Question

A solid cylinder attached to horizontal massless spring can roll without slipping along horizontal surface. Find time period of oscillation.

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

Answer 1

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

The total energy of the system (rotational and translational kinetic energy + potential energy) remains constant
$K{E}_{trans.}=\frac{1}{2}M{\left(\frac{dx}{dt}\right)}^2$
and $K{E}_{rot.}=\frac{1}{2}I{\left(\frac{d\theta }{dt}\right)}^2$
and $PE=\frac{1}{2}k{x}^2$

$E_{total.}=\frac{1}{2}M{\left(\frac{dx}{dt}\right)}^2+\frac{1}{2}\left(\frac{1}{2}M{R}^2\right)$
${\left(\frac{1}{R}\frac{dx}{dt}\right)}^2+\frac{1}{2}k{x}^2$
$\frac{d}{dt}(KE+PE)=0$
or $(\frac{3}{2}M\frac{d^{2}x}{d{t}^2}+kx)\frac{dx}{dt}=0$
since $\frac{dx}{dt}$ is not zero, the equation becomes
$\frac{3}{2}M\frac{d^{2}x}{d{t}^2}+kx=0$
So, $\omega =\sqrt{\frac{2}{3}\frac{k}{M}}\therefore T=2\pi \sqrt{\frac{3M}{2k}}$