The description of nonuniform circular motion of a particle is lot simplified, if the time rate of the change in angular velocity (angular acceleration) is constant. We have seen such simplification, while dealing with motion along a straight line. It is observed that the constant acceleration presents an additive frame work for motion along straight line. The velocity of the particle changes by a fixed value after every second (i.e. unit time). In translational motion along a straight line, we say, for example, that :
v
=
v
0
+
a
t
v
=
v
0
+
a
t
where “v”, “
v
0
v
0
” and “a” are attributes of motion along straight line with directional property. We can recall here that we are able to treat these vector quantities as signed scalars, because there are only two possible directions along a straight line. Now the moot question is : are these simplifications also possible with nonuniform circular motion, when angular acceleration is constant ?
The answer is yes. Let us see how does it work in the case of circular motion :
1: First, the multiplicity of directions associated with the velocity of the particle rotating about an axis is resolved to be equivalent to one directional angular velocity,"ω", acting along the axis of rotation. See the figure. If we stick to translational description of motion, then velocity vector is tangential to path. The direction of velocity vector changes as the particle moves along the circular path. On the other hand, angular velocity vector aligns in a single direction.
2: Second, the acceleration resulting from the change in the speed of the particle is represented by angular acceleration,"α". Like angular velocity, it is also one directional.
3: Third, there are only two possible directions as far as directional angular quantities are considered. The direction can be either clockwise (negative) or anticlockwise (positive). This situation is similar to linear consideration where motion in the reference direction is considered positive and motion in opposite direction is considered negative.
4: Fourth, the change in either angular displacement or angular speed is basically translated to a change in the magnitude of one directional angular quantities. This allows us to treat the change in angular quantities as a consideration along straight line as in the case of motion along straight line.
Clearly, it is possible to covert a motion along a circular path into an equivalent system along straight line. As the angular quantities are represented along a straight line, it is then logical that relations available for linear motion for constant acceleration are also available for circular motion with appropriate substitution of linear quantities by angular quantities.
For example, the equation "
v
=
v
0
+
a
t
v =
v
0
+ a t
" has its corresponding relation in the circular motion as :
ω
=
ω
0
+
α
t
ω
=
ω
0
+
α
t
As a matter of fact, there exists one to one correspondence between two types of equation sets. Importantly, we can treat angular vector quantities as signed scalars in the equations of motion, dispensing with the need to use vector notation. The similarity of situation suggests that we need not derive equations of motion again for the circular motion. We, therefore, proceed to simply write equation of angular motion with appropriate substitution.
In this section of circular motion kinematics, our interest or domain of study is usually limited to the motion or acceleration in tangential direction. We may not refer to the requirement of motion in the radial direction in the form of centripetal acceleration, unless stated specifically.
The correspondence goes like this : the angular position is “θ” for linear “x”; the angular displacement is “Δθ” for linear “Δx”; the angular velocity is “ω” for linear “v” and the angular acceleration is “α” for linear “a”.
The different angular quantities corresponding to their linear counterparts are listed here fore ready reference :
Quantities
Linear variables
Angular variables
Initial position
x
0
θ
0
Final position
x
θ
Displacement
Δ
x
Δ
θ
Initial velocity
v
0
ω
0
Final velocity
v
ω
Acceleration
a
α
Time interval
t
t
Quantities
Linear variables
Angular variables
Initial position
x
0
θ
0
Final position
x
θ
Displacement
Δ
x
Δ
θ
Initial velocity
v
0
ω
0
Final velocity
v
ω
Acceleration
a
α
Time interval
t
t
The corresponding equations for the two types of motion are :
S.N.
Linear equation
Angular equation
1:
v
=
v
0
+
a
t
ω
=
ω
0
+
α
t
2:
v
avg
=
(
v
0
+
v
)
2
θ
ω
avg
=
(
ω
0
+
ω
)
2
3:
Δ
x
=
x

x
0
=
v
0
t
+
1
2
a
t
2
Δ
θ
=
θ

θ
0
=
ω
0
t
+
1
2
α
t
2
S.N.
Linear equation
Angular equation
1:
v
=
v
0
+
a
t
ω
=
ω
0
+
α
t
2:
v
avg
=
(
v
0
+
v
)
2
θ
ω
avg
=
(
ω
0
+
ω
)
2
3:
Δ
x
=
x

x
0
=
v
0
t
+
1
2
a
t
2
Δ
θ
=
θ

θ
0
=
ω
0
t
+
1
2
α
t
2
The derived equations for the two types of motion are :
S.N.
Linear equation
Angular equation
1:
v
2
=
v
0
2
+
2
a
(
x

