Final answer:
By using the conservation of angular momentum, the merry-go-round's angular velocity after a boy jumps on is calculated to be 0.8 s⁻¹.
Step-by-step explanation:
Conservation of Angular Momentum and Merry-Go-Rounds
When a child jumps onto a stationary merry-go-round, the system's total angular momentum must be conserved if no external torques act on the system. We can use the conservation of angular momentum to find the angular velocity after the boy jumps on. The moment of inertia I of the merry-go-round is provided, as well as the radius r, the boy's mass m, and his tangential speed v with which he jumps on.
The initial angular momentum L_initial is zero because the merry-go-round is at rest. When the boy jumps on, he adds his angular momentum to the system. This is equal to
L = mvr where m is his mass, v is his tangential speed and r is the radius of the merry-go-round. The total moment of inertia of the system I_total after the boy jumps on is the sum of the moment of inertia of the merry-go-round I and the boy's moment of inertia I_boy = mr². Therefore, the angular velocity \(ω) of the merry-go-round after the boy jumps on can be calculated using L = I_total\(ω).
To solve this problem, we first need to calculate the boy's moment of inertia. Given his mass and the radius of the merry-go-round, the boy's moment of inertia is I_boy = (50 kg)(2.0 m)² = 200 kg·m². The total moment of inertia of the system is thus I_total = I + I_boy = 300 kg·m² + 200 kg·m² = 500 kg·m². Applying the conservation of angular momentum:
mvr = I_total\(ω), we find the angular velocity \(ω) to be \(\(ω) = \frac{mvr}{I_total} = \frac{(50 kg)(4.0 m/s)(2.0 m)}{500 kg·m²} = 0.8 s⁻¹ which is the final angular velocity of the merry-go-round after the boy jumps on.