Question

Figure shows a solid uniform cylinder of radius $R$ and mass $M$, which is free to rotate about a fixed horizontal axis O which passes through center of the cylinder. One end of an ideal spring of force constant $k$ is fixed and the other end is hinged to the cylinder at A. Distance OA is equal to $\frac{R}{2}$. An inextensible thread is wrapped round the cylinder and passes over a smooth, small pulley. A block of equal mass $M$ and having cross sectional area $A$ is suspended from free end of the thread. The block is partially immersed in a non-viscous liquid of density $\rho$. If in equilibrium, spring is horizontal and line OA is vertical, calculate frequency of small oscillations of the system.

• A. $\frac{1}{2\pi }\sqrt{\frac{k+4\rho Ag}{6M}}$
• B. $\frac{1}{2\pi }\sqrt{\frac{k-4\rho Ag}{6M}}$
• C. $\frac{1}{2\pi }\sqrt{\frac{k+2\rho Ag}{6M}}$
• D. $\frac{1}{2\pi }\sqrt{\frac{k-2\rho Ag}{6M}}$

Answer 1

The correct option is A $\frac{1}{2\pi }\sqrt{\frac{k+4\rho Ag}{6M}}$

In equilibrium,
$T+F=Mg$ (i)
When the block is further depressed by $x$, weight $Mg$ remains unchanged, upthrust $F$ increases by $\rho Axg$ and let $\Delta T$ be the increase in tension.
If $a$ is the acceleration of block then,
$\Delta T+\rho Axg=Ma$ (ii)
Restoring torque on the cylinder,
$\tau =[\frac{kx}{2}\frac{R}{2}-\Delta TR]=[\frac{kxR}{4}-(Ma-\rho Axg)R]$
$\frac{1}{2}M{R}^2\alpha =[\frac{k{R}^2\theta }{4}-(MR\alpha -\rho AgR\theta )R]$
or $\frac{3}{2}M{R}^2\alpha =[\frac{k{R}^2}{4}+\rho Ag{R}^2]\theta$
or $\alpha =\frac{-[\frac{k}{4}+\rho Ag]}{\frac{3}{2}M}\theta$
Here negative sign has been used for restoring nature of torque.
$\therefore$ $f=\frac{1}{2\pi }\sqrt{\left|\frac{\alpha }{\theta }\right|}$
$=\frac{1}{2\pi }\sqrt{\frac{k+4\rho Ag}{6M}}$