College Physics
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Transcribed Image Text:**Problem: Rotating Rod and Moving Bug**
A thin rod has a length of 0.537 m and rotates in a circle on a frictionless tabletop. The axis is perpendicular to the length of the rod at one of its ends. The rod has an angular velocity of 0.778 rad/s and a moment of inertia of \(1.10 \times 10^{-3} \, \text{kg} \cdot \text{m}^2\). A bug standing on the axis decides to crawl out to the other end of the rod. When the bug (whose mass is \(5 \times 10^{-3} \, \text{kg}\)) gets where it's going, what is the change in the angular velocity of the rod?
**Solution Approach:**
To find the change in angular velocity, we can use the principle of conservation of angular momentum. Initially, the angular momentum is only due to the rod. When the bug reaches the end, the angular momentum is distributed between the rod and the bug.
1. **Calculate Initial Angular Momentum:**
- \( L_{\text{initial}} = I_{\text{rod}} \cdot \omega_{\text{initial}} \)
- Where \( I_{\text{rod}} = 1.10 \times 10^{-3} \text{ kg} \cdot \text{m}^2 \) and \( \omega_{\text{initial}} = 0.778 \text{ rad/s} \).
2. **Calculate Final Angular Momentum:**
- \( L_{\text{final}} = (I_{\text{rod}} + I_{\text{bug}}) \cdot \omega_{\text{final}} \)
- Moment of inertia of the bug when at the end \( I_{\text{bug}} = m_{\text{bug}} \cdot r^2 \)
- Where \( r = 0.537 \text{ m} \) and \( m_{\text{bug}} = 5 \times 10^{-3} \text{ kg} \).
3. **Apply Conservation of Angular Momentum:**
- \( L_{\text{initial}} = L_{\text{final}} \)
- Use this to solve for \( \omega_{\text{final}} \).
4. **Calculate the Change in Angular Velocity:**
- \( \Delta \omega = \omega_{\text
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