CALCULATE ANGULAR MOTION
Angular motion is the motion of a body about a fixed point. The motion of
pendulums, rotation of fly wheel, motor vehicle going round a bend and planets are
examples of angular motion. Angular motion can also be used to describe the angle
made during the motion of a body about a fixed axis. Angles are measured in :
(a) Degrees
(b) Radians
(c) Revolutions
(d) Steradians
Consider a wheel rotating about an axis O
In moving from position A to B, the wheel has turned through the angle θ.
The value of θ I radians (rad) is just a ratio of S to the wheels radius.
𝑆
θ=𝑟
S = r𝜃 , where
S = arc length
r = radius of circle or wheel
Θ = angle (radians)
One radian is the angle subtended by arch length equal to the radius of the circle.
RELATIONSHIP BETWEEN THE DEGREES & THE RADIANS
1rev = 360o,
1rev = 2𝜋rad
NOTE: Degrees, revolutions and radians are dimensionless
𝜋𝑟𝑎𝑑 = 180o,
1 rad = 57.3o
Example 1. Convert 70o to radians and revolutions
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Solution
𝜋 rads = 180o
70o =
70 𝑥𝜋
= 1.22rads
180
1rev =
70o =
360o
1𝑟𝑒𝑣𝑥 70
360
=
0.194rev
Example 2. A vehicles wheel of diameter 0.56m turns through an angle of 180o.
Calculate the distance moved by a point on the tyre thread of the wheel.
Solution
𝑑
0.56
s = rθ, r = 2 =
= 0.28m, θ = 180o = 𝜋rads
2
s = rθ = 0.28 x 𝜋rads,
s ≈0.88rads
ANGULAR VELOCITY
Angular velocity is defined as the rate of change of angular displacement. Angular
displacement is the angle turned about an axis.
Angular velocity =
𝐴𝑛𝑔𝑙𝑒𝑡𝑢𝑟𝑛𝑒𝑑
𝑇𝑖𝑚𝑒𝑇𝑎𝑘𝑒𝑛
𝜃
𝜔 = 𝑡 , Typical units for 𝜔 are radians per second (rad/s),degrees per sec (θ/s) and
revolutions per min (rev/min). The SI unit is radians per second (rad/s).
Example 3. A wheel turns through 1800rev in 1.0min.
Find ;
a.) Its average angular velocity in rad/s.
b.)Through how many radians does the wheel turn in 15 seconds.
Solution
Θ = 180rev =
𝜔 =
𝜃
b.)
𝜃 = 𝜔t
𝑡
=
3600𝜋
60
2𝜋𝑟𝑎𝑑
1𝑟𝑒𝑣
x1800rev = 3600𝜋, t = 1min = 60sec
, 𝜔= 188.495rads/s, 𝜔= 188.50rad/s
=188.495 x 15 =2827.425rad
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ANGULAR ACCELERATION
Angular acceleration is defined as the rate of change of angular velocity
If the angle of a point moving in a particular circular path changes uniformly from 𝜔i to
𝜔f, then
𝜔𝑓−𝜔
𝛼 =
𝑖
𝑡
,
𝜔f = 𝜔i + 𝛼t
Where 𝜔i = initial angular velocity, 𝜔f = final angular velocity, 𝛼 = angular
acceleration. If angular acceleration is uniform, we know that, as with linear motion, the
average angular velocity is given by;
𝜔
̅=
1
2
(𝜔f + 𝜔i )
.
Example 4. A wheel starts from rest and attains a rotational velocity of 240rev/s in
2.0minutes. Determine;
a.)Its angular acceleration
b.)The wheels’ angular speed (in radians per second) 130 seconds after starting from
rest.
Solution
a.)𝜔i =0m/s, 𝜔f = 240rev/s, t = 2.0m = 120sec
𝜔𝑓−𝜔
240−0
2 rev X 2π
𝑖
𝛼 =
= 120 =2rev/s2 = 1rev
=12.566rad/s2 ≈12.57rad/s2
𝑡
𝜔f = 𝜔i + 𝛼t =0 +12.566 X 130 =1633.58rad/s
Have you seen that the equations of motion are the ones we are using in angular
motion? Table 3.42 shows the similarities between linear motion and angular motion.
:Similarities between linear motion and angular motion.
Linear motion
Angular motion
S = 𝑣̅ t
𝑣 = 𝑢 + 𝑎𝑡
Displacement, 𝑠 = 𝑢𝑡 +
1
2
𝜃 = 𝜔t
𝜔𝑓 = 𝜔𝑖 + 𝛼𝑡
1
𝜃 = 𝜔𝑖 𝑡 + 𝛼𝑡 2
2
𝜔𝑓 2 = 𝜔𝑖 2 + 2𝛼𝜃
𝑎𝑡 2
𝑣 2 = 𝑢2 + 2𝑎𝑠
Average velocity, 𝑣 = (
𝑢+𝑣
2
𝝎𝒇 =
)
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(𝜔𝑓 + 𝜔𝑖 )
𝟐
RELATIONSHIP BETWEEN LINEAR AND ANGULAR MOTION
There are many cases, such as a ball rolling on the ground, where both rotational and
linear motion occurs.