x
0
)
ω
2
=
ω
0
2
+
2
α
(
θ

θ
0
)
2:
Δ
x
=
(
x

x
0
)
=
(
v
0
+
v
)
t
2
Δ
θ
=
(
θ

θ
0
)
=
(
ω
0
+
ω
)
t
2
3:
Δ
x
=
x

x
0
=
v
t

1
2
a
t
2
Δ
θ
=
θ

θ
0
=
ω
t

1
2
α
t
2
S.N.
Linear equation
Angular equation
1:
v
2
=
v
0
2
+
2
a
(
x

x
0
)
ω
2
=
ω
0
2
+
2
α
(
θ

θ
0
)
2:
Δ
x
=
(
x

x
0
)
=
(
v
0
+
v
)
t
2
Δ
θ
=
(
θ

θ
0
)
=
(
ω
0
+
ω
)
t
2
3:
Δ
x
=
x

x
0
=
v
t

1
2
a
t
2
Δ
θ
=
θ

θ
0
=
ω
t

1
2
α
t
2
The sign of angular quantities represents direction. A positive sign indicates anticlockwise direction, whereas a negative sign indicates clockwise direction.
In the measurement of angle, a typical problem arises from the fact that circular motion may continue to rotate passing through the reference point again and again. The question arises, whether we keep adding angle or reset the measurement from the reference point ? The answer is that angle measurement is not reset in rotational kinematics. This means that we can have measurements like 540° and 20 rad etc.
This convention is not without reason. Equations of motion of circular motion with constant acceleration treats motion in an equivalent linear frame work, which considers only one reference position. If we reset the measurements, then equations of motion would not be valid.
Problem : The angular velocity – time plot of the circular motion is shown in the figure. (i) Determine the nature of angular velocity and acceleration at positions marked A, B, C, D and E. (ii) In which of the segments (AB, BC, CD and DE) of motion, the particle is decelerated and (iii) Is angular acceleration constant during the motion ?
Solution :
(i) Angular velocity :
The angular velocities at A and E are positive (anticlockwise). The angular velocities at B and D are each zero. The angular velocities at C is negative (clockwise).
Angular acceleration :
The angular acceleration is equal to the first differential of angular velocity with respect to time.
α
=
đ
ω
đ
t
α
=
đ
ω
đ
t
The sign of the angular acceleration is determined by the sign of the slope at different positions. The slopes at various points are as shown in the figure :
The angular accelerations at point A and B are negative (angular speed decreases with the passage of time). The angular accelerations at C is zero. The angular accelerations at point D and E are positive (angular speed increases with the passage of time).
(ii) Deceleration :
In the segments AB and CD, the magnitude of angular velocity i.e. angular speed decreases with the passage of time. Thus, circular motions in these two segments are decelerated. This is also confirmed by the fact that angular velocity and angular acceleration are in opposite directions in these segments.
(iii) The slopes on angular velocity  time plot are different at different points. Thus, angular accelerations are different at these points. Hence, angular acceleration of the motion is not constant.
The angular velocity increases by a constant value at the end of every unit time interval, when angular velocity and acceleration act in the same direction. On the other hand, angular velocity decreases, when angular velocity and acceleration act in the opposite direction.
To appreciate this, we consider a circular motion of a particle whose initial angular velocity is 0.3 rad/s. The motion is subjected to an angular acceleration of magnitude 0.1
rad
s
2
rad
s
2
in the opposite direction to the initial velocity. In the table here, we calculate angular velocity of the particle, using relation,
ω
=
ω
0
+
α
t
ω =
ω
0
+ α t
, at the end of every second and plot the data (for first 5 seconds) to understand the variation of angular velocity with time.

Time Angular acceleration Angular velocity
(s) (rad/s.s) (rad/s)

0 0.1 0.3
1 0.1 0.3  0.1 x 1 = 0.2
2 0.1 0.3  0.1 x 2 = 0.1
3 0.1 0.3  0.1 x 3 = 0.0
4 0.1 0.3  0.1 x 4 = 0.1
5 0.1 0.3  0.1 x 5 = 0.2

