Question

A wheel is at rest in horizontal position. Its moment of inertia about vertical axis passing through its center is \( I \). A constant torque \( \tau \) acts on it for \( t \) seconds. The change in rotational kinetic energy is

• A. \( \frac{\tau^2 t^2}{2 I} \)
• B. \( \frac{\tau t}{2 I} \)
• C. \( \frac{\tau t}{I} \)
• D. \( \frac{\tau t^2}{I} \)

Hint:

The change in rotational kinetic energy is related to the square of the angular velocity. Use the torque and time to find the angular velocity and then compute the change in kinetic energy.

Answer 1

The Correct Option is A

Step 1: Rotational kinetic energy.
The rotational kinetic energy of a body is given by: \[ K.E. = \frac{1}{2} I \omega^2 \] where \( \omega \) is the angular velocity and \( I \) is the moment of inertia.
Step 2: Angular acceleration.
The angular acceleration \( \alpha \) due to the torque \( \tau \) is: \[ \alpha = \frac{\tau}{I} \]
Step 3: Final angular velocity.
The final angular velocity \( \omega \) after time \( t \) is given by: \[ \omega = \alpha t = \frac{\tau}{I} \cdot t = \frac{\tau t}{I} \]
Step 4: Change in rotational kinetic energy.
The initial rotational kinetic energy is zero (since the wheel starts from rest). The change in rotational kinetic energy is: \[ \Delta K.E. = \frac{1}{2} I \omega^2 = \frac{1}{2} I \left( \frac{\tau t}{I} \right)^2 = \frac{\tau^2 t^2}{2 I} \]
Step 5: Conclusion.
The change in rotational kinetic energy is \( \frac{\tau^2 t^2}{2 I} \), which is option (A).