Question

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

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

Answer 1

The correct option is C

$\frac{\pi }{3}$

Magnitude of the restoring torque = Force $\times$ Perpendicular distance
$=mg\times AB=mg\times Rsin\theta$
Since $\theta$ is small, $sin\theta \simeq \theta$, here $\theta$ is expressed in radian.
The equation of motion of the scale is $I\frac{d^2\theta }{d{t}^2}=-mgR\theta$
$\Rightarrow \frac{d^2\theta }{d{t}^2}=(-\frac{mgR}{I})\theta$
$\therefore \omega =\sqrt{\frac{mgR}{I}}\Rightarrow \frac{2\pi }{T}=\sqrt{\frac{mgR}{I}}$
$\Rightarrow T=2\pi \sqrt{\frac{I}{mgR}}$
Now, $I=\frac{m{L}^2}{12}$
Hence, $T=\frac{\pi L}{\sqrt{3gR}}$
Using the values L = 1 m, g $=10m{s}^{-2}$ and R = 0.3 m, we get $T=\frac{\pi }{3}$ s
Hence, the correct choice is (c).