A body having a moment of inertia about its axis of rotation equal to `3 "kg-m"^(2)` is rotating with angular velocity of `3 "rad s^(-1)`. Kinetic energy of this rotating body is same as that of a body of mass `27 kg` moving with a velocity `v`. The value of `v` is

Options

A. `1 ms^(-1)`
B. `0.5 ms^(-1)`
C. `2 ms^(-1)`
D. `1.5 ms^(-1)`

Explanation

To solve the problem, we need to find the velocity \( v \) of a body of mass \( 27 \, \text{kg} \) such that its kinetic energy is equal to the kinetic energy of a rotating body with a moment of inertia \( I = 3 \, \text{kg-m}^2 \) and an angular velocity \( \omega = 3 \, \text{rad/s} \).

### Step-by-Step Solution:

1. **Calculate the Kinetic Energy of the Rotating Body:**
The formula for the rotational kinetic energy \( KE_{\text{rot}} \) is given by:
\[ KE_{\text{rot}} = \frac{1}{2} I \omega^2 \]
Substituting the values:
\[ KE_{\text{rot}} = \frac{1}{2} \times 3 \, \text{kg-m}^2 \times (3 \, \text{rad/s})^2 \]
\[ KE_{\text{rot}} = \frac{1}{2} \times 3 \times 9 = \frac{27}{2} \, \text{J} \]

2. **Calculate the Kinetic Energy of the Translating Body:**
The formula for the translational kinetic energy \( KE_{\text{trans}} \) is given by:
\[ KE_{\text{trans}} = \frac{1}{2} m v^2 \]
Where \( m = 27 \, \text{kg} \). Thus:
\[ KE_{\text{trans}} = \frac{1}{2} \times 27 \, \text{kg} \times v^2 \]
\[ KE_{\text{trans}} = \frac{27}{2} v^2 \, \text{J} \]

3. **Set the Kinetic Energies Equal:**
Since the kinetic energies are equal, we can set the two equations equal to each other:
\[ \frac{27}{2} = \frac{27}{2} v^2 \]

4. **Solve for \( v^2 \):**
We can simplify this equation by cancelling \( \frac{27}{2} \) from both sides (as long as it is not zero):
\[ 1 = v^2 \]

5. **Find \( v \):**
Taking the square root of both sides gives:
\[ v = \sqrt{1} = 1 \, \text{m/s} \]

### Final Answer:
The value of \( v \) is \( 1 \, \text{m/s} \).

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