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 $600 \mathrm {~kg} \cdot \mathrm { m } ^ { 2 }$ 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) 0.261 rad/s
B) 3.14 rev/s
C) 2.00 rad/s
D) 6.28 rev/s
E) 0.788 rad/s

Question

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 $600 \mathrm {~kg} \cdot \mathrm { m } ^ { 2 }$ 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)  0.261 rad/s
B)  3.14 rev/s
C)  2.00 rad/s
D)  6.28 rev/s
E)  0.788 rad/s

Quizplus · Accepted Answer

The angular speed of the merry-go-round after the child jumps on can be found using the conservation of angular momentum. The initial angular momentum (L_initial) is the child's linear momentum (m*v) times the radius (r), and there's no initial angular momentum of the merry-go-round since it's stationary. So, L_initial = m_child * v_child * r. After the child jumps on, the total moment of inertia (I_total) is the sum of the merry-go-round's moment of inertia (I_merry-go-round) and the child's moment of inertia (m_child * r^2), and the angular speed (ω) can be found by setting the initial angular momentum equal to the final angular momentum (I_total * ω). Solving for ω gives the answer.

A 400-Kg Child Running at 3 600 kg⋅m2600 \mathrm {~kg} \cdot \mathrm { m } ^ { 2 }600 kg⋅m2

A 400-Kg Child Running at 3 $600 \mathrm {~kg} \cdot \mathrm { m } ^ { 2 }$