Question

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. $\frac{MR}{T}$

Answer 1

Answer: (B)

$\frac{2T}{MR}$

Answer 2

The torque exerted on the disc is given by $\tau=T R \ldots$(1) $\tau=I \alpha \ldots$(2) From eqs. (1) and (2), we get $I \alpha =T R$ $\alpha =\frac{T R}{I}$ or $\alpha=\frac{2 T R}{M R^{2}}$ or $\alpha=\frac{2 T}{m R}$