CHAPTER 17 MOTION IN A CIRCLE
EXERCISE 90, Page 199
1. A locomotive travels around a curve of 500 m radius. If the horizontal thrust on the outer rail is
1
of the locomotive weight, determine the speed of the locomotive. The surface that the rails
50
are on may be assumed to be horizontal and the horizontal force on the inner rail may be
assumed to be zero.
Centrifugal force on outer rail =
mg
50
mv 2 mg
=
50
r
Hence,
v2 =
from which,
i.e.
v=
gr
9.81 500
=
= 98.1 m 2 /s 2
50
50
98.1 = 9.9 m/s
= 9.9
m 1km 3600s


s 1000 m
1h
i.e. the speed of the locomotive, v = 35.64 km/h
2. If the horizontal thrust on the outer rail of Problem 1 is
1
of the locomotive’s weight,
100
determine its speed.
Centrifugal force on outer rail =
Hence,
from which,
i.e.
mg
100
mv 2 mg
=
100
r
v2 =
v=
gr
9.81 500
=
= 49.05 m 2 /s 2
100
100
49.05 = 7 m/s
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= (7  3.6) km/h
i.e. the speed of the locomotive, v = 25.2 km/h
3. What angle of banking of the rails of Problem 1 is required for the outer rail to have a zero value
of outward thrust? Assume the speed of the locomotive is 15 km/h.
 v2 
Angle of banking,  = tan 1  
 rg 
v = 15 km/h = 15
15
km
1h
1000 m


=
= 4.1667 m/s
3.6
h 3600s 1km
 4.1667 2 m 2 / s 2 
Hence,  = tan 
= tan 1 (0.0035395)
2 
 500 m  9.81m / s 
1
i.e. angle of banking,  = 0.203
4. What angle of banking of the rails is required for Problem 3, if the speed of the locomotive is
30 km/h ?
 v2 
Angle of banking,  = tan 1  
 rg 
v = 30 km/h = 30
30
km
1h
1000 m


=
= 8.3333 m/s
3.6
h 3600s 1km
 8.33332 m 2 / s 2 
Hence,  = tan 
= tan 1 (0.014158)
2 
 500 m  9.81m / s 
1
i.e. angle of banking,  = 0.811
226
© John Bird & Carl Ross Published by Taylor and Francis
EXERCISE 91, Page 201
1. A conical pendulum rotates about a horizontal circle at 100 rpm. If the speed of rotation of the
mass increases by 5%, how much does the mass of the pendulum rise ?
Angular velocity,  =
2n 2100

= 10.472 rad/s
60
60
g
h
=
from which, height, h =
or  2 =
g
h
g
9.81
=
= 0.089456 m
2

10.472 2
When the speed of rotation rises by 5%, n 2 = 100  1.05 = 105 rpm
Hence,  2 =
2 =
Hence,
2n 2 2105

= 10.996 rad/s
60
60
g
h2
or  2 2 =
h2 =
g
h2
9.81
g
=
2
10.996 2
2
i.e. the new value of height, h 2 = 0.081133 m
Rise in height of the pendulum mass = ‘old’ h – ‘new’ h
= h - h 2 = 0.089456 – 0.081133
= 0.00832 m = 8.32 mm
2. If the speed of rotation of the mass of Problem 1 decreases by 5%, how much does the mass
fall ?
From Problem 1, h = 0.089456 m
When the speed of rotation decreases by 5%, n 2 = 100  0.95 = 95 rpm
Hence,  2 =
2n 2 2 95

