A nail is located below the point of suspension of a simple pendulum of length ‘l’. The bob is released from horizontal position. If the bob loops a verticle circle with nail as centre, (l) find the distance of nail from point of suspension(2). Find the angle ‘ $θ$ ’ with the lower vertical at which the resultant acceleration of the bob is along the horizontal, when the pendulum is released from horizontal position.

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$\frac{3 l}{5}, {Tan}^{- 1} \sqrt{2}$

$\frac{2 l}{5}, {Tan}^{- 1} \sqrt{2}$

$\frac{3 l}{5}, {Tan}^{- 1} (2)$

$\frac{2 l}{3}, {Tan}^{- 1} (1 / 2)$

answer is D.

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Detailed Solution

For it to loop in a vertical circle the velocity of bob should be = $\sqrt{5 g R} w h e r e R i s t h e l e n g t h o f s t r i n g a f t e r l o o p i n g w i t h n a i l$

v is the velocity of the simple pendulum at the bottommost position

According to Q, v= $\sqrt{5 g R}$

Now, mgl= $\frac{1}{2} m v^{2}$

$\sqrt{2 g l} = \sqrt{5 g R} R = \frac{2}{5} l H e n c e, d i s \tan c e o f n a i l = \frac{3 l}{5} f r o m p o i n t o f s u s p e n s i o n (2) T a n g e n t i a l a c c e l e r a t i o n = g \sin θ C e n t r i p e t a l a c c e l e r a t i o n = \frac{v^{2}}{l} N o w, m g l = \frac{1}{2} m v^{2} + m g l (1 - \cos θ) v^{2} = 2 g l \cos θ T h u s c e n t r i p e t a l a c c l e r a t i o n = 2 g \cos θ N o w a c c o r d i n g t o q u e s t i o n; \tan θ = \frac{2 g \cos θ}{g \sin θ} \tan^{2} θ = 2 \tan θ = \sqrt{2}$