Question

A rod PQ of mass M and length L is hinged at end P. The rod is kept horizontal by a massless string tied to point Q as shown in figure. When string is cut, the initial angular acceleration of the rod is

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

Answer 1

The Correct Option is A

 To determine the initial angular acceleration of the rod when the string is cut, we need to consider the torque and moment of inertia involved in the system.

1. **Concepts Involved:**

• The rod PQ is hinged at point P, and so it rotates about this point when the string is cut.
• The gravitational force acting on the rod will cause a torque about point P. This torque will be responsible for the angular acceleration of the rod.
• The moment of inertia of a rod of mass \(M\) and length \(L\) about an end is given by \(\frac{1}{3}ML^2\).

 

2. **Calculating Torque:**

• The gravitational force acts at the center of mass of the rod, which is at a distance \(\frac{L}{2}\) from point P.
• The torque \(\tau\) due to the gravitational force is given by: \(\tau = Mg \cdot \frac{L}{2}\)

 

3. **Relating Torque and Angular Acceleration:**

• Using the relation between torque and angular acceleration \(\alpha\): \(\tau = I\alpha\)
• Substitute the moment of inertia \(\frac{1}{3}ML^2\) and the torque \(Mg \cdot \frac{L}{2}\):
• \(Mg \cdot \frac{L}{2} = \left(\frac{1}{3}ML^2\right) \alpha\)
• Solve for \(\alpha\): \(\alpha = \frac{3g}{2L}\)

 

4. **Conclusion:**

• The initial angular acceleration of the rod is \(\frac{3g}{2L}\).
• This matches with option (a): \(\frac{3g}{2L}\).