(^576) „„„„„ A Textbook of Engineering Mechanics
28.5 CENTRIFUGAL FORCE ACTING ON A BODY MOVING ALONG A
CIRCULAR PATH
Consider a body moving along a circular path with a constant velocity.
Let m = Mass of the body,
r = Radius of the circular path, and
v = Linear velocity of the body.
We know that the centrifugal acceleration of the body,
v^2
a
r

and centrifugal force, Pc = Mass × Centrifugal acceleration
vmv^22
m
rr
=× = ...(when v is given)
= m ω^2 r ...(when ω is given)
Example 28.6. A railway engine of mass 60 tonnes, is moving in a circular track of radius
200 metres with a velocity of 36 km.p.h. Find the force exerted on the rails towards the centre of the
circle.
Solution. Given : Mass of railway engine (m) = 60 t ; Radius of circular path (r) = 200 m
and velocity of engine (v) = 36 km.p.h. = 10 m/s.
We know that the force exerted on the rails,
(^22) 60 (10)
30 kN
200
c
mv
P
r
== = Ans.
Example 28.7. An automobile of mass 1·5 t travelling at of 54 km.p.h.traverses a sag in the
road. The sag is a part of a circle of radius 25 m. Find the reaction between the automobile and road
while travelling at the lowest part of the sag.
Solution. Given : Mass of automobile (m) = 1·5 t; Velocity of automobile (v) = 54 km.p.h
= 15 m/s and radius of the sag (r) = 25 m
We know that reaction between the automobile and the load while travelling at the lowest
part of the sag
(^22) 1·5 (15)
1.5 9.8 kN
25
mv
mg
r
=+= +×
= 13·5 + 14·7 = 28·2 kN Ans.
28.6. SUPERELEVATION
In our day to day life, we see that whenever a roadway (or railway) is laid on a curve or curved
path, then both the edges are not at the same level. In such a case, the outer edge is always higher than
the inner one. This is done to keep the vehicle in equilibrium while going on the curved path.
We know that the body, moving along a curved path, is subjected to the following forces :

1. Its own weight, and


2. Centrifugal force.