Question

A uniform disc of radius $ r $ is rotating about an axis passing through its diameter with angular speed 800 rpm. A torque of magnitude $ 25\pi \, \text{Nm} $ is applied on the disc for 40 sec. If the final angular speed of the disc is 2100 rpm, find the diameter of the disc if its mass is 1 kg.

• A. \( \frac{40}{3} \)
• B. \( 40 \sqrt{\frac{3}{13}} \)
• C. \( 20 \sqrt{\frac{2}{13}} \)
• D. \( 10 \sqrt{\frac{3}{2}} \)

Hint:

In problems involving angular motion, use the relationship between torque, moment of inertia, and angular acceleration. Make sure to convert angular velocity to rad/s before applying the formulas.

Answer 1

The Correct Option is B

Step 1: Use the equation for angular velocity.
We are given that the initial angular speed is 800 rpm and the final angular speed is 2100 rpm.
The equation for angular velocity is given by: \[ \omega_f = \omega_i + \alpha t \] where \( \omega_f \) is the final angular velocity, \( \omega_i \) is the initial angular velocity, \( \alpha \) is the angular acceleration, and \( t \) is the time. Converting rpm to rad/s: \[ \omega_i = 800 \times \frac{2\pi}{60} = \frac{16\pi}{3} \, \text{rad/s}, \quad \omega_f = 2100 \times \frac{2\pi}{60} = 35\pi \, \text{rad/s} \] Now, substitute the values into the equation: \[ 35\pi = \frac{16\pi}{3} + \alpha \times 40 \] Solving for \( \alpha \): \[ \alpha = \frac{13\pi}{12} \, \text{rad/s}^2 \]
Step 2: Use the equation for torque.
We are given a torque of \( \tau = 25\pi \, \text{Nm} \).
The torque is related to the angular acceleration and moment of inertia \( I \) by: \[ \tau = I \alpha \] For a disc rotating about its diameter, the moment of inertia is: \[ I = \frac{1}{4} m r^2 \] Substitute the values into the torque equation: \[ 25\pi = \frac{1}{4} m r^2 \times \frac{13\pi}{12} \] Simplifying: \[ \frac{25\pi}{13\pi} = \frac{m r^2}{4 \times 12} \] \[ \frac{25}{13} = \frac{m r^2}{48} \] Since \( m = 1 \, \text{kg} \), this simplifies to: \[ \frac{25}{13} = \frac{r^2}{48} \] Solving for \( r^2 \): \[ r^2 = \frac{25 \times 48}{13} = \frac{1200}{13} \] \[ r = \sqrt{\frac{1200}{13}} \approx 10.53 \, \text{m} \]
Step 3: Calculate the Diameter.
The diameter \( D \) is twice the radius: \[ D = 2r = 2 \times 10.53 \approx 20.87 \, \text{m} \] Thus, the correct answer is \( \boxed{40 \sqrt{\frac{3}{13}}} \).