# Physics Tutorial: Dynamics of Rotational Motion

In this Physics tutorial, you will learn:

• The relationship between force and torque in translational and rotational motion respectively
• The meaning of moment of inertia and how to calculate it in various cases
• How to write the Newton's Second Law in rotational motion
• The concept of angular momentum and its relevant equation
• Work, kinetic energy and power in rotational motion and how they are related between each other and their analogue translational quantities

## Introduction

In the previous tutorial "Kinematics of Rotational Motion", we explained that there is a close analogy between linear and rotational quantities both in concept and in formulae. Logically, this analogy must extend in the Dynamics-related quantities as well. Therefore, in this tutorial we will explain Dynamics of Rotational Motion relying on Dynamics of Linear Motion, obviously making the proper arrangements to fit the current topic. Hence, let's discuss all dynamic quantities of rotational motion one by one.

## Moment of Force as Analogue of Force in Translational World

In our Physics tutorial "Moment of Force. Conditions of Equilibrium", we explained that moment of force (otherwise known as "turning effect of force") occurs when an object or system tends to rotate around a fixed point, which we called "pivot" or "turning point". Also, we explained that in a system, there are two possible conditions of equilibrium: one for the translational (linear) motion (resultant force must be zero) and the other for rotational motion (the resultant moment of force must be zero). From here, it is easy to conclude that the analogue quantity of force in rotational motion must be the moment of force, M, which - like force in translational motion - is a quantity that determines the equilibrium of the system.

Moment of force is obtained by the cross product of distance from turning point, r and the perpendicular force to the axis of rotation, F.

Mathematically, we have

M = r × F

We have also stated that the unit of moment of force is Newton × metre, [N × m].

If the system is moving, we use the concept of torque τ instead of that of moment of force M, i.e.

τ = r × F

Moment of force was extensively discussed in tutorial "Moment of Force. Conditions of Equilibrium" and our Physics tutorial on "Torque", so there is no need to dwell too much on this concept here.

## Moment of Inertia

Moment of inertia, I in the rotational motion is the analogue of mass. It is a quantity expressing a body's tendency to resist angular acceleration, i.e. to change in the actual state of rotation. Moment of inertia is the sum of the products of the mass of each particle in the body with the square of its distance from the axis of rotation. Moment of inertia therefore is quantitative measure of the rotational inertia of a body.

Mathematically, we have:

I = m × r2

Obviously, the unit of moment of inertia is [kg × m2] as mass is measured in kg and radius in metre (radius here is raised at power two).

Look at the figure:

### Example 1

A 2 kg object rotates around a fixed axis at 3 m away from it as shown in the figure above. What is the moment of inertia of this object?

### Solution 1

Applying the equation of moment of inertia, we obtain:

I = m × r2
= 2 kg × (3 m)2
= 18 kg × m2

## Moment of Inertia in Different Systems of Rotational Motion

The main equation of moment of inertia I = m × r2 can take other forms based on the structure and shape of the rotating object involved. The formulae for all cases involved are found by integration. However, here it is not the case to deal so much with the method how these equations are obtained. We will simply provide the formulae for some common situations without losing time with the procedure.

### a. A long bar of length L rotating around its centre

Moment of inertia of a bar of mass m and length L when rotating around its centre as shown in the figure, is:

I = 1/12 m × L2

### b. A long bar of length L rotating around its end

Moment of inertia of a bar of mass m and length L when rotating around its end as shown in the figure, is:

I = 1/3 m × L2

### c. A solid cylinder or disc rotating around its symmetry axis

Moment of inertia of a cylinder or disc (a disc is a cylinder with a very short height compared to its base radius) of mass m and base radius R when rotating around its axis of symmetry, which passes along the height as shown in the figure, is:

I = 1/2 m × R2

### d. A solid cylinder or disc rotating around its central diameter

Moment of inertia of a cylinder or disc of mass m, height h and base radius R when rotating around its central diameter as shown in the figure, is:

I = 1/4 m × R2 + 1/12 m × h2

### e. A ring of radius R rotating around its symmetry axis

Moment of inertia of a ring of mass m and radius R when rotating around its axis of symmetry, which passes along the centre as shown in the figure, is:

