Question

A uniform disc of mass $M$ and radius $R$ is free to rotate about its fixed horizontal axis without friction. There is sufficient friction between the inextensible light string and the disc to prevent slipping of the string over the disc. At the shown instant, extension in the light spring is $\frac{3mg}{K}$, where $m$ is the mass of the block, $g$ is acceleration due to gravity and $K$ is spring constant. Then, the acceleration of block just after it is released is

• A. $\frac{2g}{3}$
• B. $\frac{4g}{3}$
• C. $\frac{5g}{3}$
• D. $\frac{g}{3}$

Answer 1

The correct option is B $\frac{4g}{3}$

For pulley disc, from torque equation
$kx\times R-TR=3mgR-TR=\frac{m{R}^2}{2}\alpha ....\left(1\right)$
By application of Newton's second law on block, we get,
$T-mg=ma.....\left(2\right)$
where $a=R\alpha ......\left(3\right)$
On solving (1), (2) and (3),
$a=\frac{4g}{3}$