Here, the particle stops at the end of 3 seconds. The particle then reverses its direction (clockwise from anticlockwise) and continues to move around the axis. The angular velocity – time plot is as shown here :
We observe following aspects of the illustrated motion with constant acceleration :
 The angular velocity decreases at uniform rate and the angular velocity – time plot is straight line. Positive angular velocity becomes less positive and negative angular velocity becomes more negative.
 The slope of the plot is constant and negative.
Problem : A particle at the periphery of a disk at a radial distance 10 m from the axis of rotation, uniformly accelerates for a period of 5 seconds. The speed of the particle in the meantime increases from 5 m/s to 10 m/s. Find angular acceleration.
Solution : The initial and final angular velocities are :
ω
0
=
5
10
=
0.5
rad
/
s
ω
=
10
10
=
1
rad
/
s
t
=
5
s
ω
0
=
5
10
=
0.5
rad
/
s
ω
=
10
10
=
1
rad
/
s
t
=
5
s
Using equation of motion,
ω
=
ω
0
+
α
t
ω =
ω
0
+ α t
, we have :
⇒
1
=
0.5
+
α
X
5
⇒
α
=
0.5
5
=
0.1
rad
/
s
2
⇒
1
=
0.5
+
α
X
5
⇒
α
=
0.5
5
=
0.1
rad
/
s
2
The angular displacement is given as :
⇒
Δ
θ
=
θ

θ
0
=
ω
0
t
+
1
2
α
t
2
⇒
Δ
θ
=
θ

θ
0
=
ω
0
t
+
1
2
α
t
2
If we start observation of motion as t = 0 and θ° = 0, then displacement (θ) is :
⇒
θ
=
ω
0
t
+
1
2
α
t
2
⇒
θ
=
ω
0
t
+
1
2
α
t
2
The particle covers greater angular displacement for every successive time interval, when angular velocity and acceleration act in the same direction and the particle covers smaller angular displacement, when angular velocity and acceleration act in the opposite direction.
To appreciate this, we reconsider the earlier case of a circular motion of a particle whose initial angular velocity is 0.3 rad/s. The motion is subjected to an angular acceleration of magnitude 0.1
rad
/
s
2
rad /
s
2
in the opposite direction to the initial velocity. In the table here, we calculate angular displacement of the particle, using relation,
θ
=
ω
0
t
+
1
2
α
t
2
θ =
ω
0
t +
1
2
α
t
2
, at the end of every second and plot the data (for first 7 seconds) to understand the variation of angular displacement with time.

Time wo t Angular displacement (θ)
(s) (rad) (rad)

0 0.0 0.00
1 0.3 x 1 0.3 – 0.5 x 0.1 x 1 = 0.3 – 0.05 = 0.25
2 0.3 x 2 0.6 – 0.5 x 0.1 x 4 = 0.6 – 0.20 = 0.40
3 0.3 x 3 0.9 – 0.5 x 0.1 x 9 = 0.9 – 0.45 = 0.45
4 0.3 x 4 1.2 – 0.5 x 0.1 x 16 = 1.2 – 0.80 = 0.40
5 0.3 x 5 1.5 – 0.5 x 0.1 x 25 = 1.5 – 1.25 = 0.25
6 0.3 x 6 1.8 – 0.5 x 0.1 x 36 = 1.8 – 1.80 = 0.00
7 0.3 x 7 2.1 – 0.5 x 0.1 x 49 = 2.1 – 2.50 = 0.40

