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

To determine the power delivered by the applied torque, we use the relationship \( P = \tau \cdot \omega \), where \( P \) is power, \( \tau \) is torque, and \( \omega \) is angular velocity.

First, calculate the angular velocity \( \omega(t) \) by differentiating the angular position function \( \theta(t) = 5t^2 - 8t \) with respect to time \( t \).

\(\omega(t) = \frac{d\theta(t)}{dt} = \frac{d}{dt}(5t^2 - 8t) = 10t - 8\)

At \( t = 2 \) s, the angular velocity is:

\(\omega(2) = 10(2) - 8 = 20 - 8 = 12 \text{ rad/s}\)

Next, determine the torque \( \tau \). This requires the angular acceleration \( \alpha(t) \), which is the derivative of \( \omega(t) \).

The angular acceleration is:

\(\alpha(t) = \frac{d\omega(t)}{dt} = \frac{d}{dt}(10t - 8) = 10\)

For a circular disk rotating about an axis perpendicular to its plane, the moment of inertia is \( I = \frac{1}{2} M R^2 \).

Torque is calculated as \( \tau = I \cdot \alpha \).

Substituting the values:

\(\tau = (\frac{1}{2} M R^2) \cdot 10 = 5 M R^2\)

Finally, calculate the power:

\(P = \tau \cdot \omega = (5 M R^2) \cdot 12 = 60 M R^2\)

The power delivered by the applied torque at \( t = 2 \) s is \( 60 M R^2 \).