Higher Engineering Mathematics, Sixth Edition

130 Higher Engineering Mathematics

n=

1500

π

rev/min=

1500

60 π

=rev/s, then

angular velocityω= 2 π

(

1500

60 π

)

=50rad/s

The linear velocity of a point on the rim,v=ωr,where

ris the radius of the wheel, i.e.

540

2

mm=

0. 54

2

m= 0 .27m.

Thuslinear velocity v=ωr=( 50 )( 0. 27 )

=13.5m/s

Problem 19. A car is travelling at 64.8km/h and

has wheels of diameter 600mm.

(a) Find the angular velocity of the wheels in both

rad/s and rev/min.

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

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

=18m/s.

The radius of a wheel=

600

2

=300mm

= 0 .3m.

From equation (5),v=ωr, from which,

angular velocityω=

v

r

=

18

0. 3

=60rad/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

=573rev/min

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

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

speed of 18m/s is given by:

timet=

s

v

=

1440m

18m/s

=80s

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

80/60minutes it makes

573rev/min×

80

60

min=764revolutions

Now try the following exercise

Exercise 59 Further problems on linear

and angular velocity

1. A pulley driving a belt has a diameter of
   300mm and is turning at 2700/πrevolutions
   per minute. Find the angular velocity of the
   pulley and the linear velocity of the belt
   assuming that no slip occurs.
   [ω=90rad/s,v= 13 .5m/s]


2. Abicycleistravellingat36km/handthediam-
   eter of the wheels of the bicycle is 500mm.
   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=10m/s,ω=40rad/s]


3. Atrainistravellingat108km/handhaswheels
   of diameter 800mm.
   (a) Determine the angular velocity of the
   wheels in both rad/s and rev/min.
   (b) If the speed remains constant for 2.70km,
   determine the number of revolutions
   made by a wheel, assuming no slipping
   occurs. [
   (a)75rad/s, 716 .2rev/min
   (b)1074revs

]

13.7 Centripetal force

When an object moves in a circular path at constant

speed, its direction of motion is continually changing

and hence its velocity (which depends on both magni-

tude and direction) is also continually changing. Since

acceleration is the (change in velocity)/(timetaken), the

object has an acceleration. Let the object be moving

with a constant angular velocity ofωand a tangential

velocity of magnitudevand let the change of veloc-

ity for a small change of angle ofθ(=ωt)beV in

Fig. 13.17. Thenv 2 −v 1 =V. The vector diagram is

shown in Fig. 13.17(b) and since the magnitudes ofv 1

andv 2 are the same, i.e.v, the vector diagram is an

isosceles triangle.