Question

A body of mass 5 kg at rest is rotated for 25 s with a constant moment of force 10 Nm. Find the work done if the moment of inertia of the body is 5 kg m$^2$.

• A. 625 J
• B. 125 J
• C. 6250 J
• D. 1250 J

Hint:

In rotational motion, the work done is related to the torque and angular displacement. If the body starts from rest, the angular displacement can be calculated using the formula \( \theta = \frac{1}{2} \alpha t^2 \), where \( \alpha \) is found from \( \alpha = \frac{\tau}{I} \).

Answer 1

The Correct Option is C

In rotational motion, the work done \( W \) by a constant torque \( \tau \) is given by: \[ W = \tau \theta, \] where: - \( \tau \) is the constant torque (10 Nm), - \( \theta \) is the angular displacement.
Step 1: Find angular displacement
Since the body starts from rest and is rotated for 25 seconds under a constant moment of force (torque), we can use the equation for angular displacement under constant angular acceleration: \[ \theta = \omega_0 t + \frac{1}{2} \alpha t^2, \] where: - \( \omega_0 \) is the initial angular velocity (which is 0 since the body starts from rest), - \( t \) is the time (25 s), - \( \alpha \) is the angular acceleration. Since \( \tau = I \alpha \) (where \( I \) is the moment of inertia), the angular acceleration is: \[ \alpha = \frac{\tau}{I}. \] Substituting \( \tau = 10 \, \text{Nm} \) and \( I = 5 \, \text{kg m}^2 \), we get: \[ \alpha = \frac{10}{5} = 2 \, \text{rad/s}^2. \] Now, substitute \( \alpha = 2 \, \text{rad/s}^2 \) and \( t = 25 \, \text{s} \) into the equation for \( \theta \): \[ \theta = 0 \times 25 + \frac{1}{2} \times 2 \times 25^2 = \frac{1}{2} \times 2 \times 625 = 625 \, \text{rad}. \]
Step 2: Calculate the work done
Now, substitute the values of \( \tau \) and \( \theta \) into the work equation: \[ W = 10 \times 625 = 6250 \, \text{J}. \] Thus, the work done is 6250 J, which corresponds to option (C).