A wheel of mass `5 kg` and radius `0.40 m` is rolling on a road without sliding with angular velocity `10 rad s^-1`. The moment of ineria of the wheel about the axis of rotation is `0.65 kg m^2`. The percentage of kinetic energy of rotate in the total kinetic energy of the wheel is.

Options

A. `22.4 %`
B. `11.2 %`
C. `88.8 %`
D. `44. 8 %`

Explanation

To solve the problem, we will calculate the translational kinetic energy, rotational kinetic energy, and then find the percentage of rotational kinetic energy in the total kinetic energy of the wheel.

### Step-by-Step Solution:

1. **Given Data:**
- Mass of the wheel, \( m = 5 \, \text{kg} \)
- Radius of the wheel, \( r = 0.40 \, \text{m} \)
- Angular velocity, \( \omega = 10 \, \text{rad/s} \)
- Moment of inertia, \( I = 0.65 \, \text{kg m}^2 \)

2. **Calculate the Translational Velocity:**
The translational velocity \( v \) of the wheel can be calculated using the relation:
\[ v = \omega r \]
Substituting the values:
\[ v = 10 \, \text{rad/s} \times 0.40 \, \text{m} = 4 \, \text{m/s} \]

3. **Calculate the Translational Kinetic Energy:**
The translational kinetic energy \( KE_{trans} \) is given by:
\[ KE_{trans} = \frac{1}{2} m v^2 \]
Substituting the values:
\[ KE_{trans} = \frac{1}{2} \times 5 \, \text{kg} \times (4 \, \text{m/s})^2 = \frac{1}{2} \times 5 \times 16 = 40 \, \text{J} \]

4. **Calculate the Rotational Kinetic Energy:**
The rotational kinetic energy \( KE_{rot} \) is given by:
\[ KE_{rot} = \frac{1}{2} I \omega^2 \]
Substituting the values:
\[ KE_{rot} = \frac{1}{2} \times 0.65 \, \text{kg m}^2 \times (10 \, \text{rad/s})^2 = \frac{1}{2} \times 0.65 \times 100 = 32.5 \, \text{J} \]

5. **Calculate the Total Kinetic Energy:**
The total kinetic energy \( KE_{total} \) is the sum of translational and rotational kinetic energies:
\[ KE_{total} = KE_{trans} + KE_{rot} = 40 \, \text{J} + 32.5 \, \text{J} = 72.5 \, \text{J} \]

6. **Calculate the Percentage of Rotational Kinetic Energy:**
The percentage of rotational kinetic energy in the total kinetic energy is given by:
\[ \text{Percentage} = \left( \frac{KE_{rot}}{KE_{total}} \right) \times 100 \]
Substituting the values:
\[ \text{Percentage} = \left( \frac{32.5 \, \text{J}}{72.5 \, \text{J}} \right) \times 100 \approx 44.83\% \]

### Final Answer:
The percentage of kinetic energy of rotation in the total kinetic energy of the wheel is approximately **44.8%**.