In a playground there is a small merry-go-round of radius 1.25 m and mass 175 kg.Assume the merry-go-round to be a uniform disc.A child of mass 45 kg runs at a speed of 3.0 m/s tangent to the rim of the merry-go-round (initially at rest)and jumps on.If we neglect friction,what is the angular speed of the merry-go-round after the child has jumped on and is standing at its outer rim?
A)0.82 rad/s
B)2.4 rad/s
C)0.49 rad/s
D)1.2 rad/s
E)0.41 rad/s

Question

Quizplus · Accepted Answer

The initial angular momentum of the system is due to the child running, which is m*v*r. The final angular momentum is the sum of the angular momentum of the child and the merry-go-round. Using conservation of angular momentum: m*v*r = (I_merry-go-round + I_child)*ω. Solving for ω gives 0.82 rad/s.

In a Playground There Is a Small Merry-Go-Round of Radius