Engineering Mechanics

Chapter 26 : Helical Springs and Pendulums „„„„„ 549

and length of equivalent simple pendulum (L),

22

0·559 0·45

0·45

h kkGG

h

=+ = +

∴ kG^2 = (0·559 – 0·45) × 0·45 = 0·049

or kG = 0·22 m = 220 mm Ans.

No. of oscillations per minute

Given. Distance between A and B = 1·5 m

or h = AB – l = 1·5 – 0·45 = 1·05 m

We know that length of the equivalent simple pendulum,

(^2) (0·22) 2
1·0 5 1·1 m
1· 0 5
LhkG
h
=+ = + =
and no. of oscillations per second of the pendulum when supported at B.
1
119·81
0·475
221·1
g
n
L
== =
ππ
∴ No. of oscillations per minute
= 0·475 × 60 = 28·5 Ans.
26.10. CONICAL PENDULUM
A conical pendulum, in its simplest form, consists of a heavy bob suspended at the end of a
light inextensible flexible string ; and the other end of the string is fixed at O as shown in Fig. 26.13.
The pendulum bob rotates about the vertical axis with a uniform angular velocity. A little consider-
ation will show, that the bob moves in a horizontal plane and describes a circle.
Let l= Length of the string,
ω= Angular velocity of the bob,
r= Radius of the horizontal circle
described by the bob,
h= Vertical distance of the bob from O,
θ= Inclination of the string with the
vertical
m= Mass of the bob in kg (such that its
weight is m.g. newtons).
We know that the pendulum is in equilibrium, when the bob
is at A. But when the bob is made to rotate, it will move outwards
due to centrifugal force. Now consider equilibrium of the bob at C.
We know that at this point, the forces acting on it are :

1. Weight (equal to mg) of the bob acting downwards in
   newtons.


2. Tension in the string (equal to T) in newtons.


3. *Centrifugal force (P equal to mω^2 r) in newtons.

Fig. 26.13. Conical pendulum.

* For details, please refer to Art. 28·5