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: Connect energy, speed, and time.
The rotational kinetic energy is $\tfrac12 I\omega^2$, and since it starts from rest, $\omega = \alpha t$. So the energy is $\tfrac12 I(\alpha t)^2$.

Step 2: Plug in the values.
\[ 1500 = \tfrac12(1.2)(25t)^2 = 0.6\times625\,t^2 = 375\,t^2. \]

Step 3: Solve for $t^2$.
$t^2 = \tfrac{1500}{375} = 4$.

Step 4: Take the root.
$t = 2$. \[ \boxed{2\ \text{s}} \]