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 as shown in figure below. If we hang a body of mass $m$ with the cord, it has acceleration $a$ as shown in figure. Find the angular acceleration of the disc $\left(\alpha \right)$.

• A. $\frac{2m(a+g)}{MR}$
• B. $\frac{2M(g-a)}{mR}$
• C. $\frac{2m(g-a)}{MR}$
• D. $\frac{2M(a+g)}{mR}$

Answer 1

The correct option is C $\frac{2m(g-a)}{MR}$

Let acceleration of the mass $m=a$ .
Therefore tangential acceleration of disc $a_t$ is also equal to $a$ due to string constraints.

F.B.D

Force balance on mass ${}^{\prime }{m}^{\prime }$
$mg-T=ma\Rightarrow T=m(g-a)...\left(i\right)$

Torque balance on disc,
$\tau =R\times T=I\alpha ...\left(ii\right)$
where, $I=\frac{M{R}^2}{2}$ for uniform disc.
From equation $\left(i\right)$ and $\left(ii\right)$
$R\times m(g-a)=\left(\frac{M{R}^2}{2}\right)\alpha$
So, angular acceleration $\alpha =\frac{R\times m(g-a)}{\left(\frac{M{R}^2}{2}\right)}=\frac{2m(g-a)}{MR}$