To solve this problem, we'll use the principle of conservation of angular momentum.

Step-by-Step Solution:

1. Angular Momentum Conservation: The angular momentum of the system before the boy jumps on is equal to the angular momentum of the system after the boy jumps on, because there are no external torques.

   Before the boy jumps on, the merry-go-round is at rest, so its angular momentum is zero. The boy, however, has linear momentum that can be converted to angular momentum when he jumps on.

   The formula for angular momentum $L$ is: $L = m v r$ where:

   • $m$ is the mass of the boy,
   • $v$ is the linear velocity of the boy,
   • $r$ is the radius of the merry-go-round.

2. Moment of Inertia of the System:

   • The moment of inertia of the merry-go-round is given as $I_{\text{merry-go-round}} = 525 \, \text{kg} \cdot \text{m}^2$.
   • The boy's mass can be treated as a point mass at the radius $r = 2.3 \, \text{m}$. His moment of inertia $I_{\text{boy}}$ is: $I_{\text{boy}} = m r^2$ where $m = 52 \, \text{kg}$.

3. Conservation of Angular Momentum: The total angular momentum before and after the boy jumps on must be conserved: $L_{\text{before}} = L_{\text{after}}$ Before the boy jumps on, the angular momentum is entirely due to the boy's motion, and after the boy jumps on, it includes both the merry-go-round's angular momentum and the boy's angular momentum.

   The angular momentum before the boy jumps on is: $L_{\text{before}} = m v r$

   The angular momentum after the boy jumps on is the sum of the merry-go-round's angular momentum and the boy's angular momentum: $L_{\text{after}} = (I_{\text{merry-go-round}} + I_{\text{boy}}) \omega$ where $\omega$ is the angular velocity of the system after the boy jumps on.

4. Setting up the equation: Since angular momentum is conserved: $m v r = (I_{\text{merry-go-round}} + I_{\text{boy}}) \omega$ Substituting values: $(52 \, \text{kg}) (4.2 \, \text{m/s}) (2.3 \, \text{m}) = (525 \, \text{kg} \cdot \text{m}^2 + 52 \, \text{kg} (2.3 \, \text{m})^2) \omega$

5. Calculating the values:

   • First, calculate the boy's moment of inertia: $I_{\text{boy}} = 52 \times (2.3)^2 = 52 \times 5.29 = 274.08 \, \text{kg} \cdot \text{m}^2$

   • Now substitute the values into the equation: $52 \times 4.2 \times 2.3 = (525 + 274.08) \omega$ $502.56 = 799.08 \omega$

6. Solving for $\omega$: $\omega = \frac{502.56}{799.08} = 0.63 \, \text{rad/s}$

Final Answer:

The angular velocity of the system after the boy jumps on is $\boxed{0.63 \, \text{rad/s}}$.

Would you like further details or have any questions? Here are some related questions to explore:

1. How does the angular velocity change if the boy runs faster before jumping on?
2. What would happen if the merry-go-round had a smaller moment of inertia?
3. How does the moment of inertia affect the final angular velocity in such problems?
4. If the boy's mass were greater, how would it influence the system's angular velocity?
5. What is the concept of angular momentum and how is it related to linear momentum?

Tip: Angular momentum is conserved in systems where no external torques are applied. This principle is useful for analyzing rotational motion in many physical scenarios.