A merry-go-round with a moment of inertia of 6.78  10³ kg · m² is coasting at 2.20 rad/s. When a 72.6-kg man steps onto the rim, the angular velocity decreases to 2.0 rad/s. The radius of the merry-go-round is

Question

Quizplus · Accepted Answer

Using conservation of angular momentum, L_initial = L_final. The initial angular momentum is I_initial * ω_initial. The final angular momentum is (I_initial + m*r^2) * ω_final. Solving for r gives r = sqrt((I_initial * (ω_initial - ω_final)) / (m * ω_final)). Substituting the given values results in r ≈ 3.06 m.

A Merry-Go-Round with a Moment of Inertia of 6