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 \[\alpha =\frac{2T}{mR}\]