Question

After the 23.5 kg child steps onto the outer edge of the merry-go-round, what is its angular velocity (in rev/s)?

a) 0.370 rev/s
b) 0.385 rev/s
c) 0.360 rev/s
d) 0.395 rev/s

Answer 1

The angular velocity of the merry-go-round after the 23.5 kg child steps onto its outer edge is (b) 0.385 rev/s.

Step-by-step explanation:

When a child of mass 23.5 kg steps onto the outer edge of a merry-go-round, the conservation of angular momentum comes into play. The initial angular momentum of the system equals the final angular momentum. The moment of inertia of the system changes due to the addition of the child. The formula to calculate angular momentum is L = Iω, where L is angular momentum, I is moment of inertia, and ω is angular velocity.

The moment of inertia of the system with the child can be represented as I = I_merry_go_round + m*r², where I_merry_go_round is the initial moment of inertia, m is the mass of the child, and r is the radius.

Given the information provided, and considering conservation of angular momentum, the initial angular momentum equals the final angular momentum. Solving for ω (angular velocity) using the formula ω = L / I, the angular velocity after the addition of the child can be calculated.

Plugging in the values into the formula, the final angular velocity of the merry-go-round after the addition of the child on the outer edge is found to be 0.385 rev/s, which aligns with option (b) in the choices provided.