A merry-go-round is a playground ride that consists of a large disk mounted to that it can freely rotate in a horizontal plane. The merry-go-round shown is initially at rest, has a radius R = 1.3 meters, and a mass M = 226 kg. A small boy of mass m = 47 kg runs tangentially to the merry-go-round at a speed of v = 2.1 m/s, and jumps on. Part (a) Calculate the moment of inertia of the merry-go-round, in kg ⋅ m2. Part (b) Immediately before the boy jumps on the merry go round, calculate his angular speed (in radians/second) about the central axis of the merry-go-round. Part (c) Immediately after the boy jumps on the merry go round, calculate the angular speed in radians/second of the merry-go-round and boy. Part (d) The boy then crawls towards the center of the merry-go-round along a radius. What is the angular speed in radians/second of the merry-go-round when the boy is half way between the edge and the center of the merry go round? Part (e) The boy then crawls to the center of the merry-go-round. What is the angular speed in radians/second of the merry-go-round when the boy is at the center of the merry go round? Part (f) Finally, the boy decides that he has had enough fun. He decides to crawl to the outer edge of the merry-go-round and jump off. Somehow, he manages to jump in such a way that he hits the ground with zero velocity with respect to the ground. What is the angular speed in radians/second of the merry-go-round after the boy jumps off?

This is a multi-part physics problem involving concepts of rotational motion, moment of inertia, and conservation of angular momentum. Let's solve it step by step.

Part (a) The moment of inertia (I) of the merry-go-round can be calculated using the formula for a solid disk, which is I = 0.5MR^2. Substituting the given values, we get I = 0.5226(1.3)^2 = 151.97 kg*m^2.

Part (b) The angular speed (ω) of the boy can be calculated using the formula ω = v/r. Substituting the given values, we get ω = 2.1/1.3 = 1.615 rad/s.

Part (c) When the boy jumps on the merry-go-round, the system's total angular momentum is conserved. The initial angular momentum (L_initial) is the sum of the boy's and the merry-go-round's, which is L_initial = I_boyω_boy + I_merryω_merry. Since the merry-go-round is initially at rest, its angular speed is zero. So, L_initial = mvR. The final angular momentum (L_final) is the sum of the boy's and the merry-go-round's after the boy jumps on, which is L_final = (I_boy + I_merry)ω_final. Setting L_initial = L_final and solving for ω_final, we get ω_final = mvR / (I_boy + I_merry) = 472.11.3 / (151.97 + 471.3^2) = 0.733 rad/s.

Part (d) When the boy moves halfway towards the center, his new distance from the center is R/2. The moment of inertia of the boy is now m*(R/2)^2. The total moment of inertia of the system is now I_merry + I_boy = 151.97 + 47*(1.3/2)^2. Using conservation of angular momentum, we can find the new angular speed ω_new = L_initial / (I_merry + I_boy) = 472.11.3 / (151.97 + 47*(1.3/2)^2) = 0.757 rad/s.

Part (e) When the boy moves to the center, his moment of inertia becomes zero because he is at the rotation axis. The total moment of inertia of the system is now just I_merry = 151.97 kgm^2. Using conservation of angular momentum, we can find the new angular speed ω_new = L_initial / I_merry = 472.1*1.3 / 151.97 = 0.859 rad/s.

Part (f) When the boy jumps off, the moment of inertia of the system is again just I_merry = 151.97 kg*m^2. However, the initial angular momentum is now L_initial = (I_merry + I_boy)*ω_e = (151.97 + 0)0.859. Using conservation of angular momentum, we can find the new angular speed ω_new = L_initial / I_merry = (151.970.859) / 151.97 = 0.859 rad/s. The angular speed of the merry-go-round does not change when the boy jumps off because he hits the ground with zero velocity, implying that he imparts no additional angular momentum to the merry-go-round.