Question

A circular disk of radius \( R \) meter and mass \( M \) kg is rotating around the axis perpendicular to the disk. An external torque is applied to the disk such that \( \theta(t) = 5t^2 - 8t \), where \( \theta(t) \) is the angular position of the rotating disk as a function of time \( t \). How much power is delivered by the applied torque, when \( t = 2 \) s?

• A. \( 60 M R^2 \)
• B. \( 72 M R^2 \)
• C. \( 108 M R^2 \)
• D. \( 8 M R^2 \)

Hint:

Power delivered by torque can be calculated as \( P = \tau \cdot \omega \), where \( \tau \) is the torque and \( \omega \) is the angular velocity.

Answer 1

The Correct Option is A

The power delivered by the applied torque is given by: \[ P = \tau \cdot \omega, \] where \( \tau \) is the torque and \( \omega \) is the angular velocity. We know that: \[ \omega = \frac{d\theta}{dt}. \] Differentiating \( \theta(t) = 5t^2 - 8t \), we get: \[ \omega(t) = 10t - 8. \] At \( t = 2 \, \text{s} \), \( \omega(2) = 10 \times 2 - 8 = 12 \, \text{rad/s}. \) Now, the torque is related to the angular acceleration \( \alpha \) by: \[ \tau = I \cdot \alpha, \] where \( I \) is the moment of inertia of the disk. For a disk rotating about its central axis: \[ I = \frac{1}{2} M R^2. \] The angular acceleration \( \alpha \) is: \[ \alpha = \frac{d\omega}{dt} = 10. \] Thus, the torque at \( t = 2 \, \text{s} \) is: \[ \tau = \frac{1}{2} M R^2 \cdot 10. \] Finally, the power is: \[ P = \tau \cdot \omega = \frac{1}{2} M R^2 \cdot 10 \cdot 12 = 60 M R^2. \]