The moment of inertia of a rigid body is `1.5 kgxxm//s^(2)` and its initial angular velocity is zero. It start rotating with uniform angular acceleration `alpha=20rad//sec^(2))` to achieve a rotational `KE=1200J` find the time requried for this:

Options

A. 20 sec
B. 200 sec
C. 2 sec
D. 0.2 sec

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the relationship between rotational kinetic energy and angular velocity.
The rotational kinetic energy (KE) of a rigid body is given by the formula:
\[ KE = \frac{1}{2} I \omega^2 \]
where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.

### Step 2: Substitute the known values into the kinetic energy formula.
Given:
- Moment of inertia, \( I = 1.5 \, \text{kg m}^2 \)
- Rotational kinetic energy, \( KE = 1200 \, \text{J} \)

Substituting these values into the kinetic energy formula:
\[ 1200 = \frac{1}{2} \times 1.5 \times \omega^2 \]

### Step 3: Solve for \( \omega^2 \).
Rearranging the equation:
\[ 1200 = 0.75 \omega^2 \]
Multiplying both sides by 2:
\[ 2400 = 1.5 \omega^2 \]
Now, divide by 1.5:
\[ \omega^2 = \frac{2400}{1.5} = 1600 \]

### Step 4: Calculate \( \omega \).
Taking the square root of both sides:
\[ \omega = \sqrt{1600} = 40 \, \text{rad/s} \]

### Step 5: Use the angular motion equation to find time.
We know:
- Initial angular velocity, \( \omega_0 = 0 \, \text{rad/s} \)
- Final angular velocity, \( \omega = 40 \, \text{rad/s} \)
- Angular acceleration, \( \alpha = 20 \, \text{rad/s}^2 \)

Using the equation:
\[ \omega = \omega_0 + \alpha t \]
Substituting the known values:
\[ 40 = 0 + 20t \]

### Step 6: Solve for \( t \).
Rearranging gives:
\[ t = \frac{40}{20} = 2 \, \text{s} \]

### Final Answer:
The time required for the rigid body to achieve a rotational kinetic energy of 1200 J is **2 seconds**.

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