Question

The moment of inertia of a body about a given axis is \( 1.2 \, \text{kg m}^2 \). Initially the body is at rest. In order to produce rotational kinetic energy of \( 1500 \, \text{J} \), an angular acceleration of \( 25 \, \text{rad/s}^2 \) must be applied about an axis for a time duration of

• A. \( 8 \, \text{s} \)
• B. \( 2 \, \text{s} \)
• C. \( 4 \, \text{s} \)
• D. \( 1 \, \text{s} \)

Hint:

For rotational motion from rest: \(\omega = \alpha t\) and KE = \(\frac{1}{2} I \alpha^2 t^2\). Solve directly for \(t\) without finding \(\omega\) separately.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We have \(I = 1.2\ \text{kg m}^2\), initial angular velocity \(\omega_0 = 0\), rotational KE \(= 1500\ \text{J}\), angular acceleration \(\alpha = 25\ \text{rad/s}^2\). Find time \(t\).

Step 2: Key Formula or Approach:
Rotational KE = \(\frac{1}{2} I \omega^2\). Using \(\omega = \alpha t\) (since \(\omega_0 = 0\)), we get KE = \(\frac{1}{2} I (\alpha t)^2\). Solve for \(t\).

Step 3: Detailed Explanation:
\[ 1500 = \frac{1}{2} \times 1.2 \times (25 t)^2 = 0.6 \times 625 t^2 = 375 t^2. \] Thus \(t^2 = \frac{1500}{375} = 4\) ⇒ \(t = 2\ \text{s}\).

Step 4: Final Answer:
The required time is \(2\ \text{s}\), option (B).