question_answer

A disc of mass \[M=2m\]and radius R is pivoted at its centre. The disc is free to rotate in the vertical plane about its horizontal axis through its centre O. A particle of mass m is stuck on the periphery of the disc. Find the frequency of small oscillations of the system about its equilibrium position.

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 \[\theta \])             \[I\alpha =\tau \]             \[\left[ \frac{1}{2}(2m){{R}^{2}}+m{{R}^{2}} \right]\alpha =-mgR\theta \]             \[\therefore \,\,\,a=-\left( \frac{g}{2R}\theta  \right)\]           \[\therefore \,\,\,\,f=\frac{1}{2\pi }\sqrt{\frac{g}{2R}}\]