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
- 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] - 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] - 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.