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 disc to prevent slipping of string over disc. At the shown instant extension in light spring is $\frac{3mg}{2K}$, where $m$ is mass of block, $g$ is acceleration due to gravity and $K$ is spring constant.

• A. Acceleration of block just after it is released is $\frac{g}{3}$
• B. Tension in the string continuously increases till extension in the spring reaches maximum value.
• C. maximum extension in the spring is $\frac{2mg}{K}$
• D. Maximum extension in the spring is $\frac{mg}{K}$

Answer 1

Answer: (A), (B), (C)

The correct options are
A Acceleration of block just after it is released is $\frac{g}{3}$
B Tension in the string continuously increases till extension in the spring reaches maximum value.
C maximum extension in the spring is $\frac{2mg}{K}$

By application of Newton's second law on block we get,
$T-mg=ma$ .... (1)
For disc, from torque equation
$\frac{3}{2}mgR-TR=\frac{m{R}^2}{2}\alpha ....\left(1\right)$
or
$\frac{3}{2}mgR-(mg+ma)R=\frac{m{R}^2}{2}\frac{a}{R}$
where $a$ $=$ $R$$a$ .....(3)
solving $a=$$\frac{g}{3}$
Also when you consider this as an SHM we see that mg/k is the extension at the equilibrium position and now the velocity is maximum. Thus the maximum extension would be there when the velocity is zero. Thus using energy balance we have $\frac{1}{2}k{x}^2=mgx$ (velocity at both the extremes is zero thus change in kinetic energy is zero).
Thus we get maximum extension from the unstretched length as $x=\frac{2mg}{k}$.