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Question
Draw a neat labelled diagram of conical pendulum. State the expression for its periodic time in terms of length.
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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),
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")`
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