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This problem can be solved using the principle of conservation of angular momentum because there are no external torques acting on the merry-go-round system. The initial angular momentum of the system must equal the final angular momentum after the child gets on the merry-go-round.

### Given:
- Mass of the merry-go-round \( M = 120 \, \text{kg} \)
- Radius of the merry-go-round \( R = 1.80 \, \text{m} \)
- Initial angular velocity of the merry-go-round \( \omega_i = 0.500 \, \text{rev/s} = 0.500 \times 2\pi = 3.14 \, \text{rad/s} \)
- Mass of the child \( m = 22.0 \, \text{kg} \)
- The child is initially at rest, so initial angular velocity of the child is \( 0 \, \text{rad/s} \)

We assume the merry-go-round is a solid disk, so its moment of inertia is:

\[
I_{\text{merry-go-round}} = \frac{1}{2} M R^2
\]

The child can be considered as a point mass, and their moment of inertia after they grab the edge is:

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

### Step 1: Calculate the initial angular momentum
The initial angular momentum is only due to the merry-go-round since the child is not yet on it:

\[
L_{\text{initial}} = I_{\text{merry-go-round}} \cdot \omega_i = \left( \frac{1}{2} M R^2 \right) \cdot \omega_i
\]

### Step 2: Calculate the final angular momentum
After the child gets on, the total moment of inertia becomes the sum of the moments of inertia of the merry-go-round and the child:

\[
I_{\text{total}} = I_{\text{merry-go-round}} + I_{\text{child}} = \frac{1}{2} M R^2 + m R^2
\]

The final angular momentum is:

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

Since angular momentum is conserved:

\[
L_{\text{initial}} = L_{\text{final}}
\]

This gives us:

\[
\left( \frac{1}{2} M R^2 \right) \cdot \omega_i = \left( \frac{1}{2} M R^2 + m R^2 \right) \cdot \omega_f
\]

### Step 3: Solve for the final angular velocity \( \omega_f \)
Now, solving for \( \omega_f \):

\[
\omega_f = \frac{\left( \frac{1}{2} M R^2 \right) \cdot \omega_i}{\frac{1}{2} M R^2 + m R^2}
\]

Simplify the expression:

\[
\omega_f = \frac{\left( \frac{1}{2} M \right) \cdot \omega_i}{\frac{1}{2} M + m}
\]

### Step 4: Plug in the values
Now substitute the values:

\[
\omega_f = \frac{\left( \frac{1}{2} \times 120 \right) \cdot 3.14}{\frac{1}{2} \times 120 + 22}
\]

Let's calculate it step by step.

The final angular velocity of the merry-go-round after the child gets on is approximately \( 2.30 \, \text{rad/s} \).

Would you like further details on this solution or clarification on any step?

Here are 5 related questions for deeper understanding:
1. How does the moment of inertia change when mass is added at the edge?
2. Why is angular momentum conserved in this scenario?
3. How would the result change if the child had a different mass?
4. What is the physical significance of angular velocity in rotational motion?
5. How would the problem differ if the child started with an initial velocity?

**Tip**: Always check if there are any external torques acting on the system before applying conservation of angular momentum.

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 0.500 rev/s. What is its angular velocity after a 22.0-kg child gets onto it by grabbing its outer edge? The child is initially at rest.

Solve a physics problem involving a rotating merry-go-round and a child using conservation of angular momentum. Learn how to calculate the final angular velocity when a 22.0-kg child grabs onto a 120-kg merry-go-round rotating at 0.500 rev/s. This problem covers rotational motion and moment of inertia concepts, suitable for Grades 11-12 or college-level physics.

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