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Question: In a simple pendulum, the breaking strength of the string is double the weight of the bob. The bob i...

In a simple pendulum, the breaking strength of the string is double the weight of the bob. The bob is released from rest when the string is horizontal. The string breaks when it makes an angle θ\theta with the vertical

A

θ 6mu= 6mucos⁡−1(1/3)\theta\mspace{6mu} = \mspace{6mu}\cos^{- 1}(1/3)

B

θ=60o\theta = 60^{o}

C

θ 6mu= 6mucos⁡−1(2/3)\theta\mspace{6mu} = \mspace{6mu}\cos^{- 1}(2/3)

D

θ 6mu= 6mu0o\theta\mspace{6mu} = \mspace{6mu} 0^{o}

Answer

θ 6mu= 6mucos⁡−1(2/3)\theta\mspace{6mu} = \mspace{6mu}\cos^{- 1}(2/3)

Explanation

Solution

Let the string breaks at point B.

Tension =mgcos⁡θ+mvB2r== mg\cos\theta + \frac{mv_{B}^{2}}{r} =Breaking strength

=mgcos⁡θ+mvB2r=2mg= mg\cos\theta + \frac{mv_{B}^{2}}{r} = 2mg ….(i)

If the bob is released from rest (from point A) then velocity acquired by it at point B

vB=2ghv_{B} = \sqrt{2gh}

vB=2grcos⁡θv_{B} = \sqrt{2gr\cos\theta} ....(ii) [As h= r cosθ\theta]

By substituting this value in equation (i)

mgcos⁡θ+mr(2grcos⁡θ)=2mgmg\cos\theta + \frac{m}{r}(2gr\cos\theta) = 2mgor 3mgcos⁡θ=2mg3mg\cos\theta = 2mg ⇒\Rightarrow cos⁡θ=23∴θ=cos⁡−1(23)\cos\theta = \frac{2}{3}\therefore\theta = \cos^{- 1}\left( \frac{2}{3} \right)