A 25.0 -kg boy stands \(2.00 \mathrm{~m}\) from the center of a frictionless playground merry-go-round, which has a moment of inertia of \(200 . \mathrm{kg} \mathrm{m}^{2} .\) The boy begins to run in a circular path with a speed of \(0.600 \mathrm{~m} / \mathrm{s}\) relative to the ground. a) Calculate the angular velocity of the merry-go-round. b) Calculate the speed of the boy relative to the surface of the merry-go- round.

Initially, the merry-go-round is at rest and the boy is not moving relative to it. So, the initial angular momentum of the system (merry-go-round and the boy) is zero.

When the boy starts running, he will have an angular momentum relative to the center of the merry-go-round. The final angular momentum of the boy can be found using the formula: \(L = r \times p\), where \(L\) is the angular momentum, \(r\) is the distance from the center, and \(p\) is the linear momentum of the boy (mass times velocity). $$ L_{boy} = r \times m_{boy} \times v_{boy} $$Plug in the given values: $ L_{boy} = 2.00\,\mathrm{m} \times 25.0\,\mathrm{kg} \times 0.600\,\mathrm{m/s} $$Solve for \(L_{boy}\): $ L_{boy} = 30.0\,\mathrm{kg\,m^{2}/s} $

By the conservation of angular momentum principle, the final angular momentum of the system equals the initial angular momentum. As the initial angular momentum was zero, the final angular momentum of the merry-go-round should be equal in magnitude but opposite in direction to the final angular momentum of the boy: $ L_{merry-go-round} = -L_{boy} = -30.0\,\mathrm{kg\,m^{2}/s} $

Now, we can use the relationship between angular momentum and angular velocity to find the angular velocity of the merry-go-round: $ L = I \times \omega \(Where \)L\( is the angular momentum, \)I\( is the moment of inertia, and \)\omega\( is the angular velocity. Rearrange the formula to solve for \)\omega\(: \) \omega = -\frac{L_{merry-go-round}}{I} \(Plug in the given values: \) \omega = -\frac{-30.0\,\mathrm{kg\,m^{2}/s}}{200\,\mathrm{kg\,m^{2}}} \(Solve for \)\omega\(: \) \omega = 0.15\,\mathrm{rad/s} \(So, the angular velocity of the merry-go-round is \)0.15\,\mathrm{rad/s}$.

To find the speed of the boy relative to the surface of the merry-go-round, we have to find the difference in their velocities. First, find the linear velocity of the merry-go-round surface relative to the center: $ v_{merry-go-round} = r \times \omega $ Plug in the given values: $ v_{merry-go-round} = 2.00\,\mathrm{m} \times 0.15\,\mathrm{rad/s} $ Calculate the linear velocity of the merry-go-round surface relative to the center: $ v_{merry-go-round} = 0.30\,\mathrm{m/s} $ Now, we can find the relative speed of the boy with respect to the merry-go-round surface by subtracting the linear velocity of the merry-go-round surface from the speed of the boy relative to the ground: $ v_{relative} = v_{boy} - v_{merry-go-round} $ Substitute the values: $ v_{relative} = 0.600\,\mathrm{m/s} - 0.30\,\mathrm{m/s} $ Calculate the relative speed: $ v_{relative} = 0.300\,\mathrm{m/s} $ So, the speed of the boy relative to the surface of the merry-go-round is \(0.300\,\mathrm{m/s}\).