Question

A uniform rod AB of mass \( m \) and length \( l \) is at rest on a smooth horizontal surface. An impulse \( P \) is applied to the end B. The time taken by the rod to turn through a right angle is

• A. \( \frac{\pi m l}{12 P} \)
• B. \( \frac{\pi P}{m l} \)
• C. \( \frac{2 \pi m l}{P} \)
• D. \( \frac{\pi P}{m l} \)

Hint:

For a rod under an impulse, use the rotational dynamics formulas for torque and moment of inertia to calculate the time taken for rotation.

Answer 1

The Correct Option is A

Step 1: Using rotational dynamics.
When an impulse \( P \) is applied to the end of the rod, it creates a torque that causes rotational motion. The torque is given by \( \tau = P \times l \). The angular acceleration \( \alpha \) is given by \( \alpha = \frac{\tau}{I} \), where \( I \) is the moment of inertia of the rod about the axis of rotation.
Step 2: Moment of inertia of the rod.
For a uniform rod rotating about one end, the moment of inertia is: \[ I = \frac{1}{3} m l^2 \] Step 3: Calculating time for a right angle rotation.
The time to rotate through \( 90^\circ \) (a right angle) is given by: \[ \theta = \frac{1}{2} \alpha t^2 \Rightarrow t = \frac{\pi m l}{12 P} \] Step 4: Conclusion.
The correct answer is (A), \( \frac{\pi m l}{12 P} \).