Question

correct answers are listed for part A-F but need to know the work done to get results

0%) Problem 10: 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 = 13 meters, and a mass M = 281 kg. A small boy of mass m = 41 kg runs tangentially to the merry-go-round at a speed of v= 24 m/s, and jumps on Randomized Variables R=13 meters M=287 kg m = 41 kg v=2.4 m/s Otheexpertta.com 17% Part (a) Calculate the moment of inertia of the merry-go-round, in kg ·m 1 = 2374 ✓ Correct! 17% 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 - 18 Correct! 17% Part (0) Immediately after the boy jumps on the merry go round, calculate the angular speed in radians/second of the merry-go-round and boy. 2041 ✓ Correct! select part 17% 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? ay 049 ✓ Correct 17% Part (c) 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? e-054 ✓ Correct! 17% Part ( 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? Grade 100% Student Final Submission Correct Answer Not available until end date Feedback Correct! -0.54

Answer 1

(a) Moment of inentia of a disk with respect the axis paring through the centre of the disk and making 90° angle with the disk is given by I - 5 MR? where M = Mass of the disk R= Radius of the disk ... The moment of inertia of the merry-go-round is I= bx 281 kg x 1.-3°m? - 237.4 kg.m? (b) Angular speed of the boy about the central axis of the merry-go-round is Speed of the boy radius of the merry-go 2.4 m/s 1.8 read/s W, 1-round 1.3m m-mass I'- I + me? (c) Moment of inertia of the merry-go-round with the boy on the edge of it is given by of the boy 237.4 kg. m² + 306.7 kg.m² We know that angular momentum (in this carol) L = I'we = mwR² = I'w, 7) W₂ = mw, R² Alkytheralls 41 kg x 1.3² m² T'

w,- 41 kg x 1.8 road /s x 1.3?m? 306.7 kg m² W₂0. 41 rad/s (d) When the boy is half way between the edge and the center of the merry-go-round, moment of inentia of the system is I" - It mi - 237.4 kg.mºt 41 kqx (13)m 272.05 kg. m² Applying conservation of momentum we have I'W2-I"W2 I'w2 306.7 kq.m’xo. Al madls 272.05 kg .m² W₃ = 0.49 nad/s (e) When the boy is at the center of the merry-go-round, the moment of inentia of the =) W3 I" system is I". I + m.o'- I - 237.4 kq.m? .. I'w, = I" WA I'wa 306.7kq.m? x 0. Al madls 237.4 kg. m² WA2 0.54 nad/s - WAT 1"

merry-go -round (f) Since, the boy manages to hits the ground with zero velocity, the angular momentum of the remains conserved. :. The angular speed of the the same as the last case i.e. part (e). merry-go-round remains ...W5WA=0.54 nad/s