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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 Î¸ of the particle is determined by an angle (defined relative to some arbitrary reference line), Figure 5.1a. In time Ît the particle changes its angular position by sweeping an angle ÎÎ¸, as shown in Figure 5.1b. So ÎÎ¸ is the change in angular position and is called the angular displacement.

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 of the particle to be
Ï=ÎÎ¸Ît
and the instantaneous angular velocity to be
Ï=limÎtâ0ÎÎ¸Ît
The unit of angular velocity is rad sâ1.
The angular velocity will increase or decrease in the presence of angular acceleration. So, we define the average angular acceleration to be
Î±Â¯=ÎÏÎt
and the instantaneous angular acceleration to be
Î±=limÎtâ0ÎÏÎt
The unit of angular acceleration is rad sâ2.
These definitions are analogous to definitions of linear quantities: we have a âtranslationâ dictionary between linear and angular quantities:
Linear quantity
Angular quantity
Position, s
Angle, Î¸
Linear velocity, v
Angular velocity, Ï
Acceleration, a
Angular acceleration, Î±
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
Îs=ut+12at2	
ÎÎ¸=Ï1t+12Î±t2
Îs=u+v2t	
ÎÎ¸=Ï1+Ï12t
v=u+at	
Ï=Ï1+Î±t
v2=u2+2aÎs	
Ï2=Ï21+2Î±ÎÎ¸
We already know from ChapterÂ 2 that the angular velocity Ï and the linear velocity v are related by (Figure 5.2)
v=Ïr
where r is the radius of the circle.
A similar relation holds between angular and linear accelerations:
a=Î±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, 180Â° = Ï rad.
So, 90Â°â¡Ï2Â rad,60Â°â¡Ï3Â rad,45Â°â¡Ï4Â radâ¦ and so on.
It is important to understand that the angular acceleration Î± 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:
âin a graph of angle Î¸ against time t, the gradient (slope) is the angular velocity Ï
âin a graph of angular velocity Ï against time t, the gradient (slope) is the angular acceleration and the area is the angle turned (the angular displacement), ÎÎ¸
âin a graph of angular acceleration Î± against time t, the area is the change in angular velocity ÎÏ.
WORKED EXAMPLE 5.1
The initial angular speed of a rotating disc is 24 rad sâ1. The disc suffers an angular deceleration of 3.0 rad sâ2.
aCalculate how many full revolutions the disc will make before stopping.
bDetermine when the disc will stop rotating.
Answer
aFrom Ï2=Ï2i+2Î±ÎÎ¸ we find
0=242+2Ã(â3.0)ÃÎÎ¸âÎÎ¸=2422Ã3.0=96 rad
This corresponds to
962Ï=15.3â15 revolutions.
bFrom Ï = Ïi + Î±t we find
0=24+(â3.0)t
t=8.0Â s
WORKED EXAMPLE 5.2
A wheel of radius 0.80 m is rotating about its axis with angular speed 2.5 rad sâ1. Find the linear speed of a point of the circumference of the wheel.
Answer
v=ÏR=2.5Ã0.80=2.0Â mÂ sâ1
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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 R2 from the axis. Calculate the ratios:
aÏxÏY
bvXvY.
Answer
aAll points on the disc have the same angular velocity so ÏxÏY=1.
bFrom v = Ïr we find vXvY=ÏXRÏYR2=ÏRÏR2=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 Î± = â 3.0 rad sâ2.
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 sâ1 and after 5.0 s the angular velocity increases to 15 rad sâ1. 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 sâ1. The angular acceleration is 2.5 rad sâ2. 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 sâ1. The angular velocity increases to 12.4 rad sâ1 after the body has made 20 full revolutions. Calculate the angular acceleration.
4A rotating disc has an initial angular speed of 4.0 rad sâ1 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 sâ2.
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.
