Question

A uniform disc of mass $M$ and radius $R$ is released from the shown position. $PQ$ is a string, $OP$ is a horizontal line, $O$ is the centre of the disc and distance $OP$ is $\frac{R}{2}$. Then tension in the string just after the disc is released will be:

• A. $\frac{Mg}{2}$
• B. $\frac{Mg}{3}$
• C. $\frac{2Mg}{3}$
• D. None of these

Answer 1

The correct option is C

$\frac{2Mg}{3}$

Let a is the cceleration of center of mass & α is the angular acceleration of body about center of mass. Applying Newton's law on centre of mass O

$Mg-T=ma$ {$a=$acceleration of centre of mass}
$\tau =I\alpha$, about centre of mass
$T\frac{R}{2}=\frac{M{R}^2}{2}\alpha$
also $a=\frac{R}{2}\alpha$ $[\therefore$ The acceleration of point $P$ along the length of string solving would be zero$]$
solving above equation we get, $T=\frac{2mg}{3}$