Question

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 acceleration of the disc is

• A. $\frac {MR}{2T}$
• B. $\frac {2T}{MR}$
• C. $\frac {T}{MR}$
• D. $\frac {MR}{T}$

Answer 1

Answer: (B)

$\frac {2T}{MR}$

Answer 2

The torque exerted on the disc is given, by \hspace15mm \tau=TR \hspace15mm ...(i) Also \hspace15mm \tau=1 \alpha \hspace15mm ...(ii) From Eqs. (i) and (ii), we get \hspace15mm I\alpha=TR \hspace15mm \alpha=\frac {TR}{I}= \frac {2TR}{MR^2} or \hspace15mm \alpha=\frac {2T}{MR}