Question

A solid cylinder of mass $M$ and radius $R$ is connected to a spring as shown in the figure. It is placed on a rough horizontal surface. All the parts, except the wall are displaced slightly from their mean positions and released, so that they perform pure rolling back and forth about their equilibrium. Determine the time period of oscillation.
[Assume there is no slipping between the cylinder and the surface]

• A. $2\pi \sqrt{\frac{M}{k}}$
• B. $2\pi \sqrt{\frac{3M}{2k}}$
• C. $2\pi \sqrt{\frac{2M}{k}}$
• D. None of these

Answer 1

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

Let us say when displaced, the axis of cylinder is at a distance $x$ from its mean position, its velocity of centre of mass be $v$ and angular speed be $\omega$.
Since, the cylinder is in pure rolling motion, speed of the cylinder $v=R\omega$. From this we can say that, the spring elongates by the same amount $x$.
So, the whole system oscillates with angular frequency $\left(\omega \right)$.
Total energy of oscillation is
$E=\frac{I{\omega }^2}{2}+\frac{M{v}^2}{2}+\frac{k{x}^2}{2}$
Moment of inertia of solid cylinder about its axis $I=\frac{M{R}^2}{2}$,
Using this along with $v=R\omega$,
$E=\frac{3}{4}M{v}^2+\frac{k{x}^2}{2}$
$\Rightarrow v^2+\left(\frac{2}{3}\frac{k}{M}\right)x^2=\text{constant}$
Differentiating with respect to $x$ on both sides,
$2v\frac{dv}{dx}+2\left(\frac{2}{3}\frac{k}{M}\right)x=0\Rightarrow a=-\frac{2k}{3M}x$
Comparing this with $a=-{\omega }^{2}x$, we get
$\omega =\sqrt{\frac{2k}{3M}}$
Time period of oscillation $T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{3M}{2k}}$
Thus, option (b) is the correct answer.