Final answer:
The angular velocity of the merry-go-round after the boy jumps on is found using conservation of angular momentum, resulting in an angular velocity of 0.8 rad/s.
Step-by-step explanation:
To find the angular velocity of the merry-go-round after the boy jumps on, conservation of angular momentum must be applied. The boy running tangent to the rim can be treated as a point mass, so the angular momentum he contributes when jumping on is his linear momentum times the radius of the merry-go-round. Before the boy jumps on, the angular momentum of the system is zero because the merry-go-round is at rest.
Thus, the total angular momentum before and after the event will remain the same. The linear momentum of the boy is the product of his mass and velocity (50 kg × 4.0 m/s = 200 kg·m/s), which, multiplied by the radius (2.0 m), gives us the angular momentum he contributes. To calculate the angular velocity, we divide this angular momentum by the total moment of inertia of the system, which is the sum of the merry-go-round and the boy's moment of inertia about the axis of rotation (which since he is treated as a point mass, is just his mass times the radius squared).
The boy's moment of inertia will be 50 kg × (2.0 m)2 = 200 kg·m2. Adding this to the merry-go-round's moment of inertia (300 kg·m2 + 200 kg·m2 = 500 kg·m2), we use the formula:
L = Iω
Where L is the angular momentum, I is the total moment of inertia, and ω is the angular velocity. Solving for ω gives us:
ω = L / I = (200 kg·m/s × 2.0 m) / 500 kg·m2 = 0.8 rad/s
Therefore, the angular velocity after the boy jumps on the merry-go-round is 0.8 rad/s.