I = m × R2

### f. A ring of radius R rotating around its diameter

Moment of inertia of a ring of mass m and radius R when rotating around its diameter as shown in the figure, is:

I = 1/2 m × R2

### g. A solid sphere of radius R rotating around its central diameter

Moment of inertia of a sphere of mass m and radius R when rotating around its central diameter as shown in the figure, is:

I = 2/5 m × R2

### h. A thin spherical shell of radius R rotating around its central diameter

Moment of inertia of a spherical shell of mass m and radius R when rotating around its central diameter as shown in the figure, is:

I = 2/3 m × R2

### Example 2

Calculate the moment of inertia for the following objects:

1. A 12 m long metal wire of cross section diameter equal to 1 mm rotating around its centre. The mass of metal wire is 2 kg. Take the metal bar as not bendable.
2. A 80 cm long wooden rod of diameter equal to 12 cm and mass equal to 400 g rotating around its central diameter.
3. A 600 g hollow sphere of radius equal to 20 cm rotating around its central diameter.

### Solution 2

a. Since the diameter of cross-sectional area is much smaller than the length of wire, we consider it as a long bar instead of a cylinder. Therefore, we neglect the wire's thickness and focus only on its length, i.e. we have L = 12 m and m = 2 kg. Therefore, applying the formula of moment of inertia for a long bar rotating around its centre

I = 1/12 m × L2

we obtain after substitutions,

I = 1/12 × 2 kg × (12 m)2
= 24 kg × m2

b. In this case, we can take the wooden rod as a cylinder since its length is not much bigger than its radius. We have R = d / 2 = 12 cm / 2 = 6 cm = 0.06 m, h = 80 cm = 0.8 m and m = 400 g = 0.4 kg. Therefore, applying the equation

I = 1/4 m × R2 + 1/12 m × h2

we obtain after substitutions,

I = 1/4 × 0.4 × 0.062 + 1/12 × 0.4 × 0.82
= 0.00036 kg × m2 + 0.02133 kg × m2
= 0.02169 kg × m2

c. Moment of Inertia of a hollow sphere is calculated by the equation

I = 2/3 m × R2

where m = 600 g = 0.6 kg and R = 20 cm = 0.2 m. Substituting these values in the above equation, we obtain:

I = 2/3 × 0.6 kg × (0.2 m)2
= 0.016 kg × m2

## Newton's Second Law for Rotational Motion

From the analogy used to study rotational quantities in terms of linear ones, it is easy to derive the Newton's Second Law for Rotational Motion. Thus, since moment of inertia I in rotational motion is analogue to mass m in linear one, and angular acceleration α is analogue to the linear acceleration a, and giving that the Newton's Second Law of Motion in Linear Dynamics is F = m × a, we can write the Newton's Second Law of Motion as

τ = I × α

where τ is the torque, which is analogue to force in linear dynamics.

### Example 3

A 12 N force pushes perpendicularly a 4 m long metal rod as shown in the figure.

If the mass of the rod is 1 kg, calculate

1. Angular acceleration if the rod starts rotating from rest around its centre without any friction.
2. The number of rotation the bar makes in 10 s.

### Solution 3

a. First, we calculate the torque τ. Thus, since the bar rotates around its centre, the rotating arm is d = L / 2 = 4 m / 2 = 2 m and

τ = d × F
= 2 m × 12 N
= 24 N × m

Now, let's calculate the rod's moment of inertia I. We use the formula of the long bar rotating around its centre (case 1). Thus, we have

I = 1/12 m × L2
= 1/12 × 1 × 42
= 1.25 kg × m2

Now, applying the Newton's Second law for rotational motion

τ = I × α

we calculate the angular acceleration α (given that 1 N = 1 kg × m / s2):

α = τ/l
= 24 kg × m2/s2/1.25 kg × m2
= 19.2 s - 2

b. First let's calculate the angular displacement φ. Given that the bar starts rotating from rest, we have ω0 = 0. Therefore, applying the kinematic equation

φ = ω0 × t + α × t2/2

we obtain after substitutions (t = 10 s):

φ = 0 × 10 + 19.2 × 102/2

Therefore, giving that 1 rot = 2π rad, we obtain for the number of rotations N:

N = φ/
= 152.9 rotations

## Angular Momentum

Angular momentum, L, is a vector quantity (more precisely, a pseudo-vector) that represents the product of a body's rotational inertia and rotational velocity (in radians/sec) about a particular axis. It is the equivalent of momentum in linear motion.

