Question

A rod of length \(l\) is held vertically stationary with its lower end located at a point \(P\) on the horizontal plane. When the rod is released to topple about \(P\), the velocity of the upper end of the rod with which it hits the ground is

• A. \(\sqrt{\dfrac{g}{l}}\)
• B. \(\sqrt{3gl}\)
• C. \(\sqrt{\dfrac{3g}{l}}\)
• D. \(\sqrt{\dfrac{gl}{3}}\)

Hint:

For a falling rod about one end: use energy conservation and \(I=\frac{1}{3}ml^2\). Then \(v=\omega l\).

Answer 1

The Correct Option is B

Step 1: Use energy conservation.
Rod rotates about lower end \(P\) without slipping.
When released from vertical, centre of mass falls by height:
\[ \Delta h = \frac{l}{2} \]
Loss in potential energy:
\[ \Delta U = mg\frac{l}{2} \]
Step 2: Convert into rotational kinetic energy.
Rotational KE about point \(P\):
\[ KE = \frac{1}{2}I_P\omega^2 \]
Moment of inertia of rod about end:
\[ I_P = \frac{1}{3}ml^2 \]
So:
\[ mg\frac{l}{2} = \frac{1}{2}\left(\frac{1}{3}ml^2\right)\omega^2 \]
Step 3: Solve for angular velocity \(\omega\).
Cancel \(m\):
\[ g\frac{l}{2} = \frac{1}{6}l^2\omega^2 \]
\[ 3gl = l^2\omega^2 \Rightarrow \omega^2 = \frac{3g}{l} \Rightarrow \omega = \sqrt{\frac{3g}{l}} \]
Step 4: Velocity of upper end.
Upper end is at distance \(l\) from pivot:
\[ v = \omega l = l\sqrt{\frac{3g}{l}} = \sqrt{3gl} \]
Final Answer:
\[ \boxed{\sqrt{3gl}} \]