Question

A thin uniform rod AB of mass '\(m\)' and length '\(l\)' is hinged at one end A to the ground level. Initially the rod stands vertically and is allowed to fall freely to the ground in the vertical plane. The angular velocity of the rod when its B end strikes the ground is
(\(g\) = acceleration due to gravity)

• A. \(\sqrt{\frac{2g}{l}}\)
• B. \(\sqrt{\frac{3g}{l}}\)
• C. \(\sqrt{\frac{mg}{l}}\)
• D. \(\sqrt{\frac{mg}{3l}}\)

Hint:

Use energy conservation with I = m l^2 / 3 about the hinge.

Answer 1

The Correct Option is B

Step 1: Energy Conservation:
The rod rotates about the hinge A. Its centre of mass is at \(l/2\) and falls through \(l/2\) in going from vertical to horizontal. The loss of potential energy becomes rotational kinetic energy.

Step 2: Write the Equation:
\[ mg\frac l2=\frac12I\omega^2,\qquad I=\frac{ml^2}3 \]

Step 3: Solve:
\[ mg\frac l2=\frac12\cdot\frac{ml^2}3\omega^2\Rightarrow\omega^2=\frac{3g}l \]
\[ \omega=\sqrt{\frac{3g}{l}} \]

Step 4: Check the Options:
Option (A) \(\sqrt{2g/l}\) comes from using \(I=\tfrac{ml^2}{2}\) or point-mass reasoning. Options (C) and (D) contain \(m\) in the answer, but \(\omega\) cannot depend on mass because \(m\) cancels. So (B) is correct.

Final Answer:
The angular velocity is \(\sqrt{3g/l}\), option (B). \[ \boxed{\text{(B) } \sqrt{\frac{3g}{l}}} \]