130 Higher Engineering Mathematics
n =
1500
π
rev/min =
1500
60π
= rev/s, then
angular velocityω = 2π
1500
60π
= 50 rad/s
The linear velocity of a point on the rim, v = ωr, where
r is the radius of the wheel, i.e.
540
2
mm =
0.54
2
m = 0.27 m.
Thus linear velocity v = ωr = (50)(0.27)
= 13.5 m/s
Problem 19. 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 velocity v = 64.8 km/h
= 64.8
km
h
× 1000
m
km
×
1
3600
h
s
= 18 m/s.
The radius of a wheel =
600
2
= 300 mm
= 0.3 m.
From equation (5), v = ωr, from which,
angular velocity ω =
v
r
=
18
0.3
= 60 rad/s
From equation (4), angular velocity, ω = 2πn,
where n is in rev/s.
Hence angular speed n =
ω
2π
=
60
2π
rev/s
= 60 ×
60
2π
rev/min
= 573 rev/min
(b) From equation (1), since v = s/t then the time
taken to travel 1.44 km, i.e. 1440 m at a constant
speed of 18 m/s is given by:
time t =
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 59 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 revolutions
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.
(a) Determine the angular velocity of the
wheels in both rad/s and rev/min.
(b) If the speed remains constant for 2.70 km,
determine the number of revolutions
made by a wheel, assuming no slipping
occurs. (a) 75 rad/s, 716.2 rev/min
(b) 1074 revs
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 magnitude and direction) is also continually changing. Since
acceleration is the (change in velocity)/(time taken), the
object has an acceleration. Let the object be moving
with a constant angular velocity of ω and a tangential
velocity of magnitude v and let the change of velocity for a small change of angle of θ (=ωt ) be V in
Fig. 13.17. Then v 2 − v 1 = V . The vector diagram is
shown in Fig. 13.17(b) and since the magnitudes of v 1
and v 2 are the same, i.e. v, the vector diagram is an
isosceles triangle.
n =
1500
π
rev/min =
1500
60π
= rev/s, then
angular velocityω = 2π
1500
60π
= 50 rad/s
The linear velocity of a point on the rim, v = ωr, where
r is the radius of the wheel, i.e.
540
2
mm =
0.54
2
m = 0.27 m.
Thus linear velocity v = ωr = (50)(0.27)
= 13.5 m/s
Problem 19. 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 velocity v = 64.8 km/h
= 64.8
km
h
× 1000
m
km
×
1
3600
h
s
= 18 m/s.
The radius of a wheel =
600
2
= 300 mm
= 0.3 m.
From equation (5), v = ωr, from which,
angular velocity ω =
v
r
=
18
0.3
= 60 rad/s
From equation (4), angular velocity, ω = 2πn,
where n is in rev/s.
Hence angular speed n =
ω
2π
=
60
2π
rev/s
= 60 ×
60
2π
rev/min
= 573 rev/min
(b) From equation (1), since v = s/t then the time
taken to travel 1.44 km, i.e. 1440 m at a constant
speed of 18 m/s is given by:
time t =
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 59 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 revolutions
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.
(a) Determine the angular velocity of the
wheels in both rad/s and rev/min.
(b) If the speed remains constant for 2.70 km,
determine the number of revolutions
made by a wheel, assuming no slipping
occurs. (a) 75 rad/s, 716.2 rev/min
(b) 1074 revs
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 magnitude and direction) is also continually changing. Since
acceleration is the (change in velocity)/(time taken), the
object has an acceleration. Let the object be moving
with a constant angular velocity of ω and a tangential
velocity of magnitude v and let the change of velocity for a small change of angle of θ (=ωt ) be V in
Fig. 13.17. Then v 2 − v 1 = V . The vector diagram is
shown in Fig. 13.17(b) and since the magnitudes of v 1
and v 2 are the same, i.e. v, the vector diagram is an
isosceles triangle.