Question

Consider the rotation of a rod of mass $m$ and length $l$ from position AB to AB'. Which of the following statements are correct?

• A. Weight of the rod is lowered by $l/2$
• B. Loss of gravitational potential energy is $\frac{1}{2}mgl$
• C. Angular velocity is $\sqrt{3g/l}$
• D. Rotational kinetic energy is $\frac{m{l}^2{\omega }^2}{6}$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A Weight of the rod is lowered by $l/2$
B Loss of gravitational potential energy is $\frac{1}{2}mgl$
C Angular velocity is $\sqrt{3g/l}$
D Rotational kinetic energy is $\frac{m{l}^2{\omega }^2}{6}$

Centre of mass of the rods lie at C and C'

Let $w$ be the final angular velocity of the rod.

As $A{C}^{\prime }=\frac{l}{2}$, thus weight of the rod lowers by $\frac{l}{2}$.

$\Delta P.E=-mg\frac{l}{2}-0=-\frac{mgl}{2}$

Thus loss of gravitational P.E is $\frac{mgl}{2}$

Work-energy theorem, $\frac{mgl}{2}=\frac{1}{2}{I}_A{w}^2-0$

$\frac{mgl}{2}=\frac{1}{2}\times \frac{m{l}^2}{3}{w}^2$

$⟹w=\sqrt{\frac{3g}{l}}$

Rotational K.E, $E=\frac{1}{2}{I}_A{w}^2=\frac{1}{2}\frac{m{l}^2}{3}{w}^2$

$⟹E=\frac{m{l}^2{w}^2}{6}$