Shown below is a small particle of mass 33 g that is moving at a speed of 10 m/s when it collides with and sticks to the edge of a uniform solid cylinder. The cylinder is free to rotate about the axis through its center and perpendicular to the page. The cylinder has a mass of 0.8 kg and a radius of 20 cm and is initially at rest.

(a)
What is the angular velocity (in rad/s) of the system after the collision? (Indicate the direction with the sign of your answer.)

This problem involves the conservation of angular momentum. Since there are no external torques acting on the system, the angular momentum before the collision must equal the angular momentum after the collision.

### Given:
- Mass of the particle (\( m \)) = 33 g = 0.033 kg
- Speed of the particle before collision (\( v \)) = 10 m/s
- Mass of the cylinder (\( M \)) = 0.8 kg
- Radius of the cylinder (\( R \)) = 20 cm = 0.2 m
- Initial angular velocity of the cylinder (\( \omega_0 \)) = 0 rad/s (since it's initially at rest)

### Part (a): Finding the Angular Velocity After the Collision

1. **Initial Angular Momentum**:
   - The particle is moving linearly before the collision, so its angular momentum relative to the axis of the cylinder is given by:
   
   \[
   L_{\text{initial}} = m \cdot v \cdot R
   \]

   Substituting the values:

   \[
   L_{\text{initial}} = 0.033 \, \text{kg} \cdot 10 \, \text{m/s} \cdot 0.2 \, \text{m} = 0.066 \, \text{kg} \cdot \text{m}^2/\text{s}
   \]

2. **Moment of Inertia of the System After Collision**:
   - The moment of inertia of the cylinder about its axis is:

   \[
   I_{\text{cylinder}} = \frac{1}{2} M R^2
   \]

   - The moment of inertia of the particle (treated as a point mass) after it sticks to the edge of the cylinder is:

   \[
   I_{\text{particle}} = m R^2
   \]

   - The total moment of inertia after the collision is:

   \[
   I_{\text{total}} = I_{\text{cylinder}} + I_{\text{particle}} = \frac{1}{2} M R^2 + m R^2
   \]

   Substituting the values:

   \[
   I_{\text{total}} = \frac{1}{2} \cdot 0.8 \, \text{kg} \cdot (0.2 \, \text{m})^2 + 0.033 \, \text{kg} \cdot (0.2 \, \text{m})^2
   \]

   \[
   I_{\text{total}} = 0.016 \, \text{kg} \cdot \text{m}^2 + 0.00132 \, \text{kg} \cdot \text{m}^2 = 0.01732 \, \text{kg} \cdot \text{m}^2
   \]

3. **Final Angular Velocity**:
   - Using conservation of angular momentum:

   \[
   L_{\text{initial}} = L_{\text{final}} = I_{\text{total}} \cdot \omega
   \]

   Solving for \( \omega \) (the angular velocity after the collision):

   \[
   \omega = \frac{L_{\text{initial}}}{I_{\text{total}}} = \frac{0.066 \, \text{kg} \cdot \text{m}^2/\text{s}}{0.01732 \, \text{kg} \cdot \text{m}^2} \approx 3.81 \, \text{rad/s}
   \]

### Direction:
Since the particle sticks to the edge and moves in the same direction as its initial motion, the direction of the angular velocity will be counterclockwise, which is considered positive.

Thus, the angular velocity of the system after the collision is **+3.81 rad/s**.

---

Would you like further details or have any questions? 

Here are 5 related questions you might consider:
1. What if the particle had bounced off the cylinder instead of sticking to it?
2. How would the result change if the cylinder was not initially at rest?
3. What is the kinetic energy of the system after the collision?
4. How much kinetic energy is lost during the collision?
5. What if the mass of the particle was different? How would that affect the angular velocity?

**Tip:** In rotational dynamics problems, always ensure that you correctly identify the axis of rotation and account for all contributing moments of inertia.

Learn how to calculate the angular velocity of a system after a collision involving a particle and a rotating cylinder in rotational dynamics. Includes step-by-step solution using conservation of angular momentum. Relevant data particle mass 33g, speed 10 m/s, cylinder mass 0.8 kg, radius 20 cm.

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