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 need to find the percentage of the rotational kinetic energy in the total kinetic energy of the wheel. Let's break this down step by step.

### Step 1: Identify the given values
- 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, \( I = 0.65 \, \text{kg m}^2 \)

### Step 2: Calculate the rotational kinetic energy
The formula for rotational kinetic energy (\( KE_{rot} \)) is given by:
\[ KE_{rot} = \frac{1}{2} I \omega^2 \]
Substituting the values:
\[ KE_{rot} = \frac{1}{2} \times 0.65 \, \text{kg m}^2 \times (10 \, \text{rad/s})^2 \]
Calculating this:
\[ KE_{rot} = \frac{1}{2} \times 0.65 \times 100 = 32.5 \, \text{J} \]

### Step 3: Calculate the translational kinetic energy
The formula for translational kinetic energy (\( KE_{trans} \)) is given by:
\[ KE_{trans} = \frac{1}{2} m v^2 \]
Where \( v \) (the linear velocity) can be expressed in terms of angular velocity:
\[ v = \omega r \]
Substituting the values:
\[ v = 10 \, \text{rad/s} \times 0.40 \, \text{m} = 4 \, \text{m/s} \]
Now substituting \( v \) into the translational kinetic energy formula:
\[ KE_{trans} = \frac{1}{2} \times 5 \, \text{kg} \times (4 \, \text{m/s})^2 \]
Calculating this:
\[ KE_{trans} = \frac{1}{2} \times 5 \times 16 = 40 \, \text{J} \]

### Step 4: Calculate the total kinetic energy
The total kinetic energy (\( KE_{total} \)) is the sum of the rotational and translational kinetic energies:
\[ KE_{total} = KE_{rot} + KE_{trans} \]
Substituting the values:
\[ KE_{total} = 32.5 \, \text{J} + 40 \, \text{J} = 72.5 \, \text{J} \]

### Step 5: Calculate the percentage of rotational kinetic energy in total 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:
\[ \text{Percentage} = \left( \frac{32.5}{72.5} \right) \times 100 \approx 44.83\% \]
Rounding this, we get approximately \( 44.8\% \).

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

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