We see here that the particle moves in anticlockwise direction for first 3 seconds as determined earlier and then turns back (clockwise) retracing the path till it reaches the initial position. Subsequently, the particle continues moving in the clockwise direction.
The angular displacement – time plot is as shown here :
Problem : A particle at the periphery of a disk at a radial distance 10 m from the axis of rotation, uniformly accelerates for a period of 20 seconds. The speed of the particle in the meantime increases from 5 m/s to 20 m/s. Find the numbers of revolutions the particle completes around the axis.
Solution : We need to find angular displacement to know the numbers of revolutions made. Here,
ω
0
=
5
10
=
0.5
rad
/
s
ω
=
20
10
=
2
rad
/
s
t
=
20
s
ω
0
=
5
10
=
0.5
rad
/
s
ω
=
20
10
=
2
rad
/
s
t
=
20
s
To calculate angular displacement, we need to know angular acceleration. Here, we can calculate angular acceleration as in the earlier exercise as :
α
=
ω
−
ω
0
Δ
t
=
2.0
−
0.5
20
=
1.5
20
=
0.075
rad
∕
s
2
α
=
ω
−
ω
0
Δ
t
=
2.0
−
0.5
20
=
1.5
20
=
0.075
rad
∕
s
2
Now, using equation of motion,
Δ
θ
=
ω
0
t
+
1
2
α
t
2
Δ θ =
ω
0
t +
1
2
α
t
2
, we have :
⇒
Δ
θ
=
ω
0
t
+
1
2
α
t
2
Δ
θ
=
0.5
X
20
+
1
2
X
0.075
X
20
2
=
10
+
15
=
25
rad
⇒
Δ
θ
=
ω
0
t
+
1
2
α
t
2
Δ
θ
=
0.5
X
20
+
1
2
X
0.075
X
20
2
=
10
+
15
=
25
rad
The number of revolutions (nearest integer), n,
⇒
n
=
Δ
θ
2
π
=
25
6.28
=
3.97
=
3
(integer)
⇒
n
=
Δ
θ
2
π
=
25
6.28
=
3.97
=
3
(integer)
The angular velocity .vs. time plot of the motion of a rotating disk is shown in the figure. Then,
angular acceleration is constant.
angular acceleration is negative.
disk comes to a stop at a particular instant.
disk reverses direction during the motion.
The slope of the straight line is a negative constant. Therefore, options (a) and (b) are correct. The line crosses time axis, when angular velocity is zero. Thus option (c) is correct. The angular velocities have opposite sign across time axis. It means that the disk reverses its direction. Thus option (d) is correct.
Hence, options (a),(b),(c) and (d) are correct.
A point on a rotating disk, starting from rest, achieves an angular velocity of 40 rad/s at constant rate in 5 seconds. If the point is at a distance 0.1 meters from the center of the disk, then the distance covered (in meters) during the motion is :
(a)
5
(b)
10
(c)
20
(d)
20
(a)
5
(b)
10
(c)
20
(d)
20
We can find the distance covered, if we know the angular displacement. On the other hand, we can find angular displacement if we know the average angular speed as time is given.
ω
avg
=
ω
i
+
ω
f
2
=
40
+
0
2
=
20
rad
/
s
2
ω
avg
=
ω
i
+
ω
f
2
=
40
+
0
2
=
20
rad
/
s
2
We can consider this accelerated motion as uniform motion with average speed as calculated above. The angular displacement is :
⇒
θ
=
ω
avg
X
t
=
20
X
05
=
100
rad
⇒
θ
=
ω
avg
X
t
=
20
X
05
=
100
rad
The distance covered is :
⇒
s
=
θ
X
r
=
100
X
0.1
=
10
m
⇒
s
=
θ
X
r
=
100
X
0.1
=
10
m
A point on a rotating disk completes two revolutions starting from rest and achieves an angular velocity of 8 rad/s. If the angular velocity of the disk is increasing at a constant rate, the angular acceleration (
rad
/
s
2
rad /
s
2
) is :
(a)
8
(b)
8π
(c)
8
π
(d)
16
π
(a)
8
(b)
8π
(c)
8
π
(d)
16
π
We can use the equation of motion for constant acceleration. Here,
θ
=
2
revolution
=
2
X
2π
=
4π
θ
=
2
revolution
=
2
X
2π
=
4π
ω
i
=
0
ω
i
=
0
ω
f
=
8
rad/s
ω
f
=
8
rad/s
Looking at the data, it is easy to find that the following equation will serve the purpose,
ω
f
2
=
ω
i
2
=
2
αθ
ω
f
2
=
ω
i
2
=
2
αθ
Solving for "α" and putting values,
⇒
α
=
ω
f
2

ω
i
2
2
θ
=
8
2

0
2
2
X
4π
=
8
π
rad
/
s
2
⇒
α
=
ω
f
2

ω
i
2
2
θ
=
8
2

0
2
2
X
4π
=
8
π
rad
/
s
2
Hence, option (c) is correct.
A point on a rotating disk is accelerating at a constant rate 1
rad
/
s
2
rad /
s
2
till it achieves an angular velocity of 10 rad/s. What is the angular displacement (radian) in last 2 seconds of the motion?
(a)
18
(b)
24
(c)
28
(d)
32
(a)
18
(b)
24
(c)
28
(d)
32
Here, final angular velocity, angular acceleration and time of motion are given. We can find the angular displacement using equation of motion for angular displacement that involves final angular velocity :
θ
=
ω
f
t

1
2
α
t
2
θ
=
ω
f
t

1
2
α
t
2
Putting values, we have :
⇒
θ
=
10
X
2

1
2
1
2
2
=
20

2
=
18
rad
⇒
θ
=
10
X
2

1
2
1
2
2
=
20

2
=
18
rad
Hence, option (a) is correct.