CHAPTER 22 LINEAR AND ANGULAR MOTION
EXERCISE 97, Page 226
1. A pulley driving a belt has a diameter of 360 mm and is turning at 2700/ revolutions per
minute. Find the angular velocity of the pulley and the linear velocity of the belt assuming that
no slip occurs.
Angular velocity  = 2n, where n is the speed of revolution in revolutions per second, i.e.
n=
2700
revolutions per second.
60
 2700 
Thus, angular velocity,  = 2 
 = 90 rad/s
 60 
The linear velocity of a point on the rim, v = r, where r is the radius of the wheel, i.e.
r=
360
= 180 mm = 0.18 m
2
Thus, linear velocity, v = r = 90  0.18 = 16.2 m/s
2. A bicycle is travelling at 36 km/h and the diameter of the wheels of the bicycle is 500 mm.
Determine the angular velocity of the wheels of the bicycle and the linear velocity of a point on
the rim of one of the wheels.
Linear velocity, v = 36 km/h =
36 1000
m/s = 10 m/s
3600
(Note that changing from km/h to m/s involves dividing by 3.6)
Radius of wheel, r =
500
= 250 mm = 0.25 m
2
Since, v = r, then angular velocity,  =
v 10

= 40 rad/s
r 0.25
283
© John Bird Published by Taylor and Francis
EXERCISE 98, Page 227
1. A flywheel rotating with an angular velocity of 200 rad/s is uniformly accelerated at a rate of
5 rad/s 2 for 15 s. Find the final angular velocity of the flywheel both in rad/s and revolutions per
minute.
Angular velocity, 1 = 200 rad/s, angular acceleration,  = 5 rad/s 2 and time, t = 15 s.
Final angular velocity,  2 = 1 + t
= 200 + (5)(15) = 200 + 75 = 275 rad/s
In revolutions per minute, 275 rad/s = 275 
60
8250
=
rev/min or 2626 rev/min

2
2. A disc accelerates uniformly from 300 revolutions per minute to 600 revolutions per minute in
25 s. Determine its angular acceleration and the linear acceleration of a point on the rim of the
disc, if the radius of the disc is 250 mm.
Initial angular velocity, 1 = 300 
2
 10 rad/s
60
and final angular velocity, 2 = 600 
2
 20 rad/s
60
2 = 1 + t from which,
angular acceleration,  =
2  1 20  10 10


= 0.4 rad/s 2 or 1.257 rad/s 2
t
25
25
Linear acceleration, a = r = (0.25)(0.4) = 0.1 m/s 2 or 0.314 m/s 2
284
© John Bird Published by Taylor and Francis
EXERCISE 99, Page 229
1. A grinding wheel makes 300 revolutions when slowing down uniformly from 1000 rad/s to
400 rad/s. Find the time for this reduction in speed.
   2 
Angle turned through,    1
t
 2 
 1000  400 
hence 300  2 = 
t
2


i.e.
600 = 700t
from which,
time, t =
600
= 2.693 s
700
2. Find the angular retardation for the grinding wheel in question 1.
2 = 1 + t from which,
angular acceleration,  =
2  1 400  1000 600


= - 222.8 rad/s 2
t
2.693
2.693
i.e. angular retardation is 222.8 rad/s 2
3. A disc accelerates uniformly from 300 revolutions per minute to 600 revolutions per minute in
25 s. Calculate the number of revolutions the disc makes during this accelerating period.
Angle turned through,
 300  2 600  2 
 60  60 
 1  2 
= 
 (25) rad
t = 
2
 2 




However, there are 2 radians in 1 revolution, hence,
 300  2 600  2 


60   25  = 1  900   25  = 187.5 revolutions
number of revolutions =  60
 


2

  2  2  60 


285
© John Bird Published by Taylor and Francis
4. A pulley is accelerated uniformly from rest at a rate of 8 rad/s 2 . After 20 s the acceleration stops
and the pulley runs at constant speed for 2 min, and then the pulley comes uniformly to rest after
a further 40 s. Calculate: (a) the angular velocity after the period of acceleration,
(b) the deceleration,
(c) the total number of revolutions made by the pulley.
(a) Angular velocity after acceleration period,  2 = 1 + t = 0 + (8)(20) = 160 rad/s
(b) 3 = 2 + t from which,
angular acceleration,  =
3  2 0  160

= - 4 rad/s 2
t
40
i.e. angular deceleration is 4 rad/s 2
1600
   2 
 0  160 
(c) Initial angle turned through,  1 =  1
rev
 (20) = 1600 rad =
t = 
2
 2 
 2 
At constant speed, angle turned through,  2 = 160 rad/s  (2  60)s = 19200 rad =
19200
rev
2
3200
 160  0 
Angle turned through during deceleration,  3 = 
rev
 (40) = 3200 rad =
2
 2 
Hence, total number of revolutions made by the pulley =  1 +  2 +  3
=
=
1600 19200 3200
+
+
2
2
2
24000 12000
=
rev or 3820 rev

2
286
© John Bird Published by Taylor and Francis
EXERCISE 100, Page 231
1. A car is moving along a straight horizontal road at 79.2 km/h and rain is falling vertically
downwards at 26.4 km/h. Find the velocity of the rain relative to the driver of the car.
The space diagram is shown in diagram (a). The velocity diagram is shown in diagram (b) and the
velocity of the rain relative to the driver is given by vector rc where rc = re + ec
rc =
 79.2
2
 26.42  = 83.5 km/h
 79.2 
and   tan 1 
  71.6
 26.4 
(a)
(b)
i.e. the velocity of the rain relative to the driver is 83.5 km/h at 71.6 to the vertical.
2. Calculate the time needed to swim across a river 142 m wide when the swimmer can swim at
2 km/h in still water and the river is flowing at 1 km/h. At what angle to the bank should the
swimmer swim?
The swimmer swims at 2 km/h relative to the water, and as he swims the movement of the water
carries him downstream. He must therefore aim against the flow of the water – at an angle  shown
in the triangle of velocities shown below where v is the swimmers true speed.
v=
22  12  3 km/h =
 1000 
3
 m/min = 28.87 m/min
 60 
287
© John Bird Published by Taylor and Francis
Hence, if the width of the river is 142 m, the swimmer will take
142
= 4.919 minutes
28.87
= 4 min 55 s
In the above diagram, sin  =
1
from which,  = 30
2
Hence, the swimmer needs to swim at an angle of 60 to the bank (shown as angle  in the
diagram.
3. A ship is heading in a direction N 60 E at a speed which in still water would be 20 km/h. It is
carried off course by a current of 8 km/h in a direction of E 50 S. Calculate the ship’s actual
speed and direction.
In the triangle of velocities shown below (triangle 0AB), 0A represents the velocity of the ship in
still water, AB represents the velocity of the water relative to the earth, and 0B is the velocity of the
ship relative to the earth.
Total horizontal component of v = 20 cos 30 + 8 cos 310 = 22.46
Total vertical component of v = 20 sin 30 + 8 sin 310 = 3.87
 22.46
2
 3.87 2  = 22.79 km/h,
Hence,
v=
and
 3.87 
  tan 1 
  9.78
 22.46 
Hence, the ships actual speed is 22.79 km/h in a direction E 9.78 N
288
© John Bird Published by Taylor and Francis
EXERCISE 101, Page 231
Answers found from within the text of the chapter, pages 224 to 231.
EXERCISE 102, Page 231
1. (b)
2. (c) 3. (a) 4. (c) 5. (a) 6. (d)
7. (c) 8. (b)
9. (d)
10. (c)
11. (b) 12. (d)
13. (a)
289
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