The moment of inertia of a body about a given axis is `1.2 kg m^(2)`. Initially, the body is at rest. In order to produce a rotational `KE` of `1500 J`, for how much duration, an acceleration of `25 rad s^(-2)` must be applied about that axis ?

Options

A. `4 s`
B. `2 s`
C. `8 s`
D. `10 s`

Answer

To solve the problem step by step, we will follow these steps:

### Step 1: Understand the given data
- Moment of inertia (I) = 1.2 kg m²
- Initial angular velocity (ω_initial) = 0 rad/s (the body is at rest)
- Rotational kinetic energy (KE) = 1500 J
- Angular acceleration (α) = 25 rad/s²

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

### Step 3: Set up the equation
Substituting the known values into the kinetic energy formula:
\[ 1500 = \frac{1}{2} \times 1.2 \times \omega^2 \]

### Step 4: Solve for \( \omega^2 \)
Multiply both sides by 2 to eliminate the fraction:
\[ 3000 = 1.2 \times \omega^2 \]
Now, divide both sides by 1.2:
\[ \omega^2 = \frac{3000}{1.2} \]
Calculating the right side:
\[ \omega^2 = 2500 \]

### Step 5: Find \( \omega \)
Taking the square root of both sides gives:
\[ \omega = \sqrt{2500} = 50 \text{ rad/s} \]

### Step 6: Use the angular acceleration formula to find time
We know that:
\[ \alpha = \frac{\omega_{final} - \omega_{initial}}{t} \]
Substituting the known values:
\[ 25 = \frac{50 - 0}{t} \]

### Step 7: Solve for time (t)
Rearranging the equation gives:
\[ t = \frac{50}{25} = 2 \text{ seconds} \]

### Final Answer
The time duration for which an acceleration of \( 25 \, \text{rad/s}^2 \) must be applied is **2 seconds**.

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