question_answer

A uniform disc of mass M and radius R is mounted on an axle supported in friction less bearings. A light cord is wrapped around the rim of the disc and a steady downward pull T is exerted on the cord. The angular acceleration of the disc is:

A)  \[\frac{MR}{2T}\]

B)  \[\frac{2T}{MR}\]

C)  \[\frac{T}{MR}\]

D)  \[\frac{MR}{T}\]

Correct Answer: B

Solution :

 The torque exerted on the disc is given by \[\tau =TR\]             ...(1) Also        \[\tau =I\alpha \]                           ...(2) From eqs. (1) and (2), we get \[I\alpha =TR\] \[\alpha =\frac{TR}{I}\] Or \[\alpha =\frac{2TR}{M{{R}^{2}}}\] Or \[\alpha =\frac{2T}{mR}\]