Question

A uniform rod $A B$ of length $\ell$ and mass $m$ is free to rotate about point $A .$ The rod is released from rest in the horizontal position. Given that the moment of inertia of the rod about $A$ is $\frac{m \ell^{2}}{3}$, the initial angular acceleration of the rod will be:-

• A. $mg \frac{\ell}{2}$
• B. $\frac{3}{2} g \ell$
• C. $\frac{3 g}{2 \ell}$
• D. $\frac{2 g}{3 \ell}$

Answer 1

The Correct Option is C

To find the initial angular acceleration of the rod, we can use the principles of torque and rotational motion. Let's break down the solution step by step.

Step 1: Understanding the system

The rod is released from rest, and it rotates about point $A$. The force causing this rotation is the gravitational force acting at the center of mass of the rod, which is at a distance of $\frac{\ell}{2}$ from point $A$.

Step 2: Calculating the torque

Torque $(\tau)$ is given by the formula:

\[\tau = \text{{Force}} \times \text{{Perpendicular distance from axis of rotation}}\]

Here, the force is the weight of the rod, $mg$, and the distance is $\frac{\ell}{2}$. Thus, the torque is:

\[\tau = mg \times \frac{\ell}{2} = \frac{mg\ell}{2}\]

Step 3: Using the moment of inertia

The moment of inertia $(I)$ of the rod about point $A$ is given as $I = \frac{m \ell^2}{3}$.

Step 4: Calculating the angular acceleration

Using the relation between torque and angular acceleration $(\alpha)$:

\[\tau = I \alpha\]

We plug in the values:

\[\frac{mg\ell}{2} = \frac{m \ell^2}{3} \cdot \alpha\]

Simplifying for $\alpha$ gives:

\[\alpha = \frac{mg\ell}{2} \times \frac{3}{m\ell^2} = \frac{3g}{2\ell}\]

Conclusion

The initial angular acceleration of the rod is $\frac{3g}{2\ell}$. Hence, the correct option is:

$\frac{3g}{2\ell}$