question_answer

A chain of length t is placed on a smooth Spherical surface of radius \[R\] with one of its ends fixed at the top of the sphere. What will be the acceleration of the each element of the chain when its upper end is released? It is assumed that the length of the chain \[\ell <\left( \frac{\pi R}{2} \right)\]

A) \[\frac{gR}{l}\]                          

B) \[\frac{gR}{l}\left[ 1-\cos \left( \frac{1}{R} \right) \right]\]

C) \[\frac{gR}{l}\left[ 1-\cos \left( \frac{R}{1} \right) \right]\]

D) \[\frac{2l}{R}\]

Correct Answer: B

Solution :

The mass of the element of angular width  \[d\theta dm=\frac{m}{{{\theta }_{0}}}d\theta ,\]\[\therefore {{F}_{t}}=\int\limits_{0}^{{{\theta }_{0}}}{(dm)g\,sin}\theta =\frac{mg}{{{\theta }_{0}}}\int\limits_{0}^{{{\theta }_{0}}}{\sin \theta d\theta }\]

=\[\frac{mg}{{{\theta }_{0}}}|-\cos \theta |_{0}^{{{\theta }_{0}}}\]

=\[\frac{mg}{\ell /R}\left[ 1-\cos \left( \frac{\ell }{R} \right) \right]\]

Tangential acceleration,

\[{{a}_{t}}=\frac{{{F}_{t}}}{m}=\frac{gR}{\ell }\left[ 1-\cos \left( \frac{\ell }{R} \right) \right]\].