(^574) A Textbook of Engineering Mechanics
28.3. CENTRIPETAL FORCE
A body, moving in a circle or along a circular path, with a constant velocity, suffers a continu-
ous change in its direction at every point of its motion ; though the magnitude of its speed remains the
same. Since the velocity involves both magnitude as well as direction, and the velocity of the body is
continuously changing due to change in direction ; therefore according to *Newton’s First Law of
Motion, an external force must act continuously upon the body, to produce a change in the direction
of the moving body.
Strictly speaking the body, due to inertia, tends to move along the tangent at every point of its
motion, with the constant velocity. Therefore, some force must act at right angles to the direction of
motion at every point, which should change the direction of motion of the body, leaving the speed
uniform. Thus the force, which acts along the radius of the circle at every point, and is always directed
towards the centre of the circle along which the body moves, is known as centripetal force.
28.4. CENTRIFUGAL FORCE
According to **Newton’s Third Law of Motion, the force, which acts opposite to the centrip-
etal force, is known as centrifugal force. It may be noted that the centrifugal force always acts away
from the centre of the path, or in other words, the centrifugal force always tends to throw the body
away from the centre of circular path.
Example 28.1. A body of mass 5 kg is moving in a circle of radius of 1·5 m with an angular
velocity of 2 rad/s. Find the centrifugal force acting on the body.
Solution. Given : Mass of body (m) = 5 kg ; Radius of circle (r) = 1·5 m and angular velocity
of the body (ω) = 2 rad/s.
We know that centrifugal force acting on the body,
F = m ω^2 r = 5 × (2)^2 × 1·5 = 30 N Ans.
Example 28.2. A stone of mass 1 kg is revolving in a circle of radius 1 m with a linear
velocity of 10 m/s. What is the value of centrifugal force acting on the stone.
Solution. Given : Mass of stone (m) = 1 kg ; Radius of circle (r) = 1 m and linear velocity
of the stone (v) = 10 m/s.
We know that centrifugal force acting on the stone,
(^22) 1(10)
100 N
1
mv
F
r
×
== = Ans.
Example 28.3. A ball of mass 0·25 kg is attached to the end of a 2 m long string. The string
will break, if tension in the string is more than 25 N. What is the maximum angular velocity at which
the ball can be rotated?
Solution. Given : Mass of ball (m) = 0·25 kg ; Length of string or radius of circle (r) = 2 m
and maximum tension in the string (F) = 25 N.
Let ω = Maximum angular velocity at which the ball can be rotated.
We know that tension in the string or centrifugal force,
F = m ω^2 r = 0·25 × ω^2 × 2 = 0·5 ω^2
or^22550
0·5 0·5
F
ω= = =
∴ ω= 50 =7·07 rad/s Ans.
- Newton’s First Law of Motion states : “Everybody continues in its state of rest or of uniform motion, in
a straight line, unless it is acted upon by some external force”.
** Newton’s Third Law of Motion states, “To every action, there is always an equal and opposite reaction”.