A 41.0-kg child running at 1.05 m/s suddenly jumps onto a stationary playground merry-go-round at a distance 1.5 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 at the edge. The moment of inertia of the merry-go-round about its axis of rotation is 550 kg. m2 and you can neglect friction at its rotation axis. What is the angular speed of the merry-go-round just after the child has jumped onto it? 2.00 rad/s None of the other responses are correct. 0.13 rad/s .10 rads/s 0.12 rad/s .11 rads/s 94 revs/sec
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Step 1: The child's initial angular momentum is $L_i = I_c \omega_c = m_c r^2 \omega_c = m_c r v_c$, where $m_c$ is the child's mass, $r$ is the distance from the axis of rotation, and $v_c$ is the child's speed. Show more…
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