Mechanical Engineering Principles

16

Motion in a circle

At the end of this chapter you should be

able to:

• understand centripetal force


• understand D’Alembert’s principle


• understand centrifugal force


• solve problems involving locomotives and
  cars travelling around bends


• solve problems involving a conical
  pendulum


• solve problems involving the motion in a
  vertical circle


• understand the centrifugal clutch

16.1 Introduction

In this chapter we will restrict ourselves to the uni-
form circular motion of particles. We will assume
that objects such as railway trains and motor-
cars behave as particles, i.e. rigid body motion is
neglected. When a railway train goes round a bend,
its wheels will have to produce a centripetal accel-
eration towards the centre of the turning circle. This
in turn will cause the railway tracks to experience
a centrifugal thrust, which will tend to cause the
track to move outwards. To avoid this unwanted out-
ward thrust on the outer rail, it will be necessary to
incline the railway tracks in the manner shown in
Figure 16.1.
From Section 13.3, it can be seen that when a
particle moves in a circular path at a constant speed
v, its centripetal acceleration,

a= 2 vsin

θ

2

×

1
t
Whenθis small,θ≈sinθ,

hence a= 2 v

θ

2

×

1

t

=v

θ

t

However,ω=uniform angular velocity=

θ

t

Thereforea=vω

q

q

Railway tracks

Figure 16.1

Ifr=the radius of the turning circle, then

v=ωr

and a=ω^2 r=

v^2

r

Now force=mass×acceleration

Hence,

centripetal force=mω^2 r=

mv^2

r

( 16. 1 )

D’Alembert’s principle

Although problems involving the motion in a cir-

cle are dynamic ones, they can be reduced to static

problems through D’Alembert’s principle. In this

principle, the centripetal force is replaced by an

imaginarycentrifugal forcewhich acts equal and

opposite to the centripetal force. By using this

principle, the dynamic problem is reduced to a

static one.

If a motorcar travels around a bend, its tyres will

have to exert centripetal forces to achieve this. This

is achieved by the transverse frictional forces acting

on the tyres, as shown in Figure 16.2.

In Figure 16.2, the following notation is used:

CG=centre of gravity of the car,

CF=centrifugal force=

mv^2

r

,

m=mass of car,

g=acceleration due to gravity,

R 1 =vertical reaction of ‘inner’ wheel,

R 2 =vertical reaction of ‘outer’ wheel,

F 1 =frictional force on ‘inner’ wheel,

F 2 =frictional force on ‘outer’ wheel,