chapter 13 linear and angular...

209 © John Bird & Carl Ross Published by Taylor and Francis

CHAPTER 13 LINEAR AND ANGULAR MOTION

EXERCISE 73, 72

1. A pulley driving a belt has a diameter of 360 mm and is turning at 2700/π revolutions per

minute. Find the angular velocity of the pulley and the linear velocity of the belt assuming that

no slip occurs.

Angular velocity ω = 2πn, where n is the speed of revolution in revolutions per second, i.e.

n = 270060π

revolutions per second.

Thus, angular velocity, ω = 2π 270060

π

= 90 rad/s

The linear velocity of a point on the rim, v = ωr, where r is the radius of the wheel, i.e.

r = 3602

= 180 mm = 0.18 m

Thus, linear velocity, v = ωr = 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 10003600× m/s = 10 m/s

(Note that changing from km/h to m/s involves dividing by 3.6)

Radius of wheel, r = 5002

= 250 mm = 0.25 m

Since, v = ωr, then angular velocity, ω = v 10r 0.25= = 40 rad/s


EXERCISE 74, Page 173

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, 1ω = 200 rad/s, angular acceleration, α = 5 rad/s 2 and time, t = 15 s.

Final angular velocity, 2ω = 1ω + αt

= 200 + (5)(15) = 200 + 75 = 275 rad/s

In revolutions per minute, 275 rad/s = 275 602

×π

= 8250π

rev/min or 2626 rev/min

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, 1ω = 2300 1060π

× = π rad/s

and final angular velocity, 2ω = 2600 2060π

× = π rad/s

2ω = 1ω + αt from which,

angular acceleration, α = 2 1 20 10 10t 25 25

ω −ω π− π π= = = 0.4π rad/s 2 or 1.257 rad/s 2

Linear acceleration, a = rα = (0.25)(0.4π) = 0.1π m/s 2 or 0.314 m/s 2



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.

Angle turned through, 1 2 t2

ω +ω θ =

hence 300 × 2π = 1000 400 t2+

i.e. 600π = 700t

from which, time, t = 600700

π = 2.693 s

2. Find the angular retardation for the grinding wheel in question 1.

2ω = 1ω + αt from which,

angular acceleration, α = 2 1 400 1000 600t 2.693 2.693

ω −ω − −= = = - 222.8 rad/s 2

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,

θ = 1 2

2ω +ω

t =

300 2 600 260 60

2

× π × π +

(25) rad

However, there are 2π radians in 1 revolution, hence,

number of revolutions =

300 2 600 22560 60

2 2

× π × π + π

= ( )1 900 252 60

= 187.5 revolutions


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, 2ω = 1ω + αt = 0 + (8)(20) = 160 rad/s

(b) 3ω = 2ω + αt from which,

angular acceleration, α = 3 2 0 160t 40

ω −ω −= = - 4 rad/s 2

i.e. angular deceleration is 4 rad/s 2

(c) Initial angle turned through, θ 1 = 1 2

2ω +ω

t = 0 1602+

(20) = 1600 rad = 16002π

rev

At constant speed, angle turned through, θ 2 = 160 rad/s × (2 × 60)s = 19200 rad = 192002π

rev

Angle turned through during deceleration, θ 3 = 160 02+

(40) = 3200 rad = 32002π

rev

Hence, total number of revolutions made by the pulley = θ 1 + θ 2 + θ 3

=16002π

+ 192002π

+ 32002π

= 240002π

= 12000π

rev or 3820 rev



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 = ( )2 279.2 26.4+ = 83.5 km/h and 1 79.2tan 71.626.4

− θ = = °

(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 θ shown

in the triangle of velocities shown below where v is the swimmers true speed.

v = 2 22 1 3− = km/h = 1000360

m/min = 28.87 m/min


Hence, if the width of the river is 142 m, the swimmer will take 14228.87

= 4.919 minutes

= 4 min 55 s

In the above diagram, sin θ = 12

from which, θ = 30°

Hence, the swimmer needs to swim at an angle of 60° to the bank (shown as angle α 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

Hence, v = ( )2 222.46 3.87+ = 22.79 km/h,

and 1 3.87tan 9.7822.46

− θ = = °

Hence, the ships actual speed is 22.79 km/h in a direction E 9.78° N



Answers found from within the text of the chapter, pages 171 to 177.


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)

chapter 13 linear and angular...

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