Mathematically, we can write

L = I × ω

The unit of angular momentum is [kg × m2 / s].

### Example 4

What is the angular momentum of Earth when revolving around itself? Take the following approximate values: MEarth = 6 × 1024 kg, REarth = 6400 km and TEarth = 86400 s.

### Solution 4

Earth can be considered as a solid sphere. Therefore, its moment if inertia is

I = 2/5 m × R2

We have, R = 6400 km = 6 400 000 m = 6.4 × 106 m. Also, m in the formula is replaced by M as Earth has a very large mass. Hence, we can write:

I = 2/5 M × R2
= 2/5 × 6 × 1024 × (6.4 × 106)2
= 1.64 × 1037 kg × m2

Now, let's calculate the angular velocity of Earth. Since for one revolution φ = 2π and t = T, we have

ω = φ/t
= /T
= 2 × 3.14 rad/86400 s

Hence, Earth's angular momentum is

L = I × ω
= 1.64 × 1037 kg × m2 × 7.27 × 10-5 rad/s
= 1.19 × 1033 kg × m2/s

## Work in Rotational Motion

In linear motion, Work if the dot product of Force applied on an object and the linear distance the object moves (see the Physics tutorial "Work. Energy. Types of Energy." Similarly, in the rotational world, work is done by a torque τ applied through some angle φ. Mathematically, we have

Wrot = τ × φ

We use torque here instead of moment of force, as the process is dynamic, i.e. there is no work if the object doesn't move.

## Kinetic Energy in Rotational Motion

Again, we use the similarity between translational and rotational motion to deduce the formula of rotational kinetic energy KErot. Thus, given that kinetic energy of translational motion is

KE = m × v2/2

where m is the mass of the moving object and v is its velocity, and since moment of inertia is analogue to mass and angular velocity is analogue to linear velocity, we obtain for kinetic energy of rotational motion:

KErot = I × ω2/2

## Power in Rotational Motion

The last quantity in which the analogy between translational and rotational quantities is valid, is power. Thus, given that power in translational motion is

Power = Work/time
= F × ∆x/t
= F × v

we obtain for power in rotational motion

Prot = Wrot/t
= τ × φ/t
= τ × ω

### Example 5

The end of a 60 cm and 300 g rod start rotating from rest rotates in the horizontal plane through a small electric motor around a vertical axis as shown in the figure.

10 s after its start of rotation, the rod gains an angular velocity of 5 rad/s.

#### Calculate:

1. Work done by the electric motor to rotate the rod
2. Angular momentum at the end of 10 s
3. Rotational kinetic energy of the rod at the end of 10 s
4. Average power delivered by the motor supposing that all energy produced by it, is used to do work.

### Solution 5

a. We must calculate torque and angle of rotation to find rotational work, as he Newton's Second Law of rotational motion is

τ = I × α

Moment of inertia I of a bar rotating around its end is

I = 1/3 m × L2

where m = 300 g = 0.3 kg and L = 60 cm = 0.6 m. Thus,

I = 1/3 × 0.3 kg × (0.6 m)2
= 0.036 kg × m2

Angular acceleration α is calculated by

α = ω - ω0/t

where ω = 5 rad/s, ω0 = 0 and t = 10 s. Thus,

= 0.5 s-2

Hence,

τ = I × α
= 0.036 kg × m2 × 0.5 s-2
= 0.018 kg × m2/s2
= 0.018 N × m

(Remember that 1 N = 1 kg × m/s2)

