Question

A circular disk of mass $M$ and radius $R$ is rotating clockwise with a uniform angular velocity $\omega$ about an axis passing through the centre, normal to the disk. At time $t = 0$, a torque $T$ is applied along the same axis to oppose the rotation of the disk. What is the angular displacement $\theta$ (measured from $t = 0$ in the clockwise direction) that the disk attains before it starts rotating counterclockwise?

• A. $\theta = \frac{\omega^2 M R^2}{4T}$
• B. $\theta = \frac{\omega^2 M R^2}{8T}$
• C. $\theta = -\frac{\omega^2 M R^2}{4T}$
• D. $\theta = -\frac{\omega^2 M R^2}{8T}$

Hint:

Using the work-energy theorem for rotation: the work done by the retarding torque equals the change in rotational kinetic energy.
\[ W = \tau \theta = -T \theta \]
\[ \Delta K_{\text{rot}} = 0 - \frac{1}{2} I \omega^2 \implies -T \theta = -\frac{1}{4} M R^2 \omega^2 \]
This gives $\theta = \frac{\omega^2 M R^2}{4T}$ in one simple step!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
This question requires calculating the angular displacement of a rotating disk brought to rest by a constant retarding torque.
Step 2: Key Formulas and Approach:
1. Moment of inertia of a uniform circular disk of mass $M$ and radius $R$ about its central perpendicular axis:
\[ I = \frac{1}{2} M R^2 \]
2. Torque-angular acceleration relation:
\[ \tau = I \alpha \implies \alpha = \frac{\tau}{I} \]
3. Rotational kinematic equation relating angular velocities, angular acceleration, and displacement:
\[ \omega_f^2 = \omega_i^2 + 2\alpha \theta \]
Step 3: Detailed Explanation:

Let the clockwise direction be positive. The initial angular velocity is $\omega_i = \omega$.

The retarding torque $T$ opposes the clockwise rotation, so the applied torque is negative: $\tau = -T$.

The moment of inertia of the disk is:
\[ I = \frac{1}{2} M R^2 \]

This torque produces a constant angular acceleration (deceleration) $\alpha$:
\[ \alpha = \frac{-T}{I} = \frac{-T}{\frac{1}{2} M R^2} = -\frac{2T}{MR^2} \]

The disk will temporarily come to a halt before reversing its direction of rotation. Thus, at the maximum clockwise angular displacement, the final angular velocity is $\omega_f = 0$.

Using the rotational equation of motion:
\[ \omega_f^2 = \omega_i^2 + 2\alpha \theta \]

Substituting the known values:
\[ 0^2 = \omega^2 + 2 \left( -\frac{2T}{MR^2} \right) \theta \]
\[ 0 = \omega^2 - \frac{4T}{MR^2} \theta \]
\[ \theta = \frac{\omega^2 MR^2}{4T} \]

Step 4: Final Answer:
The angular displacement attained by the disk is $\theta = \frac{\omega^2 M R^2}{4T}$, which matches Option (A).