A 40.0-kg child running at a speed 3.00 m/s jumps on a stationary playground merry-go-round at a distance 1.50 m from the axis of rotation of the merry-go-round. The child is traveling tangential to the edge of the merry-go-round which has a 600 kg∙m² moment of inertia about its axis of rotation as she is running. What is the angular speed of the merry-go-round after the child jumps on it? A) 0.788 rad/s B) 0.261 rad/s C) 6.28 rev/s D) 2.00 rad/s E) 3.14 rev/s

Question

Quizplus · Accepted Answer

Before the child jumps on the merry-go-round, the total angular momentum of the system is zero because both the child and the merry-go-round are stationary. After the child jumps on the merry-go-round, the total angular momentum is conserved:

Iωf = Iωi + mvr

where I is the moment of inertia of the merry-go-round, ωi is its initial angular speed (zero), and ωf is its final angular speed. Solving for ωf gives:

ωf = (mvr)/(I) = (40.0 kg)(3.00 m/s)(1.50 m)/(600 kg∙m²) = 0.261 rad/s

Therefore, the correct answer is B.

A 400-Kg Child Running at a Speed 3