= 9.9484 rad/s
60
60
227
© John Bird & Carl Ross Published by Taylor and Francis
2 =
Hence,
g
h2
or  2 2 =
h2 =
g
h2
9.81
g
=
2
9.9484 2
2
i.e. the new value of height, h 2 = 0.099120 m
Fall in height of the pendulum mass = ‘old’ h – ‘new’ h
= h - h 2 = 0.089456 – 0.099120
= - 0.00966 m = 9.66 mm
3. A conical pendulum rotates at a horizontal angular velocity of 2 rad/s. If the length of the string
is 3 m and the pendulum mass is 0.25 kg, determine the tension in the string. Determine also the
radius of the turning circle.
Angular velocity,  = 2 rad/s
Tension in the string, T = m  2 L
= 0.25 kg  (2 rad/s) 2  3 m
T = 3 kg m/s 2
i.e.
However, 1 kg m/s 2 = 1 N, hence,
tension in the string, T = 3 N
T=
mg
mg
0.25 kg  9.81m / s 2
from which, cos  =
=
= 0.8175
cos 
T
3N
Hence, the cone angle,  = cos 1 (0.8175) = 35.165
sin  =
r
, from which, radius of turning circle, r = L sin  = 3 m  sin 35.165
L
= 1.728 m
228
© John Bird & Carl Ross Published by Taylor and Francis
EXERCISE 92, Page 203
1. A uniform disc of diameter 0.1 m rotates about a vertical plane at 200 rpm. The disc has a mass
of 1.5 kg attached at a point on its rim and another mass of 2.5 kg at another point on its rim,
where the angle between the two masses is 90 clockwise. Determine the magnitude of the
resultant centrifugal force that acts on the axis of the disc, and its position with respect to the
1.5 kg mass.
Angular velocity,  =
2n 2 200

= 20.944 rad/s
60
60
Force, F1  m12 r  1.5  20.9442 
0.1
= 32.90 N
2
Force, F2  m 2 2 r  2.5  20.9442 
0.1
= 54.83 N
2
From the diagram, resultant, R =
F
2
1
 F22  
32.90
2
 54.832  by Pythagoras
= 63.94 N
cos θ =
and
32.90 32.90

= 0.515
R
63.94
θ = cos 1 (0.515) = 59º clockwise
2. If a mass of 4 kg is placed on some position on the disc in Problem 1, determine the position
where this mass must be placed to nullify the unbalanced centrifugal force.
From Problem 1, resultant, R = 63.94 N = m2 r
i.e.
from which,
63.94 = 4  20.9442  r
radius, r =
63.94
= 0.03644 m
4  20.944 2
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© John Bird & Carl Ross Published by Taylor and Francis
Hence, radius, r = 36.44 mm at (180º - 59º) anticlockwise to the 1.5 kg mass
i.e.
radius r = 36.44 mm at 121º anticlockwise to the 1.5 kg mass
3. A stone of mass 0.1 kg is whirled in a vertical circle of 1 m radius by a mass-less string, so that
the string just remains taut. Determine the velocity and tension in the string at (a) the top of the
circle, (b) the bottom of the circle, (c) midway between (a) and (b).
(a) At the top, tension T = 0 N
m2 r  mg
i.e. angular velocity,  
and
g
9.81
= 3.132 rad/s

r
1
linear velocity, v = ωr = 3.132 rad/s × 1 m
i.e. velocity at the top, v 1 = 3.132 m/s and the tension = 0 N
(b) At the bottom, (kinetic energy + potential energy) top = (kinetic energy) bottom
i.e.
i.e.
mv12
mv 2 2
 mg  2 
2
2
v 2 2  v12  4g
= 3.132 2  4  9.81 = 49.049
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© John Bird & Carl Ross Published by Taylor and Francis
v2  49.049 = 7 m/s
and

v2 
Tension at the bottom, T2  mg  m r  m  g  
r 

2
49 

= 0.1 ×  9.81   = 5.88 N
1 

i.e. velocity at the bottom, v 2 = 7 m/s and the tension = 5.88 N
(c) Mid-way, (kinetic energy + potential energy) top = (kinetic energy + potential energy) mid  way
mv32
mv12
 mg  2 
 mg 1
2
2
i.e.
v32  v12  2g
i.e.
= 3.132 2  2  9.81 = 29.4294
v2  29.4294 = 5.42 m/s
and
Tension mid-way, T3 
mv32 0.1 5.422

= 2.94 N
r
1
i.e. velocity mid-way, v 3 = 5.42 m/s and the tension = 2.94 N
EXERCISE 93, Page 203
Answers found from within the text of the chapter, pages 196 to 203.
EXERCISE 94, Page 203
1. (b) 2. (d) 3. (a) 4. (b) 5. (b)
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