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

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 \(4\).