Moment of inertia of a body about a given axis is `1.5 kg m^(2)`. Intially the body is at rest. In order to produce a rotational kinetic energy iof 1200 J, the angular acceleration of 20 rad/`s^(2)` must be applied about the axis for a duration of

Options

A. 2.5 s
B. 2s
C. 5s
D. 3s

Answer

To solve the problem, we need to find the time duration for which an angular acceleration of 20 rad/s² must be applied to a body with a moment of inertia of 1.5 kg m² in order to produce a rotational kinetic energy of 1200 J.

### Step-by-Step Solution:

1. **Identify Given Values**:
- Moment of inertia, \( I = 1.5 \, \text{kg m}^2 \)
- Initial angular velocity, \( \omega_0 = 0 \, \text{rad/s} \) (the body is at rest)
- Rotational kinetic energy, \( K = 1200 \, \text{J} \)
- Angular acceleration, \( \alpha = 20 \, \text{rad/s}^2 \)

2. **Use the Formula for Rotational Kinetic Energy**:
The formula for rotational kinetic energy is given by:
\[ K = \frac{1}{2} I \omega^2 \]
Rearranging this formula to find \( \omega \):
\[ \omega^2 = \frac{2K}{I} \]

3. **Substituting the Values**:
Substitute the known values into the equation:
\[ \omega^2 = \frac{2 \times 1200 \, \text{J}}{1.5 \, \text{kg m}^2} \]
\[ \omega^2 = \frac{2400}{1.5} = 1600 \]

4. **Calculate Angular Velocity**:
Taking the square root to find \( \omega \):
\[ \omega = \sqrt{1600} = 40 \, \text{rad/s} \]

5. **Use the Angular Motion Equation**:
The relationship between angular velocity, initial angular velocity, angular acceleration, and time is given by:
\[ \omega = \omega_0 + \alpha t \]
Since \( \omega_0 = 0 \):
\[ \omega = \alpha t \]

6. **Substituting Known Values**:
Substitute \( \omega = 40 \, \text{rad/s} \) and \( \alpha = 20 \, \text{rad/s}^2 \):
\[ 40 = 20 t \]

7. **Solve for Time \( t \)**:
Rearranging gives:
\[ t = \frac{40}{20} = 2 \, \text{s} \]

### Final Answer:
The time duration for which the angular acceleration must be applied is \( t = 2 \, \text{s} \).

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