# Physics Tutorial: Kinematics of Rotational Motion

In this Physics tutorial, you will learn:

• The meaning of some quantities used as a background in the study of rotation, such as radius of curvature, period and frequency of rotation.
• Quantities involved in kinematics of uniform rotational motion, such as angular displacement and angular velocity.
• Quantities involved in kinematics of non-uniform rotational motion, such as angular acceleration and change in angular velocity.
• The relationship between linear and rotational quantities
• The meaning and formula of centripetal acceleration.
• Equations of all abovementioned quantities (there are more than one equation for each quantity)

## Introduction

Do you think the periperal points of a minute hand move at the same speed for a wall clock and a hand clock? Why?

Do you think the two abovementioned minute hands rotate at the same rate? Why?

In this tutorial you will understand why it is not enough to know only the moving velocity or speed for objects revolving around a fixed point. There are other paramaters to take into consideration if we want to make a thorough study of rotational motion. So, let's discuss them one by one.

## Understanding Some Useful Quantities of Rotational Motion as a Background

Before starting with Kinematics of Rotational Motion, it is better to discuss some quantities, somehow (albeit not always directly) involved in this kind of motion.

As explained in Physics tutorial "Motion. Types of Motion", a circular (rotational) motion involves the rotation of a particle about a fixed point in space by always keeping the same distance from this fixed point. Therefore, a rotational motion implies moving around a circle whose centre and radius are fixed as shown in the figure below.

Therefore, radius of curvature is a fundamental parameter that appears in most formulae of rotational motion, as it is an indicator of the object's position at any instant.

### b. Period of rotation

The time neccessary to make one complete revolution around a fixed point is called Period, T. It is measured by the unit of time, i.e. second, [s]. Period is a very important quantity as it also helps us limit the study of a rotational motion to a single rotation if we are sure that the parameters won't change during the entire process. It is just like studying a small group of people regarding a certain phenomenon and then generalize the outcome for a wider group or society.

For example, period of the second hand of a clock is T = 60 s for all kinds of clocks, from the smallest to the biggest, as it always takes 60 s for such a hand to make one complete rotation.

Another example: the period of Earth revolution around the Sun is about 365 days as this is the time needed for the Earth to make a complete revolution around the Sun.

Period is very suitable to use when a revolution process occurs slowly.

### c. Frequency of rotation

When an object rotates very fast around a fixed point, period results in a very small number. Therefore, to avoid the use of decimals, in such cases it is more suitable using the inverse of period, known as Frequency, f, to represent the time-related phenomena. Frequency is measured in revolutions per second but this unit is widely recognized as Hertz [Hz].

Thus, we have

f = 1/T

and

[1 Hz] = [1 1/s] = [1 s-1 ]

For example, if an object makes 10 revolutions per second around a fixed point, its period is

T = 1/f = 1/10 s-1 = 0.1 s

This result means it takes 0.1 s to this objet to make a complete revolution.

#### Example 1

The drum of a washing machine makes 1200 rpm (rotations per minute). Calculate:

1. Frequency of drum's rotation
2. Period of drum's rotation

#### Solution 1

a. To calculate the frequency, we have to convert rpm into rps (revolution per second). We have

f = 1200 rev/min = 1200 rev/60 s = 20 rev/s = 20 s-1 = 20 Hz

b. Period of drum's rotation therefore is

T = 1/f = 1/20 s-1 = 0.05 s

## Kinematics of Uniform Circular Motion

The simplest case of rotational motion is the uniform circular motion which represents objects moving at the same speed (not velocoty) around a fixed point by maintaining a constant distance from it. The reason why velocity is not constant although its magnitude doesn't change, is because the object's moving direction changes continously when it follows a circular path. Remember from Section 2 "Vectors and Scalars" that two vectors are equal only if they have both the same magnitude and direction, not just only the same magnitude or the same direction. Look at the figure below.

In this figure, velocites of the moving object in the two given instants are different despite having the same magnitude (velocity vectors have the same length). Such velocities of objects moving according peripheral points of a circle (i.e. according the circle's circumference) are known as "tangential velocity, v " as velocity vector lies according the circle's tangent at every instant.

