Question

A uniform rod of length \( 2L \) is placed with one end in contact with the earth and is then inclined at an angle \( \alpha \) to the horizontal and allowed to fall without slipping at the contact point. When it becomes horizontal, its angular velocity will be:

• A. \( \sqrt{\frac{3g \sin \alpha}{2L}} \)
• B. \( \sqrt{\frac{2L}{3g \sin \alpha}} \)
• C. \( \sqrt{\frac{6g \sin \alpha}{L}} \)
• D. \( \sqrt{\frac{L}{g \sin \alpha}} \)

Hint:

For a freely rotating rigid body, use energy conservation to relate potential energy to rotational kinetic energy. The moment of inertia about the pivot is crucial in determining the angular velocity.

Answer 1

The Correct Option is A

Step 1: Understanding the Motion - The rod is hinged at one end and allowed to fall due to gravity.
- The rod rotates about the hinge point without slipping. - The goal is to find the angular velocity \( \omega \) when the rod reaches the horizontal position.
Step 2: Applying the Energy Conservation Principle Using conservation of energy: \[ \text{Initial Potential Energy} = \text{Final Rotational Kinetic Energy} \] Initial Potential Energy:
- The center of mass of the rod is located at a distance \( \frac{2L}{2} = L \) from the hinge.
- The vertical height of the center of mass from the horizontal position is: \[ h_{\text{initial}} = L \sin \alpha. \] - Thus, the initial potential energy is: \[ PE_{\text{initial}} = Mg h_{\text{initial}} = MgL \sin \alpha. \] Final Energy (Rotational Kinetic Energy): Since the rod rotates about a fixed hinge, its kinetic energy is purely rotational: \[ KE_{\text{rot}} = \frac{1}{2} I \omega^2. \]
Step 3: Moment of Inertia About the Hinge For a uniform rod of length \( 2L \) pivoted at one end, the moment of inertia about the hinge is: \[ I = \frac{1}{3} M (2L)^2 = \frac{4}{3} M L^2. \]
Step 4: Equating Energies By conservation of mechanical energy: \[ PE_{\text{initial}} = KE_{\text{rot}}. \] \[ MgL \sin \alpha = \frac{1}{2} \times \frac{4}{3} ML^2 \omega^2. \] Cancel \( M \) from both sides: \[ gL \sin \alpha = \frac{2}{3} L^2 \omega^2. \]
Step 5: Solving for \( \omega \) Rearrange the equation: \[ \omega^2 = \frac{3g \sin \alpha}{2L}. \] Taking the square root: \[ \omega = \sqrt{\frac{3g \sin \alpha}{2L}}. \] Thus, the correct answer is: \[ \boxed{\sqrt{\frac{3g \sin \alpha}{2L}}}. \]

Answer 2

Step 1: Describing the System’s Dynamics

- A uniform rod of length \( 2L \) is hinged at one end and falls under gravity.
- As it falls, it rotates about the hinge without slipping.
- We aim to calculate the angular velocity \( \omega \) when the rod becomes horizontal.

Step 2: Using the Principle of Energy Conservation

According to energy conservation:
\[ \text{Initial Potential Energy} = \text{Final Rotational Kinetic Energy} \]
- The center of mass of the rod lies at a distance \( L \) from the hinge (since the rod's length is \( 2L \)).
- Its vertical displacement from the final (horizontal) position is:
\[ h_{\text{initial}} = L \sin \alpha \]
- So, initial potential energy is:
\[ PE_{\text{initial}} = MgL \sin \alpha \]
The rod's kinetic energy is purely rotational as it pivots about a fixed hinge:
\[ KE_{\text{rot}} = \frac{1}{2} I \omega^2 \]
Step 3: Determining Moment of Inertia About the Pivot

- The moment of inertia of a uniform rod of length \( 2L \) about one end is:
\[ I = \frac{1}{3} M (2L)^2 = \frac{4}{3} M L^2 \]
Step 4: Applying Energy Conservation Equation

Equating potential energy and rotational kinetic energy:
\[ MgL \sin \alpha = \frac{1}{2} \cdot \frac{4}{3} M L^2 \omega^2 \]
Simplifying by canceling \( M \):
\[ gL \sin \alpha = \frac{2}{3} L^2 \omega^2 \]
Step 5: Solving for Angular Velocity \( \omega \)

Isolate \( \omega \):
\[ \omega^2 = \frac{3g \sin \alpha}{2L} \]
Take square root on both sides:
\[ \omega = \sqrt{\frac{3g \sin \alpha}{2L}} \]
Final Answer:
\[ \boxed{\sqrt{\frac{3g \sin \alpha}{2L}}} \]