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5.1Kinematics of rotational motion

Consider a particle that rotates around a circle about an axis through the centre. The angular position the angle defining the position of a particle relative to some arbitrary reference line; measured in rad θ of the particle is determined by an angle (defined relative to some arbitrary reference line), Figure 5.1a. In time Δt the particle changes its angular position by sweeping an angle Δθ, as shown in Figure 5.1b. So Δθ is the change in angular position and is called the angular displacement the change in angular position; measured in rad.

Figure 5.1: A particle that rotates about an axis that is perpendicular to the plane of the paper.

We define the average angular velocity the rate of change of the angular position; measured in rad s−1 of the particle to be

ω=Δ θ Δ t

and the instantaneous angular velocity to be

ω=lim Δ t→0 Δ θ Δ t

The unit of angular velocity is rad s−1.

The angular velocity will increase or decrease in the presence of angular acceleration. So, we define the average angular acceleration the rate of change of angular velocity; measured in rad s−2 to be

α¯=Δ ω Δ t

and the instantaneous angular acceleration to be

α=lim Δ t→0 Δ ω Δ t

The unit of angular acceleration is rad s−2.

These definitions are analogous to definitions of linear quantities: we have a ‘translation’ dictionary between linear and angular quantities:

Linear quantity | Angular quantity
Position, s | Angle, θ
Linear velocity, v | Angular velocity, ω
Acceleration, a | Angular acceleration, α

Because of this correspondence between linear and angular quantities, if a formula applies to the linear quantities, a similar formula will apply to the angular quantities. In other words, we have the following relations (the subscript i refers to initial quantities at t = 0):

Linear quantity | Angular quantity
Δ s=u t+1 2 a t 2 | Δ θ=ω 1 t+1 2 α t 2
Δ s=u+v 2 t | Δ θ=ω 1+ω 1 2 t
v=u+a t | ω=ω 1+α t
v 2=u 2+2 a Δ s | ω 2=ω 1 2+2 α Δ θ

We already know from Chapter 2 that the angular velocity ω and the linear velocity v are related by (Figure 5.2)

v=ω r

where r is the radius of the circle.

A similar relation holds between angular and linear accelerations:

a=α r

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Figure 5.2: Linear and angular velocities.

Angular velocity and angular acceleration are vector quantities, but we will not make use of their vector nature in this course.

When using the formulae, always express the angle in radians (rad). Remember, 180° = π rad.

So, 90°≡π 2 rad,60°≡π 3 rad,45°≡π 4 rad… and so on.

It is important to understand that the angular acceleration α is related to the linear acceleration a. This linear acceleration is the rate of change of speed and is not the centripetal acceleration of circular motion.

Similarly, whatever applies to graphs involving linear quantities also applies to graphs of angular quantities. For example:

●in a graph of angle θ against time t, the gradient (slope) is the angular velocity ω

●in a graph of angular velocity ω against time t, the gradient (slope) is the angular acceleration and the area is the angle turned (the angular displacement), Δθ

●in a graph of angular acceleration α against time t, the area is the change in angular velocity Δω.

Worked example 5.1

The initial angular speed of a rotating disc is 24 rad s−1. The disc suffers an angular deceleration of 3.0 rad s−2.

aCalculate how many full revolutions the disc will make before stopping.

bDetermine when the disc will stop rotating.

Answer

aFrom ω 2=ω i 2+2 α Δ θ we find

0=24 2+2×(−3.0)×Δ θ⇒Δ θ=24 2 2×3.0=96 rad

This corresponds to

96 2 π=15.3≈15 revolutions.

bFrom ω = ω i + α t we find

0=24+(−3.0)t

t=8.0 s

Worked example 5.2

A wheel of radius 0.80 m is rotating about its axis with angular speed 2.5 rad s−1. Find the linear speed of a point of the circumference of the wheel.

Answer

v=ω R=2.5×0.80=2.0 m s−1

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Worked example 5.3

A disc of radius R rotates about its axis with constant angular velocity (Figure 5.3).

Figure 5.3: For worked example 5.3.

A point X is at the circumference of the disc and another point Y is at a distance R 2 from the axis. Calculate the ratios:

a ω x ω Y

b v X v Y.

Answer

aAll points on the disc have the same angular velocity so ω x ω Y=1.

bFrom v = ω r we find v X v Y=ω X R ω Y R 2=ω R ω R 2=2.

Worked example 5.4

The graph in Figure 5.4 shows the variation of the angular velocity of a rotating body. Calculate:

athe angular acceleration

bthe total angle the body turned in 4.0 s.

Figure 5.4: For worked example 5.4.

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Answer

aThe angular acceleration is the gradient of the curve and so α = – 3.0 rad s–2.

bThe angle turned is the area under the curve and so equals 24 rad.

TEST YOUR UNDERSTANDING

1A disc has initial angular velocity 3.5 rad s–1 and after 5.0 s the angular velocity increases to 15 rad s–1. Determine the angle though which the disc has turned in the 5.0 s.

2A body rotates about an axis with an angular velocity of 5.0 rad s–1. The angular acceleration is 2.5 rad s–2. Calculate the angular velocity after the body turned through an angle of 54 rad.

3A body rotates with an initial angular velocity of 3.2 rad s–1. The angular velocity increases to 12.4 rad s–1 after the body has made 20 full revolutions. Calculate the angular acceleration.

4A rotating disc has an initial angular speed of 4.0 rad s–1 and rotates with constant angular acceleration. After 6.0 s the disc has rotated by an angle of 35 rad. What is the angular acceleration?

5A rod begins to rotate from rest with constant angular acceleration 1.5 rad s–2.

aWhat is the angular velocity after five complete revolutions?

bDraw graphs to show the variation with time of the angular velocity and the angular position.