Engineering Mechanics

Chapter 28 : Motion Along a Circular Path „„„„„ 573

Let r = Radius of the circular path in m, and
v = Linear velocity of the particle in m/s.
Now consider an instant when A be the position of the particle. After a small interval of time
(δt), let its position be changed from A to B. When the body is at A its linear velocity (v) will be
along the dotted line AD i.e., tangent at A. Similarly, when the body is at B, its linear velocity (again
equal to v) will be along the dotted line BE i.e, tangent at B as shown in Fig. 28.1 (a).

Fig. 28.1. Centrifugal acceleration.
First of all, take any suitable point o, draw oa parallel to AD and let it represent the velocity
v at A, in direction and magnitude to some scale. Similarly, draw ob parallel to BE and let it represent
the velocity v at B, in direction and magnitude to the scale. Join ab as shown in Fig. 28.1 (b).

Now in triangle aob, oa represents the initial velocity ; ob represents the final velocity and
ab represents the change in velocity.

We know that the acceleration,

Change in velocity

Time

ab

a

t

==

δ

...(i)

Since the interval of time (δt) is considered to be very small, therefore chord AB may be
considered to be equal to arc AB. Now consider two triangles AOB and aob, which are similar.

∴

ab oa

ABOA

=

Substituting the values of AB = v × δt, oa = v and OA = r, in the above equation,

ab v

vtr

=

×δ

or

ab v^2

tr

=

δ

Therefore acceleration,

ab v^2

a

tr

==

δ

...(ii)

Now substituting the value v = ωr in equation (ii) where ω is the angular velocity,

22

arr^2

r

ω

==ω

Note. The same formula holds good for centrifugal acceleration also.