A 50 kg rider on a moped of mass 75 kg is traveling with a speed of 20 m/s. Each of the two wheels of the moped has a radius of 0.2 m and a moment of inertia of `0.2kg*m^(2)`. What is the total rotational kinetic energy of the wheels?

Options

A. 500 J
B. 4000 J
C. 100 J
D. 2000 J

Answer

To find the total rotational kinetic energy of the wheels of the moped, we can follow these steps:

### Step 1: Understand the Formula for Rotational Kinetic Energy
The rotational kinetic energy (RKE) of a rotating object is given by the formula:
\[ RKE = \frac{1}{2} I \omega^2 \]
where:
- \( I \) is the moment of inertia of the object,
- \( \omega \) is the angular velocity in radians per second.

### Step 2: Determine the Moment of Inertia
Given that each wheel has a moment of inertia \( I = 0.2 \, \text{kg} \cdot \text{m}^2 \).

### Step 3: Calculate the Angular Velocity
The angular velocity \( \omega \) can be calculated from the linear speed \( v \) of the moped and the radius \( r \) of the wheels using the relationship:
\[ \omega = \frac{v}{r} \]
Given:
- \( v = 20 \, \text{m/s} \)
- \( r = 0.2 \, \text{m} \)

Substituting the values:
\[ \omega = \frac{20 \, \text{m/s}}{0.2 \, \text{m}} = 100 \, \text{rad/s} \]

### Step 4: Calculate the Rotational Kinetic Energy of One Wheel
Now we can calculate the rotational kinetic energy for one wheel:
\[ RKE_{\text{one wheel}} = \frac{1}{2} I \omega^2 \]
Substituting the values:
\[ RKE_{\text{one wheel}} = \frac{1}{2} \times 0.2 \, \text{kg} \cdot \text{m}^2 \times (100 \, \text{rad/s})^2 \]
Calculating:
\[ RKE_{\text{one wheel}} = \frac{1}{2} \times 0.2 \times 10000 = 1000 \, \text{J} \]

### Step 5: Calculate the Total Rotational Kinetic Energy for Both Wheels
Since there are two wheels, the total rotational kinetic energy is:
\[ RKE_{\text{total}} = 2 \times RKE_{\text{one wheel}} = 2 \times 1000 \, \text{J} = 2000 \, \text{J} \]

### Final Answer
The total rotational kinetic energy of the wheels is:
\[ \boxed{2000 \, \text{J}} \]