question_answer

A uniform disc of mass M and radius R is mounted on an axle supported in frictionless 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 acceleratio- 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 (r) acting on body is given by                 \[\tau =T\,R\]                    ... (i)                 \[\tau =I\alpha \]                             ... (ii) where I is moment of inertia and a the angular acceleration. From Eqs. (i) and (ii), we get                 \[TR=I\alpha \] \[\Rightarrow \]               \[\alpha =\frac{TR}{I}\]                 Also,      \[I=\frac{M{{R}^{2}}}{2}\] \[\therefore \]  \[\alpha =\frac{2TR}{M{{R}^{2}}}=\frac{2T}{MR}\]