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: Determine angular acceleration.
The initial angular speed is 800 rpm, and the final angular speed is 2100 rpm. The equation for angular velocity is \( \omega_f = \omega_i + \alpha t \).

Convert rpm to rad/s:
\[\omega_i = 800 \times \frac{2\pi}{60} = \frac{16\pi}{3} \, \text{rad/s}, \quad \omega_f = 2100 \times \frac{2\pi}{60} = 35\pi \, \text{rad/s}\]

Substitute known values into the angular velocity equation:\[35\pi = \frac{16\pi}{3} + \alpha \times 40\]

Solve for angular acceleration \( \alpha \):\[\alpha = \frac{13\pi}{12} \, \text{rad/s}^2\]

Step 2: Calculate the moment of inertia.
A torque of \( \tau = 25\pi \, \text{Nm} \) is applied. The relationship between torque, moment of inertia \( I \), and angular acceleration is \( \tau = I \alpha \).

For a disc rotating about its diameter, the moment of inertia is \( I = \frac{1}{4} m r^2 \).

Substitute values into the torque equation:\[25\pi = \frac{1}{4} m r^2 \times \frac{13\pi}{12}\]

Simplify the equation:\[\frac{25}{13} = \frac{m r^2}{48}\]

Given that mass \( m = 1 \, \text{kg} \), the equation becomes:\[\frac{25}{13} = \frac{r^2}{48}\]

Solve for \( r^2 \):\[r^2 = \frac{25 \times 48}{13} = \frac{1200}{13}\]

Calculate the radius \( r \):\[r = \sqrt{\frac{1200}{13}} \approx 10.53 \, \text{m}\]

Step 3: Compute the Diameter.
The diameter \( D \) is twice the radius:\[D = 2r = 2 \times 10.53 \approx 20.87 \, \text{m}\]

The correct answer is \( \boxed{40 \sqrt{\frac{3}{13}}} \).