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The sphere at P is given a downward velocity $v_{0}$ and swings
in a vertical plane at the end of a rope of length l=1m
attached to a support at O. The rope breaks at angle $30^{0}$
from horizontal, knowing that it can withstand a maximum
tension equal to three times the weight of the sphere.
Then the value of $v_{0}$ will be :

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$\frac{g}{2} m s^{- 1}$

$\sqrt{\frac{2 g}{3}} m s^{- 1}$

$\sqrt{\frac{3 g}{2}} m s^{- 1}$

$\frac{g}{3} m s^{- 1}$

answer is C.

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Detailed Solution

when it travels $30^{0}$ from the initial position the loss in potential energy is mgl sin $30^{0}$ = mg/2 $(l = 1 m)$

$t e n s i o n i n t h e s t r i n g i n t h e d i s p l a c e d p o s i t i o n T = \frac{m v^{2}}{l} + m g \cos 60^{0} = 3 m g$

$m v^{2} = 3 m g - m g \cos 60^{0} = \frac{5 m g}{2}$ ( $(\frac{1}{2} m v^{2} = \frac{5 m g}{4})$

applying conservation of energy $\frac{1}{2} m v_{0}^{2} + \frac{m g l}{2} = \frac{1}{2} m v^{2} = \frac{5 m g}{4}$

$m v_{0}^{2} = \frac{5 m g}{2} - m g = \frac{3 m g}{2} v_{0} = \sqrt{\frac{3 g}{2}}$

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