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 mass with the help of a cord, the tension in the cord is

• A. $\frac{mg}{M+m}$
• B. $\frac{mMg}{M+2m}$
• C. $\frac{2mMg}{Mmg}$
• D. none of these

Answer 1

Answer: (B)

$\frac{mMg}{M+2m}$

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'

$Tr=I\alpha where,I=\frac{M{r}^2}{2}\&a=\alpha rm(g-a)r=\frac{Mra}{2}2mg=(M+2m)aa=\frac{2mg}{M+2m}$

Tension from above equations:

$\therefore T=m(g-\frac{2mg}{M+2m})=\frac{mMg}{M+2m}$

Please note that gravitational force on mass 'M' is supported by the bearings.