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 need to use the relationship between torque, angular velocity, and power. The power \( P \) delivered by the torque is given by: 

\(P = \tau \cdot \omega\)

where:

• \(\tau\) is the torque,
• \(\omega\) is the angular velocity.

First, we need to find the angular velocity \( \omega(t) \), which is the derivative of the angular position function \( \theta(t) \) with respect to time \( t \).

Given:

\(\theta(t) = 5t^2 - 8t\)

Find \(\omega(t)\):

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

Evaluating at \( t = 2 \) s:

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

Next, we find the torque \( \tau \) from the second derivative of \( \theta(t) \), which gives the angular acceleration \( \alpha(t) \).

The angular acceleration is:

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

The moment of inertia \( I \) of a circular disk rotating about an axis perpendicular to its plane is given by:

\(I = \frac{1}{2} M R^2\)

Torque is related to angular acceleration by:

\(\tau = I \cdot \alpha\)

Substituting the values:

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

Now we can calculate the power:

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

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

Answer 2

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: \[ P = \tau \cdot \omega = \frac{1}{2} M R^2 \cdot 10 \cdot 12 = 60 M R^2. \]