Question

An $L$ shaped thin uniform rod of total length $2l$ is free to rotate in a vertical plane about a horizontal axis at $P$ as shown in the figure. The bar is released from rest. Neglect air and contact friction. The angular velocity at the instant it has rotated through ${90}^o$ and reached the dotted position shown is

• A. Zero
• B. $\sqrt{\frac{6g}{5l}}$
• C. $\sqrt{\frac{3g}{5l}}$
• D. none

Answer 1

Answer: (B)

$\sqrt{\frac{6g}{5l}}$

Moment of inertia of $L$ shaped rod about $P=I_{about}O+m{l}^2=\frac{2m{l}^2}{3}+2m{l}^2=\frac{8m{l}^2}{3}$

Where m is mass of individual rod.

Change in potential energy in $90°rotation=mgl$

$\therefore$ According to conservation of energy.

$mgl=\frac{1}{2}\frac{8m{l}^2}{3}.{\omega }^2\Rightarrow {\omega }^2=\frac{3}{4}\frac{g}{l}\Rightarrow =\sqrt{\frac{3g}{4l}}$