Question

When rod becomes horizontal find its angular velocity. It is pivoted at point A as shown. 

• A. $\sqrt{\dfrac{3g}{\ell}}$
• B. $\sqrt{\dfrac{2g}{\ell}}$
• C. $\sqrt{\dfrac{g}{\ell}}$
• D. $\sqrt{\dfrac{5g}{\ell}}$

Hint:

For rotating bodies, always calculate potential energy change of the center of mass.

Answer 1

The Correct Option is A

Step 1: Identifying the system.
A uniform rod of length $\ell$ is pivoted at point A. When released, it rotates under gravity.
Step 2: Change in potential energy.
The center of mass falls by a vertical distance of $\dfrac{\ell}{3}$.
\[ \Delta PE = mg \cdot \dfrac{\ell}{3} \]
Step 3: Rotational kinetic energy.
Moment of inertia of rod about point A is:
\[ I = \dfrac{1}{3}m\ell^2 \]
Rotational kinetic energy:
\[ KE = \dfrac{1}{2}I\omega^2 \]
Step 4: Applying energy conservation.
\[ mg\dfrac{\ell}{3} = \dfrac{1}{2}\cdot \dfrac{1}{3}m\ell^2\omega^2 \]
Step 5: Solving for angular velocity.
\[ \omega^2 = \dfrac{3g}{\ell} \Rightarrow \omega = \sqrt{\dfrac{3g}{\ell}} \]