Question

A thin uniform rod (\(X\)) of mass \(M\) and length \(L\) is pivoted at a height \( \left(\dfrac{L}{3}\right) \) as shown in the figure. The rod is allowed to fall from a vertical position and lie horizontally on the table. The angular velocity of this rod when it hits the table top is ________. (\(g\) = gravitational acceleration) 

• A. \( \sqrt{\dfrac{3g}{2L}} \)
• B. \( \dfrac{3}{\sqrt{2}} \sqrt{\dfrac{g}{L}} \)
• C. \( \sqrt{\dfrac{3g}{L}} \)
• D. \( \dfrac{1}{\sqrt{2}} \sqrt{\dfrac{g}{L}} \)

Hint:

Always use the parallel axis theorem when rotation is about a point other than the centre of mass.

Answer 1

The Correct Option is C

Step 1: Determine the change in height of centre of mass.
The centre of mass of the rod is at its midpoint. Initially, the rod is vertical, and the centre of mass is at a height \[ \frac{2L}{3} - \frac{L}{2} = \frac{L}{6} \] above the pivot. When the rod becomes horizontal, the centre of mass is at the same level as the pivot.
Hence, the vertical drop of centre of mass is \[ h = \frac{L}{6} \]
Step 2: Use conservation of mechanical energy.
Loss in potential energy = Gain in rotational kinetic energy.
\[ Mg h = \frac{1}{2} I \omega^2 \]
Step 3: Moment of inertia of the rod about pivot.
Moment of inertia about centre is \[ I_{cm} = \frac{1}{12} ML^2 \] Distance of centre of mass from pivot is \( \frac{L}{6} \). Using parallel axis theorem,
\[ I = \frac{1}{12}ML^2 + M\left(\frac{L}{6}\right)^2 = \frac{1}{9}ML^2 \]
Step 4: Substitute values and solve.
\[ Mg\left(\frac{L}{6}\right) = \frac{1}{2}\left(\frac{1}{9}ML^2\right)\omega^2 \] \[ \omega^2 = \frac{3g}{L} \Rightarrow \omega = \sqrt{\frac{3g}{L}} \]