Question

The moment of inertia of a body about a given axis is \(1.2\ \text{kg·m}^2\). To produce a rotational kinetic energy of \(1500\ \text{J}\) an angular acceleration of \(25\ \text{rad/s}^2\) must be applied for

• A. \(8.5\ \text{s}\)
• B. \(5\ \text{s}\)
• C. \(2\ \text{s}\)
• D. \(1\ \text{s}\)

Hint:

Rotational KE = \(\frac{1}{2}I\omega^2\) is analogous to translational KE = \(\frac{1}{2}mv^2\).

Answer 1

The Correct Option is C

Step 1: Concept:
Rotational KE = \(\frac{1}{2}I\omega^2\). Using \(\omega = \omega_0 + \alpha t\) with \(\omega_0 = 0\).
Step 2: Explanation:
Given KE = \(1500\ \text{J}\), \(I = 1.2\ \text{kg·m}^2\).
\(\frac{1}{2}I\omega^2 = 1500 \Rightarrow \frac{1}{2} \times 1.2 \times \omega^2 = 1500 \Rightarrow 0.6\omega^2 = 1500 \Rightarrow \omega^2 = 2500 \Rightarrow \omega = 50\ \text{rad/s}\).
Using \(\omega = \alpha t\): \(50 = 25 \times t \Rightarrow t = 2\ \text{s}\).
Step 3: Conclusion:
Thus, time = \(2\ \text{s}\).