Question

A uniform metre scale of length $1\text{m}$ is balanced on a fixed semi-circular cylinder of radius $30\text{cm}$ as shown in figure. One end of the scale is slightly depressed and released. The time period (in seconds) of the resulting simple harmonic motion is $(\text{Take}g=10{\text{ms}}^{-2})$

• A. $\pi$
• B. $\frac{\pi }{2}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{4}$

Answer 1

The correct option is C $\frac{\pi }{3}$

Let the scale be displaced by a small angle $\theta$.


From the figure, the magnitude of the restoring torque about $A$ = force $\times$ perpendicular distance.
${\tau }_{rest}=mg\times AB=mg\times R\sin \theta$
Since $\theta$ is small, $\sin \theta \simeq \theta$.
$\Rightarrow I\frac{d^2\theta }{d{t}^2}=-mg\times R\theta$
$\Rightarrow \frac{d^2\theta }{d{t}^2}=(-\frac{mgR}{I})\theta$
Comparing with SHM equation $\alpha =-{\omega }^2\theta$,
we get $\omega =\sqrt{\frac{mgR}{I}}$
$\Rightarrow \frac{2\pi }{T}=\sqrt{\frac{mgR}{I}}$
$\Rightarrow T=2\pi \sqrt{\frac{I}{mgR}}$
$\Rightarrow T=\frac{\pi L}{\sqrt{3gR}}[\because I=\frac{m{L}^2}{12}]$
Using the values $L=1\text{m},g=10{\text{ms}}^{-2}$ and $R=0.3\text{m}$, we get
$T=\frac{\pi }{3}$ sec.
Hence, option $\left(c\right)$ is the correct answer.