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

Given: - Mass of the rod: $m = 2 \, \text{kg}$ - Length of the rod: $L = 30 \, \text{cm} = 0.3 \, \text{m}$ - Impulse applied at end $B$: $J = 0.2 \, \text{Ns}$

Step 1: Calculating the Moment of Inertia

The moment of inertia of the rod about its center of mass is given by:

$I_{\text{cm}} = \frac{1}{12} m L^2$

Substituting the given values:

$I_{\text{cm}} = \frac{1}{12} \times 2 \times (0.3)^2$ $I_{\text{cm}} = \frac{1}{12} \times 2 \times 0.09 = \frac{0.18}{12} = 0.015 \, \text{kg} \times \text{m}^2$

Since the impulse is applied at the end of the rod, we use the parallel axis theorem to find the moment of inertia about point $B$:

$I_B = I_{\text{cm}} + m \left(\frac{L}{2}\right)^2$

Substituting the values:

$I_B = 0.015 + 2 \left(\frac{0.3}{2}\right)^2$ $I_B = 0.015 + 2 \times (0.15)^2$ $I_B = 0.015 + 2 \times 0.0225 = 0.015 + 0.045 = 0.06 \, \text{kg} \times \text{m}^2$

Step 2: Calculating the Angular Velocity

The angular impulse is related to the change in angular momentum by:

$J \times L = I_B \times \omega$

Rearranging to find $\omega$:

$\omega = \frac{J \times L}{I_B}$

Substituting the values:

$\omega = \frac{0.2 \times 0.3}{0.06}$ $\omega = \frac{0.06}{0.06} = 1 \, \text{rad/s}$

Step 3: Calculating the Time to Turn Through a Right Angle

The time taken to turn through a right angle ($\frac{\pi}{2}$ radians) is given by:

$t = \frac{\theta}{\omega} = \frac{\frac{\pi}{2}}{1} = \frac{\pi}{2} \, \text{s}$

Comparing with the given expression $\frac{\pi}{x} \, \text{s}$:

$x = 4$

Conclusion:

The value of $x$ is $4$.