A) \[\frac{1}{2\pi }\sqrt{\frac{g}{2R}}\]
B) \[\frac{1}{\pi }\sqrt{\frac{g}{R}}\]
C) \[\frac{1}{2\pi }\sqrt{\frac{g}{R}}\]
D) \[\frac{2}{\pi }\sqrt{\frac{g}{R}}\]
Correct Answer: A
Solution :
[a] In equilibrium the particle is at the lowest position. Consider the system at an angular position\[\theta \]. \[\tau =mgR\sin \theta \simeq mgR\theta \] (for small \[\theta \]) \[I\alpha =\tau \] \[\left[ \frac{1}{2}(2m){{R}^{2}}+m{{R}^{2}} \right]\alpha =-mgR\theta \] \[\therefore \,\,\,a=-\left( \frac{g}{2R}\theta \right)\] \[\therefore \,\,\,\,f=\frac{1}{2\pi }\sqrt{\frac{g}{2R}}\]You need to login to perform this action.
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