects are not constant. So, the objects will be in non-uniform circular motion. An object in non-uniform circular motion has another acceleration in addition to the centripetal acceleration. That acceleration is called the tangential acceleration. The tangential acceleration of an object is

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

Note that when $v$ is constant, $a_{tan}=0$ and there will only be the centripetal acceleration.

From the equation, $v=r\omega$, we can write, $\Delta v=r\Delta \omega$. Substituting this in the previous equation,

  $a_{tan}=r\dfrac{\Delta \omega}{\Delta t}$

Since, $\Delta \omega/\Delta t=\alpha$, we get,

  $\boxed{a_{tan}=r\alpha}$

The direction of tangential acceleration is tangent to the circle as shown in the figure above. Note that the tangential acceleration is same as the acceleration in a translation motion. So, we call this as linear acceleration.

Since there are two accelerations for the point object in non-uniform circular motion, the net acceleration, $\vec a_{net}$ of the object is the vector sum of the two accelerations:

$\vec a_{net} = \vec a_c+\vec a_{tan}$

If you add the two vectors, you will get the direction of the net acceleration, $\vec a_{net}$ as shown in the figure above.

And, you will get the magnitude of the net acceleration,

$a_{net}=\sqrt{a_c^2+a_{tan}^2}$

Rotational kinematics

$\Delta x=r \Delta \theta$; $v=r\omega$ ,and $a=a_{tan}=r\alpha$

After the substitution, you can cancel out the $r$ in the equations and you will get the following equations,

$\omega=\omega_0+\alpha t$

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

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

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

These equations are called the rotational kinematic equations.

In the equations, $\omega_0$ is the initial angular velocity of the rigid body, $\omega$ is the velocity at time, $t$, you can also call this the final angular velocity, and $\Delta \theta$ is the angular displacement at time $t$. Note that $t=0$ is the initial time as in the 1-D kinematic equations.

It is important to note that the rotational kinematic equations are valid when the angular acceleration, $\alpha$ is constant. This is because the 1-D kinematic equations are valid for constant linear acceleration.

Torque

To define torque, let us consider an object in the shape of a beam as shown in the figure below. Assume the object can rotate freely about the axis of rotation as in the figure. And, you are applying a force, $\vec F$ on the object to rotate it as shown.

The magnitude of torque, $\tau$ exerted on the object by the force is defined as

$\tau=rF\sin \theta$

Now, draw a perpendicular line from the pivot to the force vector (you can extend the force line if needed). The length of this line is $r_\perp$, which is called the lever arm or the moment arm.

From the right angled triangle in the figure above, we can write,

$r_\perp = r \sin\theta$

So, the torque can also be written as

$\tau=r_\perp F$

Torque is also called moment or moment of force. A net torque on an object rotates the object in rest or changes the angular velocity if the object is already in rotational motion.

Direction of torque

Sign convention for torques

Torque on a rotating point 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,

$\tau=mr^2\alpha$.

This is the equation for the torque on a point object.

Torque on a rotating rigid body

$\tau=m_1r_1^2\alpha+m_2r_2^2\alpha+...m_nr_n^2\alpha$

$\tau=(m_1r_1^2+m_2r_2^2+...m_nr_n^2)\alpha$

$\tau=I\alpha$

where $I = m_1r_1^2+m_2r_2^2+...m_nr_n^2$

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

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

The moment of inertia of objects of different shapes and axis of rotation are given in the table below. $M$ in the table is the mass of the object

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	$\dfrac{1}{2}M\left(R_1^2+R_2^2\right)$
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

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

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

Substituting $v=r\omega$, we have

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

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

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

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

With this equation, you can find the rotational kinetic energy of a rotating rigid body.

Kinetic energy of a rolling object

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

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

Work and power in rotational motion

Work in rotation

To derive the equation for work by torque, we consider a wheel with a rope around its rim. And assume, you are pulling the rope with a force of magnitude $F$, so that the wheel moves a distance $d$. Now the work done by the force is

$W=F\, d$.

We can relate the distance $d$ with the rotating angle, $\Delta \theta$ of the wheel by

$d=r\Delta \theta$

Substituting this in the above equation, you get

$W=F\, r \Delta \theta$

$F\, r$ is nothing but the torque on the wheel, so we can write

$W=\tau \Delta \theta$

Thus, if a torque on an object rotate the object, a work is done on the object by the torque.

Power in rotation

$P=\dfrac{W}{\Delta t}$

Substituting, $W=\tau \Delta \theta$,

$P=\tau \dfrac{\Delta \theta}{\Delta t}$

Since $\dfrac{\Delta \theta}{\Delta t}=\omega$, the power generated by torque is

$P=\tau \omega$

Angular momentum

$L=I\omega$

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

We have the relation between the torque and the angular acceleration of a rigid body,

$\tau=I\alpha$.

 $=I\dfrac{\Delta \omega}{\Delta t}$

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

i.e., the torque is just the rate of change of angular momentum.

When you compare the angular momentum with linear momentum, you see that force is the rate of change of linear momentum, and torque is the rate of change of angular momentum.

Conservation of angular momentum

Angular momentum is also conserved when there is no net torque acting on a rigid body as you see below.

We have,

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

Putting, $\tau=0$,

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

$\Delta L=0$

$L=$ constant

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

So, if $L_1$ and $L_2$ are the initial and final angular momenta of a rotating object or a system, then

$L_1=L_2$