(^550) „„„„„ A Textbook of Engineering Mechanics
Resolving the forces horizontally and equating the same,
T sin θ = mω^2 r ...(i)
or
22
2
sin
mr mr
Tml
r
l
ωω
===ω
θ ⎛⎞
⎜⎟
⎝⎠
...(ii)
Now taking moments about O, and equating the same,
(mω^2 r) h = mgr
∴
2 g
h
ω=
We know that the periodic time (i.e. time taken by the bob for one revolution),
2
2
h
t
g
π
==π
ω
and now resolving the forces vertically and equating the same,
T cos θ = mg ...(iii)
or
cos
mg mg mgl
T
h h
l

θ
...(iv)
Now dividing equation (i) by (iii),
sin^2
cos
Tmr
Tmg
θω

θ
∴
2
tan
r
g
ω
θ= ...(v)
Example 26.18. A sphere of 2 kg mass is attached to an inextensible string of length 1·3 m,
whose upper end is fixed to the ceiling. The sphere is made to describe a horizontal circle of radius
0·5 m (i) Calculate the time taken by the bob for one revolution ; (ii) What is the tension in the string?
Solution. Given : Mass of the sphere (m) = 2 kg ; Length of the
string (l) = 1·3 m and radius of the horizontal circle (r) = 0·5 m
(i) Time taken by the bob for one revloution
We know that vertical distance between the bob and O (i.e AO),
h= (1·3) – (0·5)^22 = 1·2 m
and time taken by the bob for one revolution.
1· 2
22s
9·8
h
t
g
=π =π
= 2·2 s Ans.
(ii) Tension in the string
We also know that tension in the string,
29·81·3
21.2 N
1· 2
mgl
T
h
××
== = Ans.
Fig. 26.14.