A uniform rod of length l and mass m is free to rotate in a vertical plane about A. The rod initially in horizontal position is released. The initial angular acceleration of the rod is : (Moment of inertia of rod about A is $\bigg(\frac{ml^2}{3}\bigg)$

Question

A charge of 2 $\mu C$ is placed in an electric field of intensity $ 4 \times 10^3 \, \text{N/C} $. What is the force experienced by the charge?

Answer 1

In this problem, we are dealing with a uniform rod of length $ l $ and mass $ m $ which is free to rotate about point $ A $ in a vertical plane. The task is to determine the initial angular acceleration of the rod given that the moment of inertia about $ A $ is $ \frac{ml^2}{3} $.

Let's analyze the forces acting on the rod when it is released from a horizontal position:

The gravitational force $ mg $ acts vertically downward at the center of mass of the rod, which is at a distance $ \frac{l}{2} $ from point $ A $.
Since the rod is free to rotate about point $ A $, we can apply the equation of rotational motion to find the initial angular acceleration $ \alpha $.

The torque $ \tau $ about point $ A $ due to the gravitational force is given by:

$\tau = mg \cdot \frac{l}{2}$

According to Newton's second law for rotation, the torque is also equal to the moment of inertia times angular acceleration, $ I \alpha $:

$\tau = I \alpha$

Substituting the given moment of inertia and torque:

$mg \cdot \frac{l}{2} = \left(\frac{ml^2}{3}\right) \alpha$

Simplifying this equation for angular acceleration $ \alpha $:

$\alpha = \frac{mg \cdot \frac{l}{2}}{\frac{ml^2}{3}}$

Finally, simplify the expression:

$\alpha = \frac{3g}{2l}$

Hence, the initial angular acceleration of the rod is $\frac{3g}{2l}$, which corresponds to the first option provided. This is the correct answer.

Answer 2