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. If we hang a body of mass $m$ with the cord, the tangential acceleration of the disc will be

• A. $\frac{mg}{M+m}$
• B. $\frac{mg}{M+2m}$
• C. $\frac{2mg}{M+2m}$
• D. $\frac{M+2m}{2mg}$

Answer 1

The correct option is C $\frac{2mg}{M+2m}$

Let the tangential acceleration of the disc be a.
Therefore acceleration of mass m is a.
Force Balance om mass 'm'
$mg-T=maT=m(g-a)$
Torque Balance on mass 'M'
$Tr=I\alpha where,I=\frac{M{r}^2}{2}\&a=\alpha rm(g-a)r=\frac{Mra}{2}2mg=(M+2m)aa=\frac{2mg}{M+2m}$
Please note that gravitational force on mass 'M' is supported by the bearings.