Question

A solid cylinder (radius $r=0.2\text{m}$) rolls without slipping in a cylindrical trough (radius $R=0.8\text{m}$). The time period of small oscillations is
[Take $g=10{\text{m/s}}^2$]

• A. $\frac{3\pi }{5}\text{sec}$
• B. $\frac{\pi }{5}\text{sec}$
• C. $\frac{2\pi }{5}\text{sec}$
• D. $\frac{\pi }{3}\text{sec}$

Answer 1

The correct option is A $\frac{3\pi }{5}\text{sec}$

Let us rotate solid cylinder from its stable equilibrium position by the small angle $\theta$. Let $v_c$ be the instantaneous speed of the COM $\left(C\right)$ of the solid cylinder.
Solid cylinder is in pure rolling, hence its angular velocity about its own centre $C$ is
$\omega =\frac{v_c}{r}......\left(1\right)$
Since $C$ moves along a circle of radius $(R-r)$, the speed of $C$ along the circular path can be written as
$v_c={\omega }^{\prime }(R-r)....\left(2\right)$
where ${\omega }^{\prime }$ is the angular velocity of cylinder about the centre of trough.
Thus from eqns. $\left(1\right)$ and $\left(2\right)$, we get
$\omega ={\omega }^{\prime }\frac{(R-r)}{r}......\left(3\right)$

Mechanical energy of oscillation of the solid cylinder is conserved, i.e. $E=K+U=$ constant
So, $\frac{1}{2}m{v}_c^2+\frac{1}{2}{I}_c{\omega }^2+mg(R-r)(1-\cos \theta )=\text{constant}$
(where $m$ is the mass of solid cylinder and $I_c$ is the moment of inertia of the solid cylinder about an axis passing through its COM and perpendicular to the plane of figure of the solid cylinder)
$\Rightarrow \frac{1}{2}m{\omega }^2{r}^2+\frac{1}{2}\frac{m{r}^2}{2}{\omega }^2+mg(R-r)(1-\cos \theta )=\text{constant}$
Using $\left(3\right)$ in the above equation, we get
$\frac{3}{4}m{\omega }^2(R-r{)}^2+mg(R-r\left)\right(1-\cos \theta )=\text{constant}$
Differentiating w.r.t. time on both sides,
$\frac{3}{2}m\omega (R-r{)}^2\alpha +mg\omega (R-r)\sin \theta =0$
Since $R\ne r$ and $\omega \ne 0$, we get
$\alpha =-\frac{2g}{3(R-r)}\sin \theta$
For small values of $\theta ,\sin \theta \approx \theta$
$\therefore \alpha =-\frac{2g}{3(R-r)}\theta$
Comparing this with $\alpha =-{\omega }_0^2\theta$, we get
${\omega }_0=\sqrt{\frac{2g}{3(R-r)}}$ is the angular frequency of oscillations.
Hence, time period $T=\frac{2\pi }{{\omega }_0}=2\pi \sqrt{\frac{3(R-r)}{2g}}$
From the data given in the question,
$R=0.8\text{m}$ and $r=0.2\text{m}$
$T=2\pi \sqrt{\frac{3\times (0.8-0.2)}{2\times 10}}=\frac{3\pi }{5}\text{sec}$
Thus, option (a) is the correct answer.