Two discs A and B are mounted coaxially on a vertical axle. The discs have moment of inertia I and 2 I respectively about the common axis. Disc A is imparted an initial angular velocity 2 \omega using the entire potential energy of a spring compressed by a distance x_1 . Disc B is imparted an angular velocity \omega by a spring having the same spring constant and compressed by a distance x_2 . Both the discs rotate in the clockwise direction. When disc B is brought in contact with disc A, they acquire a common angular velocity in time t. The average frictional torque on one disc by the other during this period is

Rotational Motion PYQ — JEE Advanced 2007 · Shift 1 · Q57 with Solution

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Correct answer: A

Initial angular velocities of the discs

From the question:

Disc $A$ has moment of inertia $I$ and is given angular velocity $2\omega$ .
Disc $B$ has moment of inertia $2I$ and is given angular velocity $\omega$ .

Both rotate in the same (clockwise) direction.

Common angular velocity after contact

When the two discs are brought into contact, friction acts between them. Since there is no external torque about the common axis, angular momentum is conserved.

Initial angular momentum: $L_i = I(2\omega) + 2I(\omega) = 2I\omega + 2I\omega = 4I\omega$

Let the common angular velocity be $\Omega$ .

Final angular momentum: $L_f = (I+2I)\Omega = 3I\Omega$

By conservation of angular momentum, $4I\omega = 3I\Omega$ $\Omega = \frac{4\omega}{3}$

Change in angular momentum of one disc

The frictional torque between the discs is internal to the two-disc system, but on each individual disc it changes angular momentum.

For disc $A$ :

Initial angular velocity $=2\omega$
Final angular velocity $=\dfrac{4\omega}{3}$

Change in angular momentum of disc $A$ :

= I\left(\frac{4\omega-6\omega}{3}\right) = -\frac{2I\omega}{3}$$ Magnitude of change: $$|\Delta L_A| = \frac{2I\omega}{3}$$ For disc $B$: - Initial angular velocity $=\omega$ - Final angular velocity $=\dfrac{4\omega}{3}$ Change in angular momentum of disc $B$: $$\Delta L_B = 2I\left(\frac{4\omega}{3}-\omega\right) = 2I\left(\frac{\omega}{3}\right) = \frac{2I\omega}{3}$$ So the magnitude is the same for both discs. 4. **Average frictional torque** Average torque is given by $$\tau_{\text{avg}} = \frac{\Delta L}{t}$$ Thus, magnitude of average frictional torque on one disc by the other is $$\tau_{\text{avg}} = \frac{2I\omega/3}{t} = \frac{2I\omega}{3t}$$ 5. **Match with options** This corresponds to: $$\boxed{\frac{2I\omega}{3t}}$$ which is **Option A**. 6. **Comparison with stored answer** Stored correct answer: **A** Our derived answer: **A** So they agree.

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View written solutionFree

Correct answer: A

Initial angular velocities of the discs

From the question:

Disc $A$ has moment of inertia $I$ and is given angular velocity $2\omega$ .
Disc $B$ has moment of inertia $2I$ and is given angular velocity $\omega$ .

Both rotate in the same (clockwise) direction.

Common angular velocity after contact

When the two discs are brought into contact, friction acts between them. Since there is no external torque about the common axis, angular momentum is conserved.

Initial angular momentum: $L_i = I(2\omega) + 2I(\omega) = 2I\omega + 2I\omega = 4I\omega$

Let the common angular velocity be $\Omega$ .

Final angular momentum: $L_f = (I+2I)\Omega = 3I\Omega$

By conservation of angular momentum, $4I\omega = 3I\Omega$ $\Omega = \frac{4\omega}{3}$

Change in angular momentum of one disc

The frictional torque between the discs is internal to the two-disc system, but on each individual disc it changes angular momentum.

For disc $A$ :

Initial angular velocity $=2\omega$
Final angular velocity $=\dfrac{4\omega}{3}$

Change in angular momentum of disc $A$ :

= I\left(\frac{4\omega-6\omega}{3}\right) = -\frac{2I\omega}{3}$$ Magnitude of change: $$|\Delta L_A| = \frac{2I\omega}{3}$$ For disc $B$: - Initial angular velocity $=\omega$ - Final angular velocity $=\dfrac{4\omega}{3}$ Change in angular momentum of disc $B$: $$\Delta L_B = 2I\left(\frac{4\omega}{3}-\omega\right) = 2I\left(\frac{\omega}{3}\right) = \frac{2I\omega}{3}$$ So the magnitude is the same for both discs. 4. **Average frictional torque** Average torque is given by $$\tau_{\text{avg}} = \frac{\Delta L}{t}$$ Thus, magnitude of average frictional torque on one disc by the other is $$\tau_{\text{avg}} = \frac{2I\omega/3}{t} = \frac{2I\omega}{3t}$$ 5. **Match with options** This corresponds to: $$\boxed{\frac{2I\omega}{3t}}$$ which is **Option A**. 6. **Comparison with stored answer** Stored correct answer: **A** Our derived answer: **A** So they agree.

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