In a playground there is a small merry-go-round of radius \(1.22 \mathrm{~m}\) and mass \(176 \mathrm{~kg} .\) The radius of gyration (see Exercise \(9-20\) ) is \(91.6 \mathrm{~cm}\). A child of mass \(44.3 \mathrm{~kg}\) runs at a speed of \(2.92 \mathrm{~m} / \mathrm{s}\) tangent to the rim of the merry-go-round when it is at rest and then jumps on. Neglect friction between the bearings and the shaft of the merry-go-round and find the angular speed of the merry-go-round and child.

Before the child jumps onto the merry-go-round, the system (child+merry-go-round) is at rest. However, the child is running at a speed of \(2.92 \mathrm{~m} / \mathrm{s}\) tangent to the rim of the merry-go-round. This motion of the child contributes to the initial angular momentum of the entire system. Angular momentum can be obtained using the formula \[ L = mvr \] where m is the mass of the running child (44.3 kg), v is the velocity (2.92 m/s), and r is the radius of the merry-go-round (1.22 m). Calculation gives that, \( L = 44.3 \times 2.92 \times 1.22 = 157.337 kg.m²/s \]

After the child jumps onto the merry-go-round, the system (child+merry-go-round) moves with an angular speed \(\omega\). The child and the merry-go-round are now moving in rotation about the same axis and hence their angular momenta can be given as \(I\omega\) and \(mr²\omega\) respectively. Here I is the moment of inertia of the merry-go-round (which can be calculated using the formula \(I=mr_{g}^{2}\) where \(m\) is mass of merry-go-round and \(r_{g}\) is radius of gyration, and r being the distance from where the child jumps on to the center of the rigid body. We need to find the total angular momentum after the child jumps onto the merry-go-round. The total angular momentum is therefore given by \(I\omega + mr²\omega\).

From the law of conservation of angular momentum, the initial angular momentum must be equal to the final angular momentum. Hence, \(mvr = I\omega + mr²\omega\). We can solve this equation to find the angular speed (\(\omega\)) after the child jumps onto the merry-go-round. By substitution all the given data and solving we find the value of \(\omega\) to be approximately 0.114 rad/s.