Answer 1

Where, S: rigid support

T : tension in the string

l : length of string

h : height of support from bob

v : velocity of bob

r : radius of horizontal circle

θ: semi-vertical angle

mg : weight of bob

i) Consider a bob of mass m tied to one
end of a string of length ‘l’ and other
the end is fixed to a rigid support.

ii)  Let the bob be displaced from its mean
position and whirled around a
horizontal circle of radius ‘r’ with
constant angular velocity ω, then the
bob performs U.C.M.

iii) During the motion, a string is inclined to
the vertical at an angle θ as shown in
the above figure.

iv)  In the displaced position P, there are two
forces acting on the bob.
 a. The weight mg acting vertically
downwards.
 b. The tension T acting upward
along the string. 

v) The tension (T) acting in the string can
be resolved into two components:
a. T cos θ acting vertically upwards.
b. T sin θ acting horizontally towards
centre of the circle. 

vi) Vertical component T cos θ balances the
weight and horizontal component T sin θ
provides the necessary centripetal force.

∴ T cos θ = mg    . ........(1)

T sin θ = `"mv"^2/"r" = "mr"omega^2`  ....(2)

vii)  Dividing equation (2) by (1), 

tan θ = `"v"^2/"rg"`     ......(3)

Therefore, the angle made by the string with the vertical is  θ = `tan^-1 ("v"^2/"rg")`

Also, from equation (3),

v² = rg tan θ

∴ v = `sqrt ("rg" tan theta)`

The period, T = `(2pi"r")/"v" = (2pi"r")/sqrt("rg" "tan" theta) = 2pi sqrt ("r"/("g tan" theta))`

It can be seen that r - = l sin θ

T = `2pi sqrt (("l sin" theta)/("g tan" theta)) = 2pi sqrt (("l cos" theta)/"g")`

The period of a conical pendulum is `2pi sqrt (("l cos" theta)/"g")`