Question

A uniform rod AB of mass 2 kg and length 30 cm is at rest on a smooth horizontal surface. An impulse of force 0.2 Ns is applied to end B. The time taken by the rod to turn through a right angle will be \[\frac{\pi}{x} \, s, \, \text{where} \, x = \_.\]

Answer 1

Correct Answer: 4

To address this problem, we must determine the duration required for the rod to rotate by a right angle following an impulse. This involves first calculating the angular velocity and subsequently determining the time needed for a rotation of \( \frac{\pi}{2} \) radians.

Step 1: Determine the Moment of Inertia

The rod is uniform. Its moment of inertia about an axis through its center, perpendicular to its length, is \( I = \frac{1}{12}ml^2 \). For rotation about an end, we apply the parallel axis theorem: \( I_{\text{end}} = I + md^2 \), where \( d = \frac{l}{2} \). Given \( m = 2\, \text{kg} \) and \( l = 0.3\, \text{m} \).

First, calculate \( I \):

\[I = \frac{1}{12}(2)(0.3)^2 = 0.003\, \text{kg m}^2\]

Next, calculate the moment of inertia about end B:

\[I_{\text{end}} = 0.003 + 2\left(\frac{0.3}{2}\right)^2 = 0.003 + 0.045 = 0.048\, \text{kg m}^2\]

Step 2: Find Angular Velocity

The angular impulse is defined as \( J = I_{\text{end}} \times \omega \). With an impulse \( J = 0.2\, \text{Ns} \), the angular velocity is:

\[\omega = \frac{J}{I_{\text{end}}} = \frac{0.2}{0.048} \approx 4.17\, \text{rad/s}\]

Step 3: Calculate Time to Rotate \(\frac{\pi}{2}\)

Assuming constant angular velocity (as no external torque acts after the impulse), the time \( t \) is calculated as:

\[t = \frac{\theta}{\omega} = \frac{\frac{\pi}{2}}{4.17} = \frac{\pi}{4.17}\, \text{s}\]

Step 4: Compare with Given Range

The provided time in the problem is \( \frac{\pi}{x} \, \text{s} \). Equating this with our calculated time to solve for \( x \):

\[\frac{\pi}{x} = \frac{\pi}{4.17}\]

This implies \( x = 4.17 \).

After rounding, \( x = 4 \).

Conclusion:
\( x = 4 \).