Answer to Question #80546 in Mechanics | Relativity for Ankit

Question

A child of mass 50 kg is standing on the edge of a merry go round of mass 250 kg and
radius 3.0 m which is rotating with an angular velocity of 3.0 rad s−1. The child then
starts walking towards the centre of the merry go round. What will be the final angular
velocity of the merry go round when the child reaches the centre?

Answer 1

We can name the system, which consists of child and merry-go-round as isolated and use the law of conservation of angular momentum. It states that “angular momentum can be exchanged between objects in a closed system, but total angular momentum before and after an exchange remains constant”. At the beginning of an experiment total angular momentum about the axis that goes vertically through the center of merry-go-round is: L_1+L_2=m_1 r^2 ω+1/2 m_2 r^2 ω=ωr^2 (m_1+1/2 m_2), where L1 – is an angular momentum of child, L2 – is an angular momentum of merry-go-round, m1 – is a mass of a child, m2 – is a mass of a merry-go-round. At the end of experiment total angular momentum about the axis that goes vertically through the center of merry-go-round is: L_1^'+L_2^'=m_1 〖r^'〗^2 ω^'++1/2 m_2 r^2 ω^'=1/2 m_2 r^2 ω^', cause, when the child reaches the center, its radius-vector is zero.
So: ωr^2 (m_1+1/2 m_2 )=1/2 m_2 r^2 ω^'; ω^'=(2ωr^2 (m_1+1/2 m_2 ))/(m_2 r^2 )=2ω(m_1+1/2 m_2 )/m_2 =(2*3(50+1/2*250))/250=(6*175)/250=4.2 rad/s