The angle φ wiped by the rod during its rotation is

φ = ω0 × t + α × t2/2
= 0 × 10 + 0.5 × 102/2

Hence, the rotational work done by the engine is

Wrot = τ × φ
= 0.018 N × m × 25 rad
= 0.45 J

b. Angular momentum at the end of 10 s is

L = I × ω

Substituting the values we found earlier, we obtain

L = 0.036 kg × m2 × 5 rad/s
= 0.18 kg × m2/s

c. Rotational kinetic energy of the rod at the end of 10 s is

KErot = I × ω2/2
= 0.036 kg × m2 × (5 rad/s)2/2
= 0.45 J

As you see, this value is equal to the work done by the motor to make the rod rotate.

d. Average power delivered by the motor supposing that all energy produced by it, is used to do work, is:

Prot = Wrot/t
= 0.45 J/10 s
= 0.045 W

The following table includes everything discussed above regarding the relationship between dynamics translational and rotational quantities and the relevant formulae:

## Summary

Moment of force in rest and especially torque when the system is in motion is the rotational equivalent of force. The equation of torque is

τ = r × F

Moment of inertia, I in the rotational motion is the analogue of mass. It is a quantity expressing a body's tendency to resist angular acceleration, i.e. to change in the actual state of rotation. The general equation of moment of inertia is

I = m × r2

The unit of moment of inertia is [kg × m2].

The above equation can take other forms based on the structure and shape of the rotating object involved. Thus,

a. Moment of inertia of a bar of mass m and length L when rotating around its centre is:

I = 1/12 m × L2

b. Moment of inertia of a bar of mass m and length L when rotating around its end is:

I = 1/3 m × L2

c. Moment of inertia of a cylinder or disc of mass m and base radius R when rotating around its axis of symmetry, which passes along the height, is:

I = 1/2 m × R2

d. Moment of inertia of a cylinder or disc of mass m, height h and base radius R when rotating around its central diameter is:

I = 1/4 m × R2 + 1/12 m × h2

e. Moment of inertia of a ring of mass m and radius R when rotating around its axis of symmetry, which passes along the centre, is:

I = m × R2

f. Moment of inertia of a ring of mass m and radius R when rotating around its diameter is:

I = 1/2 m × R2

g. Moment of inertia of a sphere of mass m and radius R when rotating around its central diameter is:

I = 2/5 m × R2

h. Moment of inertia of a spherical shell of mass m and radius R when rotating around its central diameter is:

I = 2/3 m × R2

Newton's Second Law of Motion in Linear Dynamics is F = m × a. Thus, giving the analogy between translational and rotational quantities, we can write the Newton's Second Law of Motion in rotational motion as

τ = I × α

where τ is the torque, which is analogue to force in linear dynamics and α is the angular acceleration.

Angular momentum, L, is a vector quantity (more precisely, a pseudo-vector) that represents the product of a body's rotational inertia and rotational velocity (in radians/sec) about a particular axis. It is the equivalent of momentum in linear motion.

The equation of angular momentum is

L = I × ω

and its unit is [kg × m2 / s].

In the rotational world, work, W is done by a torque τ applied through some angle φ. Mathematically, we have

Wrot = τ × φ

Since moment of inertia is analogue to mass and angular velocity is analogue to linear velocity, we obtain for kinetic energy of rotational motion:

KErot = I × ω2/2

Power in rotational motion is

Prot = τ × ω

because power in translational motion is P = F × v.

## Dynamics of Rotational Motion Revision Questions

1) What is the angular momentum of a 3 kg cylinder rotating at 2 rad/s around its central axis of symmetry as shown in the figure? Take the radius of cylinder R = 40 cm and its height h = 2 m.

1. 0.24 kg × m2 / s
2. 0.48 kg × m2 / s
3. 0.96 kg × m2 / s
4. 1.12 kg × m2 / s

2) What is the mass of a 2 m bar rotating around its centre as shown in the figure if its rotational kinetic energy is 8 J and it rotates uniformly at 0.4 rad/s?

1. 75 kg
2. 100 kg
3. 300 kg
4. 1200 kg

3) A 1.6 m bar rotates at 0.6 rot/s around its centre when a 20 N force is used horizontally as shown in the figure.

What is the power delivered by the source?

1. 9.6 W
2. 16 W
3. 26.7 W
4. 32 W