A playground merry-go-round has a mass of 120 kg and a radius of 1.80 m and it is rotating with an angular velocity of 0.550 rev/s. What is its angular velocity (in rev/s) after a 20.0 kg child gets on grabbing its outer edge? The child is initially at rest. rev/s

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### Rotational Dynamics Problem - Playground Merry-Go-Round

**Problem Statement:**
A playground merry-go-round has a mass of 120 kg and a radius of 1.80 meters. It is rotating with an angular velocity of 0.550 revolutions per second (rev/s). What is its angular velocity (in rev/s) after a 20.0 kg child gets onto it by grabbing its outer edge? The child is initially at rest.

**Solution:**

1. **Initial Data:**
   - Mass of merry-go-round \( (M) \) = 120 kg
   - Radius of merry-go-round \( (R) \) = 1.80 m
   - Initial angular velocity \( (\omega_i) \) = 0.550 rev/s
   - Mass of child \( (m) \) = 20.0 kg

2. **Moment of Inertia:**
   - The moment of inertia of the merry-go-round \( (I_{\text{merry-go-round}}) \) can be approximated by the formula for a solid disk: 
     \[ I_{\text{merry-go-round}} = \frac{1}{2} M R^2 \]
   - Upon the child grabbing the outer edge, their moment of inertia is:
     \[ I_{\text{child}} = m R^2 \]
   
3. **Total Initial Moment of Inertia:**
   \[ I_{\text{initial}} = I_{\text{merry-go-round}} \] 
   \[ I_{\text{initial}} = \frac{1}{2} M R^2 \]
   
4. **Total Final Moment of Inertia:**
   \[ I_{\text{final}} = I_{\text{merry-go-round}} + I_{\text{child}} \]
   \[ I_{\text{final}} = \frac{1}{2} M R^2 + m R^2 \]
   
5. **Applying Conservation of Angular Momentum:**
   Since there are no external torques, the angular momentum before and after the child gets onto the merry-go-round is conserved.
   \[ L_{\text{initial}} = L_{\text{final}} \]
   \[ I_{\text{initial}} \cdot \omega_i = I_{\text{final}} \cdot \omega_f \]
   
6. **Calcul
Transcribed Image Text:### Rotational Dynamics Problem - Playground Merry-Go-Round **Problem Statement:** A playground merry-go-round has a mass of 120 kg and a radius of 1.80 meters. It is rotating with an angular velocity of 0.550 revolutions per second (rev/s). What is its angular velocity (in rev/s) after a 20.0 kg child gets onto it by grabbing its outer edge? The child is initially at rest. **Solution:** 1. **Initial Data:** - Mass of merry-go-round \( (M) \) = 120 kg - Radius of merry-go-round \( (R) \) = 1.80 m - Initial angular velocity \( (\omega_i) \) = 0.550 rev/s - Mass of child \( (m) \) = 20.0 kg 2. **Moment of Inertia:** - The moment of inertia of the merry-go-round \( (I_{\text{merry-go-round}}) \) can be approximated by the formula for a solid disk: \[ I_{\text{merry-go-round}} = \frac{1}{2} M R^2 \] - Upon the child grabbing the outer edge, their moment of inertia is: \[ I_{\text{child}} = m R^2 \] 3. **Total Initial Moment of Inertia:** \[ I_{\text{initial}} = I_{\text{merry-go-round}} \] \[ I_{\text{initial}} = \frac{1}{2} M R^2 \] 4. **Total Final Moment of Inertia:** \[ I_{\text{final}} = I_{\text{merry-go-round}} + I_{\text{child}} \] \[ I_{\text{final}} = \frac{1}{2} M R^2 + m R^2 \] 5. **Applying Conservation of Angular Momentum:** Since there are no external torques, the angular momentum before and after the child gets onto the merry-go-round is conserved. \[ L_{\text{initial}} = L_{\text{final}} \] \[ I_{\text{initial}} \cdot \omega_i = I_{\text{final}} \cdot \omega_f \] 6. **Calcul
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