Question

In the arrangement shown, the solid cylinder of mass m is slightly rolled to the left and released. It starts oscillating on the horizontal surface without slipping. Then time period of oscillation is

• A. $\pi \sqrt{\frac{3K}{m}}$
• B. $2\pi \sqrt{\frac{3M}{2K}}$
• C. $2\pi \sqrt{\frac{2K}{3M}}$
• D. $2\pi \sqrt{\frac{3K}{2m}}$

Answer 1

The correct option is B $2\pi \sqrt{\frac{3M}{2K}}$

Torque=$f\left(R\right)$ $R\to$ Radius of cylinder

$I\propto =fR$ moment of inertia for solid cylinder

As no slipping $I=\frac{1}{2}m{R}^2$

$a=R\propto$

$\frac{1}{2}m{R}^2\left(\frac{a}{R}\right)=fR⟶\therefore f=\frac{1}{2}ma$

$ma=-kx-f=-kx-\frac{1}{2}ma$

$-kx=\frac{3}{2}ma\Rightarrow \frac{d^{2}x}{d{t}^2}+\frac{2}{3}\frac{k}{m}x=0$

It is of form $\frac{d^{2}x}{d{t}^2}+w^{2}x=0$

$\therefore w=\sqrt{\frac{2}{3}\frac{k}{m}}$

$T=\frac{2\pi }{w}=2\pi \sqrt{\frac{3}{2}\frac{m}{R}}$