Question

A composite rod of mass $2m$ and length $2l$ comprises two identical rods joined end to end at $P$. The composite rod hinged at one of its end is kept horizontal as shown in the figure. If it is released from rest, find its angular speed when it becomes vertical.

• A. $\omega =\sqrt{\frac{3}{2}\frac{g}{l}}$
• B. $\omega =\sqrt{\frac{5}{2}\frac{g}{l}}$
• C. $\omega =\sqrt{\frac{7}{2}\frac{g}{l}}$
• D. $\omega =\sqrt{\frac{9}{2}\frac{g}{l}}$

Answer 1

The correct option is A $\omega =\sqrt{\frac{3}{2}\frac{g}{l}}$

Let the composite rod acquire an angular speed $\omega$ at the vertical position. Its centre of mass $C$ moves from $C_1$ to $C_2$. Therefore the potential energy $\left(gr\right)$ of the composite rod decreases by $2mgh$ here $h=l$. Applying conservation of mechanical energy at horizontal and vertical position.
Using $\Delta K=\Delta U=0$
$\Rightarrow \frac{1}{2}I{\omega }^2=\left(2m\right)gl$
where $I=MI$ of the composite rod about $O=\left(2m\right){\left(2l\right)}^2/3=8m{l}^2/3$
$\Rightarrow \omega =\sqrt{\frac{3}{2}\frac{g}{l}}$