Draw a neat labelled diagram of conical pendulum. State the expression for its periodic time in terms of length.

#### Solution

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),

v^{2} = 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")`