# A bob of mass m is suspended by inextensible string of length l from a fixed point. the bob is given a speed of √6gl at the bottom point when was in equilibrium. find the tension in the string when string deflects through an angle 120 degree from vertical.

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by anishkatareani

2014-10-15T01:55:50+05:30

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I am not sure if the angle 120° is the angle θ of the string with its initial position. or, θ = 60°.   please substitute accordingly in the formula.  It is not clear to me from the last sentence of the question.

Total energy = PE + KE when θ=0° initial vetical position.
= 0 + 1/2 m (√6gl)² = 1/2 m 6 gl = 3 mgl

Total energy when the string makes angle θ with initial position = PE + KE
= m g l (1 - cos θ) + 1/2 m v²

Energy is conserved.

1/2 m v² = 3 mg l - mgl (1- cos θ)  = 2 mg l + mgl cos θ

v²/l = 4 g + 2 g cos θ

Tension in the string T = centripetal force + component of weight pointing away from the string radially.
= m v² / l + m g cos θ
= (4 m g + 2 mg cos θ ) + mg cos θ

T =  mg ( 4 + 3 cos θ)

If θ = 120°, then   T = mg ( 4 + 3 cos 120 ) = 2.5 mg
If θ = 60°      then   T = mg ( 4 + 3/2 ) = 5.5 mg