Question

A uniform square lamina of side 2a is hung up by one corner and oscillates in its own plane which is vertical. Find the length of the equivalent simple pendulum.

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

The situation is shown in the figure .Let's firs calculate the moment of interia about 0

$I_0=I_c+m(\sqrt{2}A{)}^2$[$\perp$ axis theorem]

$\Rightarrow I_0=I_c+2m{a}^2$

Since the object is plannar and $I_c$ is the moment of intertia of the square about an axis perpendicular to its

plane passing through C.Using parallel axis theorem one can write

$I_c=\frac{m(2a{)}^2}{12}+\frac{m(2a{)}^2}{12}$

$I_c=\frac{2m{a}^2}{3}$

$\therefore \frac{2m{a}^2}{3}=\frac{8m{a}^2}{3}$

Now Torque acting on the plane ,$\tau =-mgsin\theta .\left(\sqrt{2}a\right)$

$\approx -mg\sqrt{2}a\theta$

We known,

${\tau }_0=I_0\alpha$