Question

A thin uniform circular disc of mass \(\frac{10}{\pi^2}\) kg and radius 2 m is rotating about an axis passing through its centre and perpendicular to its plane. The work done to increase the angular speed of the disc from 90 rev/min to 120 rev/min is

• A. 35 J
• B. 70 J
• C. 140 J
• D. 210 J

Hint:

Always convert units to the standard SI system before calculation (e.g., rev/min to rad/s, g to kg). If your answer doesn't match any options, double-check for possible typos in the question's given values, especially with constants like \(\pi\).

Answer 1

The Correct Option is B

Step 1: Calculate Moment of Inertia (I): For a disc rotating about a perpendicular axis through the center: \( I = \frac{1}{2}MR^2 \). Mass \( M = \frac{10}{\pi^2} \, \text{kg} \). Radius \( R = 2 \, \text{m} \). \[ I = \frac{1}{2} \times \frac{10}{\pi^2} \times (2)^2 = \frac{1}{2} \times \frac{10}{\pi^2} \times 4 = \frac{20}{\pi^2} \, \text{kg m}^2 \]
Step 2: Convert Angular Velocities to rad/s: Initial speed \( \omega_1 = 90 \, \text{rpm} \). \[ \omega_1 = 90 \times \frac{2\pi}{60} = 3\pi \, \text{rad/s} \] Final speed \( \omega_2 = 120 \, \text{rpm} \). \[ \omega_2 = 120 \times \frac{2\pi}{60} = 4\pi \, \text{rad/s} \]
Step 3: Apply Work-Energy Theorem for Rotation: Work Done \( W = \Delta KE_{rot} = \frac{1}{2} I (\omega_2^2 - \omega_1^2) \). \[ W = \frac{1}{2} \left( \frac{20}{\pi^2} \right) \left[ (4\pi)^2 - (3\pi)^2 \right] \] \[ W = \frac{10}{\pi^2} [ 16\pi^2 - 9\pi^2 ] \] \[ W = \frac{10}{\pi^2} [ 7\pi^2 ] \] \[ W = 10 \times 7 = 70 \, \text{J} \]