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

Explanation

To solve the problem, we need to find the duration for which an angular acceleration must be applied to achieve a specified rotational kinetic energy. Let's break this down step by step.

### Given Data:
- Moment of Inertia (I) = 1.5 kg m²
- Rotational Kinetic Energy (K) = 1200 J
- Angular Acceleration (α) = 20 rad/s²
- Initial Angular Velocity (ω₀) = 0 rad/s (since the body is initially at rest)

### Step 1: Write the formula for rotational kinetic energy
The formula for rotational kinetic energy is given by:
\[ K = \frac{1}{2} I \omega^2 \]
where \( \omega \) is the final angular velocity.

### Step 2: Express angular velocity in terms of angular acceleration and time
Since the body starts from rest, we can use the equation of motion for rotational motion:
\[ \omega = \omega_0 + \alpha t \]
Substituting the initial angular velocity:
\[ \omega = 0 + \alpha t = \alpha t \]
Thus,
\[ \omega = 20t \]

### Step 3: Substitute ω in the kinetic energy equation
Now we substitute \( \omega \) in the kinetic energy formula:
\[ K = \frac{1}{2} I (20t)^2 \]
Substituting the value of I:
\[ 1200 = \frac{1}{2} \times 1.5 \times (20t)^2 \]

### Step 4: Simplify the equation
First, simplify the right side:
\[ 1200 = \frac{1.5}{2} \times 400t^2 \]
\[ 1200 = 0.75 \times 400t^2 \]
\[ 1200 = 300t^2 \]

### Step 5: Solve for t²
Now, divide both sides by 300:
\[ t^2 = \frac{1200}{300} \]
\[ t^2 = 4 \]

### Step 6: Take the square root to find t
Taking the square root of both sides:
\[ t = \sqrt{4} \]
\[ t = 2 \text{ seconds} \]

### Final Answer
The duration for which the angular acceleration must be applied is **2 seconds**.

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