### a. Angle of Rotation (Angular Displacement)

Since in circular motion we are more interested in how an object rotates rather than how it moves, the angle of rotation, φ,, is a very important parameter of such a motion. This angle is measured in radians (rad) and it is also known as "angular displacement". Thus, when an object makes a complete rotation, the angle φ is equal to 2π rad = 2 × 3.14 rad = 6.28 rad (we simply write 2π rad). When an object makes N rotations around a fixed point, the angle of rotation is

φ = N × 2π rad

As the angle of rotation represents the surface "wiped" by the radius during rotation, it is analogue to distance in linear motion. Here, we will denote the distance by L instead of the symbol s as used during the study of 1-D Kinemmatics. From geometry, it is known that the magnitude of central angle φ is equal to that of the corresponding arc L. Thus, since a complete arc represent the circumference of circle, where C = 2πr, we have:

φ/Compelete angle = L/Circumference
φ/ = L/2πr
φ = L/r

So,

L = r × φ

The arc length L represents the distance traveled by the object during its movement when the total angle is φ rad (the value of φ is not limited only in one rotation, i.e. it is not perodical; it depends on the number of rotation an object makes).

Look at the figure:

For example, if an object is moving around a circular path of radius r = 4 m and it makes 20 rotations (N = 20), the total distance travelled by the object is

L = r × φ
= r × (N × 2π)
= 4 × 20 × 2 × 3.14
= 502.4 m

### b. Angular Velocity

If we divide the angle of rotation by the time this process takes, we obtain another important quantity of rotational motion. It is known as the angular velocity, ω, is measured in radians per second, [rad/s]. Since radian is not a proper dimemsion in itself, we often write 1/s or s-1 for the unit of angular velocity. However, we will write rad/s here to not confuse this unit with that of frequency.

We have

ω = φ/t

This equation is very similar to the equation of uniform linear motion v = L / t where L is the distance traveled by an object during the time interval t, and v is the obect's speed.

In uniform circular motion, an object moves at the same (both linear and angular) speed. Therefore, we can limit the study of such a motion to a single rotation only. In this case, we use the period T instead of time t, Circumference C = 2πr instead of distance L and the angle of one rotation instead of the total angle φ. Look at the table below:

The last equation is obtained by dividing both sides of the previous equation by the time t. Thus,

L = r × φ
L/t = r × φ/t
L/t = r × φ/t
v = r × ω

#### Example 2

A girl has tied a stone at the edge of a 2 m long rope and she rotates it uniformly for 20 s. The stone makes in total 8 revolutions during this interval.

##### Calculate:
1. Period of rotation
2. Frequency of rotation
3. Angular displacement
4. The total distance travelled by the stone
5. Tangential speed of the stone
6. Angular velocity of the stone

#### Solution 2

a. Period is the time needed for the object to make one complete revolution. Thus, since the total time is t = 20 s and the number of revolutions is N = 8, we obtain for the period T,

T = t/N
= 20 s/8
= 2.5 s

b. Frequency represents the number or revolution in one second. It is the inverse of period. Thus,

f = 1/T
= 1/2.5 s
= 0.4 Hz

c. Angular displacement represents the total angle wiped by the rope during the entire motion. Thus, giving that the angle for one complete rotation is 2π rad (where π ≈ 3.14), for 8 rotations we obtain

φ = N × 2π rad

d. The total distance travelled by the stone is calculated by using the relationship between the linear and the corresponding circular quantities, i.e. by multiplying the circular quantities by radius to obtain the corresponding circular ones. Thus,

L = r × φ
= 2 × 16π
= 2 × 16 × 3.14 m
= 100.48 m

e. Tangential speed of stone can be calculated in many ways. One of them is by dividing the total distance by the total time. Thus,

v = L/t
= 100.48 m/20 s
= 5.024 m/s

f. Angular velocity of the stone is calculated in many ways as well. We can divide the angular displacement by time, divide the tangential speed by radius, multiplying frequency by 2π and so on. Let's use the first method for example. We have:

ω = φ/t

## Uniformly Accelerated (Decelerated) Rotational Motion

Again, we can use the analogy between the linear and circular motion to explain the uniformly accelerated (decelerated) motion, which in itself is part of non-uniform circular motion which we will not discuss here as it is very complicated for the level required in this webpage.

