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