A wheel of moment of inertia `0.500 kg-m^2` and radius 20.0 cm is rotating about its axis at an angular speed of 20.0 rad/s. It picks up a stationary particle of mass 200 g at its edge. Find the new angular speed of the wheel.

Explanation

To solve the problem, we will use the principle of conservation of angular momentum. The angular momentum before the particle is picked up by the wheel will be equal to the angular momentum after the particle is picked up, since there is no external torque acting on the system.

### Step-by-Step Solution:

1. **Identify Given Values**:
- Moment of inertia of the wheel, \( I_1 = 0.500 \, \text{kg m}^2 \)
- Radius of the wheel, \( r = 20.0 \, \text{cm} = 0.20 \, \text{m} \)
- Initial angular speed of the wheel, \( \omega_1 = 20.0 \, \text{rad/s} \)
- Mass of the particle, \( m = 200 \, \text{g} = 0.200 \, \text{kg} \)

2. **Calculate Initial Angular Momentum**:
The initial angular momentum \( L_i \) can be calculated using the formula:
\[ L_i = I_1 \cdot \omega_1 \]
Substituting the values:
\[ L_i = 0.500 \, \text{kg m}^2 \cdot 20.0 \, \text{rad/s} = 10.0 \, \text{kg m}^2/\text{s} \]

3. **Calculate the Moment of Inertia of the Particle**:
When the particle is picked up at the edge of the wheel, its moment of inertia \( I_2 \) can be calculated as:
\[ I_2 = I_1 + m \cdot r^2 \]
Substituting the values:
\[ I_2 = 0.500 \, \text{kg m}^2 + 0.200 \, \text{kg} \cdot (0.20 \, \text{m})^2 \]
\[ I_2 = 0.500 \, \text{kg m}^2 + 0.200 \, \text{kg} \cdot 0.04 \, \text{m}^2 \]
\[ I_2 = 0.500 \, \text{kg m}^2 + 0.008 \, \text{kg m}^2 = 0.508 \, \text{kg m}^2 \]

4. **Apply Conservation of Angular Momentum**:
According to the conservation of angular momentum:
\[ L_i = L_f \]
where \( L_f \) is the final angular momentum:
\[ L_f = I_2 \cdot \omega_2 \]
Setting the two equal gives:
\[ 10.0 \, \text{kg m}^2/\text{s} = 0.508 \, \text{kg m}^2 \cdot \omega_2 \]

5. **Solve for the New Angular Speed \( \omega_2 \)**:
Rearranging the equation to solve for \( \omega_2 \):
\[ \omega_2 = \frac{10.0 \, \text{kg m}^2/\text{s}}{0.508 \, \text{kg m}^2} \]
\[ \omega_2 \approx 19.68 \, \text{rad/s} \]

### Final Answer:
The new angular speed of the wheel after picking up the particle is approximately \( \omega_2 \approx 19.68 \, \text{rad/s} \).