To investigate this relationship, examine the wheel of radius r that has rolled a distance
s as shown below in figure 3.43.
A reference point (A) on to the rim of the wheel has been identified. We first observe
that the arc length traced out by our reference point is equal to the distance s that the
arc wheel moves. This distance is called the tangential distance. The relation
between s and θ has already been used in our definition of radian measure.
We will now use the s = rθ equation and the defining equation of angular velocity to
determine a relationship between linear and rotational velocity.
s = rθ
Displacement
Linear velocity is given by v =
time
We have
V =
𝑆
𝑟𝜃
=
v=
𝑡
𝑡
, but s = rθ
𝜃
𝑡
𝜃
x r, since
𝑡
Similarly, linear acceleration is given by, a =
=𝜔
v=
𝑟𝜃
𝑡
becomes v = 𝝎r
v −u
t
and angular acceleration is given
𝛼 =
ωf−ω
i
t
We have
∴a=
∆v
t
But, ω =
a=
= a =
Therefore
∆v
𝜔𝑟
t
a =
v
r
Substituting ω =
v
r
into 𝛼 =
ωf−ω
i
t
𝑁𝑜𝑤 ∆𝑣 = 𝜔𝑟
t
, we know that 𝛼 =
𝜔𝑟
t
is
ω
t
a = 𝜶r
Example 1. A car accelerates uniformly from rest to 17.5m/s in 23.6 seconds.
If the radius of the car wheel is 0.4m. Find;
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(a) the angular acceleration of one of its wheels
(b) the number of revolutions turned by the wheel in the process.
Solution
a.) U =0, v = 17.5m/s, t =23.6sec
17.5−0
a=
v−u
t
= 0.74m/s2
23.6
a = 𝜶r
a=
𝛼=
𝑎
𝑟
=
1
b.) 𝜃 = 𝜔it + 2 𝛼t2
Or
𝜃 = 517.973rad x
0.74
0.4
𝛼 = 1.86rad/s2
1
= 0 x 23.6 + 2 1.86 x(23.6) 𝜃 = 517.973rad
1𝑟𝑒𝑣
2𝜋
= 82.437rev ≈82.44rev
Example 2. A belt runs on a wheel of radius 0.44m. During the time the wheel takes
to coast to Uniform rest from an initial speed of 1.8rev/s, 29.5m of belt length passes
over the Wheel. Find;
(a) the number of revolutions the wheel turns while stopping
(b) the deceleration of the wheel
(c) Linear acceleration of the wheel
solution
ωi = 1.8rev/s, ωf = 0rev/s, s = 29.5m, r = 0.44m
from s = r𝜽
(a) 𝜃 =
s
r
29.5
= 0.44
=67.05rad
= 67.05rad x
1𝑟𝑒𝑣
2𝜋
= 10.671rev
(b) From the equation 𝜔2f = 𝜔2i + 2𝛼𝜃
𝛼 =
𝜔𝑓 2 −𝜔𝑖 2
2𝜃
= - 0.152rev/s2
Therefore, deceleration is 0.152rev/s2. Deceleration is negative acceleration, 𝑹 = −𝒂.
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SELF HELP TASKS
Question 1.
(a) What is meant by the following
(i)
Angular displacement
(ii)
Radian
(b) A motor running at 2000 rev/min for 20 seconds is switched off and the motor
comes to rest after 30 seconds. Calculate
(i)
the number of revolutions made in the first 20 seconds.
(ii)
the angular acceleration in the last 30 seconds.
Question 2. A grinding wheel of diameter 40cm rotating at 800 rev/minute increases
its speed to 1600 rev/ minutes in 20 seconds. Calculate the
(a)
(b)
(c)
(d)
(e)
Angular acceleration
Angle turned
Number of revolutions
Linear displacement covered
Linear acceleration
Question 3. The Fun in a computer is rotating at speed of 3000 rev/min and slows
down to 1000 rev/min while making 80 revolutions during the deceleration. Calculate;
(a) The angular displacement of the spindle
(b) The angular deceleration of the spindle
(c) Time taken by the spindle to slow down during this motion.
QUESTION 4. A car accelerates uniformly from rest to 17.5m/s in 23.6 seconds.
If the radius of the car wheel is 0.4m, Find;
(a) the angular acceleration of one of its wheels
(b) the number of revolutions turned by the wheel in the process.
QUESTION 5. A bicycle with 62.5cm diameter wheels is coasting at 6.6m/s. It
decelerates uniformly and stops in 38 seconds.
(a) How far does it go in this time?
(b) Through how many revolutions does each wheel turn as the bicycle comes to
stop.
QUESTION 6. A belt runs on a wheel of radius 0.44m. During the time the wheel
takes to coast to Uniform rest from an initial speed of 1.8rev/s, 29.5m of belt length
passes over the Wheel. Find;
(a) the deceleration of the wheel
(b) the number of revolutions the wheel turns while stopping
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