As examples of uniformly accelerated rotational motion we can mention objects that start rotating from rest. Their rotation accelerates until they reach a steady rate. Only then, the rotational motion can be considered as uniform. On the other hand, when a rotation is slowing down until it stops (wheels when a car is slowing down, drum of a washing machine when you turn the power off, etc), we have a uniformly decelerated rotational motion.

### a. Initial and final angular velocity

Just as in linear motion, we have an initial and final angular velocity, whose values are not equal. We can write ω0 for the initial and ω for the final angular velocity of a rotating object.

### b. Angular acceleration

It is obvious that the change in angular velocity mentioned above, takes a certain time to occur. Hence, we can speak here for a kind of acceleration, known as angular acceleration and is denoted by the symbol alpha (α). From the analogy with linear motion it is very easy to write its formula.

α = ∆ω/t = ω - ω0/t

The unit of angular acceleration obviously is [rad/s2] as angular velocity is measured in [rad/s] and the time in [s].

The abovementioned analogy can be extended for the other three equations of uniformly accelerated (deceleration) motion of both types. The following table includes all these formuleae.

(L is used here to represent the linear distance instead of s, in order to fit the topic discussed here).

#### Example 3

A platform starts rotating from rest and after 12 s it reaches an angular velocity of 24 rad/s. The platform radius is 4 m and a small object is attached at the edge of platform as shown in the figure.

##### Calculate:
1. Angular displacement of the object
2. Angular acceleration of the platform
3. Total distance travelled by the object

#### Solution 3

a. From the equation

φ = (ω + ω0) × t/2

and giving that ω0 = 0, ω = 24 rad/s and t = 12 s, we obtain for the angular displacement

φ = (0 + 24) × 12/2
= 24 × 12/2

b. Angular acceleration is calculated by the first equation or rotational accelerated motion written in terms of α, as

α = ω - ω0/t

Substituting the values, we obtain

α = 24 - 0/12

c. Distance L is calculated by the equation

L = φ × r

where r = 4 m is the radius of the platform. Therefore, we obtain

L = 144 × 4
= 576 m

## The Meaning of Centripetal Acceleration

We stated earlier that even when the rotational motion is uniform, the object has not the same velocity because of the change in direction, despite velocity may have the same magnitude. Therefore, we can assign an acceleration to this type of motion which is not as the traditional acceleration which takes place when an object sppeds up or slows down, but only because the velocity vector changes direction and thus, the difference between two such vectors in two different instants is not zero. As a result, when this difference (change in velocity) is divided by the time interval Δt it occurs, we obtain a non-zero acceleration, whose vector is directed towards the centre of circle. This is the reason why it is called "centripetal acceleration" (in symbols, aC) and not because the object itself point towards the centre. Look at the figure.

We can write:

aC = v2-v1/∆t

Obviously, the magnitude of centripetal acceleration does not depend only on the magnitude of velocity itself, but also on the radius of curvature. This is because greater the radius, larger the circle and smoother the curve. This means arcs resemble more and more to straight lines. When radius of circle is so long that arcs become straight, we cannot speak for centripetal acceleration anymore as it exists only in rotational motion. As an example in this regard, we can mention objects moving at short distances on Earth surface. We consider such paths as linear, despite the fact that the Earth is a sphere.

Therefore, we calculate the magnitude of centripetal acceleration by the formula

aC = v2/r

The unit of centripetal acceleration is [m/s2] although it is an acceleration that occurs only in rotational motion. Therefore, centripetal acceleration acts as a bridge between linear and rotational quantities.

### Example 4

What is the centripetal acceleration of a human at rest on Earth surface if he is at equator? Take radius of Earth R = 6371 km.

