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Question
Answer the following question.
Show that its time period is given by, 2π`sqrt((l cos theta)/("g"))` where l is the length of the string, θ is the angle that the string makes with the vertical, and g is the acceleration due to gravity.
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Solution
Conical pendulum
Where,
O: 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
- Consider a bob of mass m tied to one end of a string of length ‘l’ and the other end is fixed to a rigid support.
- 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.
- During the motion, string is inclined to the vertical at an angle θ as shown in the figure above.
- In the displaced position, there are two forces acting on the bob.
a. The weight mg acting vertically downwards.
b. The tension T acting upward along the string. - 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. - Since there is no net force, the vertical component T cos θ balances the weight and the 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) - 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")` - Since we know v = `(2pi"r")/"T"`
∴ tan θ = `(4pi^2"r"^2)/("T"^2"rg")` ....[From (3)]
T = `2pi sqrt("r"/("g"tan theta))`
T = `2pi sqrt((l sin theta)/("g"tan theta)) ....[because "r" = l sin theta]`
T = `2pi sqrt((l cos theta)/("g"))`
T = `2pi sqrt("h"/"g")` .....(∵ h = l cos θ)
where l is length of the pendulum and h is the vertical distance of the horizontal circle from the fixed point O.
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