Question

A circular cylinder of mass m and radius r is connected by a spring of stiffness k on an inclined plane as shown in the figure below. If it is free to roll on the rough surface which is horizontal without slipping, then its natural frequency is

• A. $\sqrt{\frac{k}{m}}$
• B. $\sqrt{\frac{2k}{m}}$
• C. $\sqrt{\frac{k}{3m}}$
• D. $\sqrt{\frac{2k}{3m}}$

Answer 1

The correct option is D $\sqrt{\frac{2k}{3m}}$

Equation of motion

$m\ddot{x}+kx-F_t=\mathrm{0.....}\left(1\right)$

$F_t=\text{friction force}$

Also,

$I\ddot{\theta }=-F_t.r$

$\frac{1}{2}m{r}^2\ddot{\theta }=-F_t.r$

$\therefore F_t=\frac{-1}{2}m{r}^2\ddot{\theta }$

$=\frac{-1}{2}mr\frac{\ddot{x}}{r}$

$=\frac{-1}{2}m\ddot{x}$

Equation (1) can be written as

$m\ddot{x}+kx+\frac{1}{2}m\ddot{x}=0$

$\frac{3}{2}m\ddot{x}+kx=0$

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