Question

A uniform metre scale of length 1 m is balanced on a fixed semi – circular cylinder of radius 30 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 $(Takeg=10m{s}^{-2})$

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

The magnitude of the restoring
Torque = force $\times$ perpendicular distance
= mg $\times$ AB
= mg $\times$ R sin $\theta$
Since $\theta$ is small, sin $\theta$ here $\theta$ is expressed in radian.
The equation of motion of the scale is

$I\frac{d^2\theta }{d{t}^2}=-mgR\theta$
or,$\frac{d^2\theta }{d{t}^2}=(-\frac{mgR}{I})\theta$
$\therefore \omega =\sqrt{\frac{mgR}{I}}or\frac{2\pi }{T}=\sqrt{\frac{mgR}{I}}orT=2\pi \sqrt{\frac{I}{mgR}}.NowI=\frac{m{L}^2}{12}$
Hence $T=\frac{\pi L}{\sqrt{3gR}}$
Using the values L = 1 m, g = 10 $m{s}^{-2}$ and R = 0.3 m, get $T=\frac{\pi }{3}$ second. Hence the correct choice is (c)