A 40.0-kg child running at 3.00 m/s suddenly jumps onto 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 just before jumping on.The moment of inertia about its axis of rotation is 600 kg ∙ m² and very little friction at its rotation axis.What is the angular speed of the merry-go-round just after the child has jumped onto 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 onto the merry-go-round, the total angular momentum is zero since the child is traveling tangentially and the merry-go-round is stationary. After the child jumps on, the total angular momentum is conserved. Therefore, the child's initial angular momentum will be equal to the final angular momentum of both the child and the merry-go-round. 
Initially, the child has an angular momentum of L = mvR, where v is the initial tangential velocity of the child and R is the distance from the axis of rotation. Thus, L = (40.0 kg) (3.00 m/s) (1.50 m) = 180 kg*m^2/s. 
Since the merry-go-round is at rest initially, its initial angular momentum is zero. Therefore, the final angular momentum is also 180 kg*m^2/s. 
Using the equation for angular momentum, L = Iω, where I is the moment of inertia and ω is the angular velocity, we can solve for ω: 
ω = L/I = 180 kg*m^2/s / 600 kg*m^2 = 0.3 rad/s 
Therefore, the angular speed of the merry-go-round just after the child has jumped onto it is 0.3 rad/s or choice B.

A 400-Kg Child Running at 3