### Solution 4

Since the object is at rest, the only velocity we can assign to it, is the velocity of Earth rotation around itself. However, it is more convenient to use the concept of speed instead of velocity here, as we can apply the equation

v = 2πR/T

where T = 24 h = 24 × 60 × 60 = 86400 s and R = 6371 km = 6 371 000 m. Therefore, we have

v = 2 × 3.14 × 6371000 m/86400 s
= 463 m/s

Therefore, centripetal acceleration of the person at equator is

aC = v2/r = 4632/6 371 000 = 0.03 m/s2

Compared to gravitational acceleration (g = 9.81 m/s2) this centripetal acceleration is very small and thus, it is often negligible during the calculations. This is the reason why it is rarely taken into consideration during the soultion of exercises.

## Summary

A circular (rotational) motion involves the rotation of a particle about a fixed point in space by always keeping the same distance from this fixed point. Therefore, a rotational motion implies moving around a circle whose centre and radius are fixed.

The time neccessary to make one complete revolution around a fixed point is called Period, T. It is measured by the unit of time, i.e. second, [s].

When an object rotates very fast around a fixed point, period results in a very small number. Therefore, to avoid the use of decimals, in such cases it is more suitable using the inverse of period, known as Frequency, f, to represent the time-related phenomena. Frequency is measured in revolutions per second, but this unit is widely recognized as Hertz [Hz] instead. Thus, we have

f = 1/T

and

[1 Hz] = [1 1/s ] = [1 s-1 ]

The simplest case of rotational motion is the uniform circular motion which represents objects moving at the same speed (not velocoty) around a fixed point by maintaining a constant distance from it. The reason why velocity is not constant although its magnitude doesn't change, is because the object's moving direction changes continously when it follows a circular path.

Angle of rotation, φ, is a very important parameter of rotational motion. This angle is measured in radians (rad) and it is also known as "angular displacement". When an object makes N rotations around a fixed point, the angle of rotation is

φ = N × 2π rad

The arc length L represents the distance traveled by the object during its movement when the total angle is φ rad. Its relationship with the angular displacement is given by the formula

L = r × φ

If we divide the angle of rotation by the time this process takes, we obtain another important quantity of rotational motion. It is known as the angular velocity, ω is measured in radians per second, [rad/s]. Its formula is

ω = φ/t

The table below gives the relationship between quantities in kinematics of both linear and rotational uniform motion.

If the rotational motion is uniformly accelerated or decelerated, there is another quantity added. It is known as angular acceleration, α and is measured in [rad/s2].

Equations of uniformly accelerated or decelerated rotational motion and their relationship with those of uniformly accelerated or decelerated linear motion are shown in the table below.

In uniform rotational motion, the velocity is not the same everywhere despite its constant magnitude. This is because the velocity vector changes direction during the rotation. Therefore, we obtain a non-zero difference of velocity vectors which when divided by time gives a non-zero acceleration. Its vector always points towards the centre of curvature. That's why it is called centripetal acceleration, aC. Its formula is:

aC = v2 - v1/∆t

We calculate the magnitude of centripetal acceleration by the formula

aC = v2/r

The unit of centripetal acceleration is [m/s2] although it is an acceleration that occurs only in rotational motion. Therefore, centripetal acceleration acts as a bridge between linear and rotational quantities.

## Kinematics of Rotational Motion Revision Questions

1) The rotor of an engine having a radius of 21 cm rotates unformly at 300 rot/min. Calculate the distance travelled by rotor's peripheral points during 10 s. Take π = 22/7.

1. 6600 m
2. 66 m
3. 1.32 m
4. 0.66 m

2) The blade of a helicopter rotating at 720 rot/min stops moving in 15 s after it lands and the pilot turns the engine off. What is the angular displacement of any blade's point during this process?

3) A boy turns uniformly around his head a 1.6 m rope at whose edge he has attached a stone. The stone makes 24 rotations in a minute.

What is the value of centripetal acceleration of the stone during this process? Take π = 3.14.

1. 1.64 m/s2
2. 1.024 m/s2
3. 10.1 π m/s2
4. 10.1 m/s2