The total kinetic energy of a body of mass 10 kg and radius 0.5 m moving with a velocity of 2 m/s without slipping is 32.8 joule. The radius of gyration of the body is

Options

A. `0.25` m
B. `0.2` m
C. `0.5` m
D. `0.4` m

Explanation

To find the radius of gyration \( k \) of the body, we will use the total kinetic energy formula for a rolling object. The total kinetic energy \( KE \) is given by the sum of translational and rotational kinetic energies.

### Step-by-Step Solution

1. **Identify Given Values**:
- Mass of the body \( m = 10 \, \text{kg} \)
- Radius of the body \( R = 0.5 \, \text{m} \)
- Velocity \( v = 2 \, \text{m/s} \)
- Total kinetic energy \( KE = 32.8 \, \text{J} \)

2. **Write the Total Kinetic Energy Formula**:
The total kinetic energy \( KE \) for a rolling object can be expressed as:
\[ KE = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 \]
where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.

3. **Relate Angular Velocity to Linear Velocity**:
For rolling without slipping, the relationship between linear velocity \( v \) and angular velocity \( \omega \) is:
\[ \omega = \frac{v}{R} \]

4. **Substitute Moment of Inertia**:
The moment of inertia \( I \) can be expressed in terms of the radius of gyration \( k \):
\[ I = m k^2 \]
Substituting this into the kinetic energy equation gives:
\[ KE = \frac{1}{2} mv^2 + \frac{1}{2} (m k^2) \left(\frac{v}{R}\right)^2 \]

5. **Simplify the Equation**:
Substitute \( \omega \) into the kinetic energy equation:
\[ KE = \frac{1}{2} mv^2 + \frac{1}{2} m k^2 \frac{v^2}{R^2} \]
Factor out \( \frac{1}{2} mv^2 \):
\[ KE = \frac{1}{2} mv^2 \left(1 + \frac{k^2}{R^2}\right) \]

6. **Plug in Known Values**:
Substitute the known values into the equation:
\[ 32.8 = \frac{1}{2} \cdot 10 \cdot 2^2 \left(1 + \frac{k^2}{(0.5)^2}\right) \]
Calculate \( \frac{1}{2} \cdot 10 \cdot 4 = 20 \):
\[ 32.8 = 20 \left(1 + \frac{k^2}{0.25}\right) \]

7. **Solve for \( k^2 \)**:
Divide both sides by 20:
\[ 1.64 = 1 + \frac{k^2}{0.25} \]
Rearranging gives:
\[ 1.64 - 1 = \frac{k^2}{0.25} \]
\[ 0.64 = \frac{k^2}{0.25} \]
Multiply both sides by 0.25:
\[ k^2 = 0.64 \times 0.25 = 0.16 \]

8. **Find the Radius of Gyration \( k \)**:
Taking the square root:
\[ k = \sqrt{0.16} = 0.4 \, \text{m} \]

### Final Answer
The radius of gyration \( k \) of the body is \( 0.4 \, \text{m} \).

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