Higher Engineering Mathematics

THE CIRCLE AND ITS PROPERTIES 143

B

The unit of angular velocity is radians per second
(rad/s). An object rotating at a constant speed of
nrevolutions per second subtends an angle of 2πn
radians in one second, i.e., its angular velocityωis
given by:

ω= 2 πnrad/s (3)

From equation (1) on page 138,s=rθand from
equation (2) on page 142,θ=ωt

hence s=r(ωt)

from which

s

t

=ωr

However, from equation (1)v=

s

t

hence v=ωr (4)

Equation (4) gives the relationship between linear
velocityvand angular velocityω.

Problem 14. A wheel of diameter 540 mm is

rotating at

1500

π

rev/min. Calculate the angular

velocity of the wheel and the linear velocity of

a point on the rim of the wheel.

From equation (3), angular velocityω= 2 πnwhere
nis the speed of revolution in rev/s. Since in this case

n=

1500

π

rev/min=

1500

60 π

=rev/s, then

angular velocityω= 2 π

(

1500

60 π

)

=50 rad/s

The linear velocity of a point on the rim,v=ωr,
whereris the radius of the wheel, i.e.
540
2

mm=

0. 54

2

m= 0 .27 m.

Thuslinear velocity v=ωr=(50)(0.27)

= 13 .5m/s

Problem 15. A car is travelling at 64.8 km/h

and has wheels of diameter 600 mm.

(a) Find the angular velocity of the wheels in

both rad/s and rev/min.

(b) If the speed remains constant for 1.44 km,

determine the number of revolutions made

by the wheel, assuming no slipping occurs.

(a) Linear velocityv= 64 .8km/h

= 64. 8

km

h

× 1000

m

km

×

1

3600

h

s

=18 m/s.

The radius of a wheel=

600

2

=300 mm

= 0 .3m.

From equation (5),v=ωr, from which,

angular velocityω=

v

r

=

18

0. 3

=60 rad/s

From equation (4), angular velocity,ω= 2 πn,

wherenis in rev/s.

Hence angular speedn=

ω

2 π

=

60

2 π

rev/s

= 60 ×

60

2 π

rev/min

=573 rev/min

(b) From equation (1), sincev=s/tthen the time

taken to travel 1.44 km, i.e., 1440 m at a constant

speed of 18 m/s is given by:

timet=

s

v

=

1440 m

18 m/s

=80 s

Since a wheel is rotating at 573 rev/min, then in

80/60 minutes it makes

573 rev/min×

80

60

min=764 revolutions

Now try the following exercise.

Exercise 67 Further problems on linear and

angular velocity

1. A pulley driving a belt has a diameter of
   300 mm and is turning at 2700/π revolu-
   tions per minute. Find the angular velocity
   of the pulley and the linear velocity of the
   belt assuming that no slip occurs.
   [ω=90 rad/s,v= 13 .5 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 linear velocity of a
   point on the rim of one of the wheels of
   the bicycle, and the angular velocity of the
   wheels.
   [v=10 m/s,ω=40 rad/s]


3. A train is travelling at 108 km/h and has
   wheels of diameter 800 mm.