(9%) Problem 10: A merry-go-round is a playground ride that consists ol a large
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(9%) Problem 10: A merry-go-round is a playground ride that consists ol 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 R1.3 meters, and a mass M- 271 kg. A small boy of mass nm = 43 kg runs tangentially to the merry-go-round at a speed of v 1.7 m/s, and jumps on Randomized Variables R-1.3 meters M- 271 kg m- 43 kg v1.7 m/s Otheexpertta.com 17 % Part (a) Calculate the moment of inertia of the merry-go-round, in kg-m2 Grade Summary Deductions Potential 0% 100% cosO tan 78 9 Submissions Sin Attempts remaining: S (1% per attempt) detailed view cotan) asin) acos0 atan) acotan)sinh(0 coshOtanh) cotanhO Degrees O Radians 0 END BACKSPACE Submit Hint Hints: 1 % deduction per hint. Hints remaining 3 Feedback: 0% deduction per feedback Submission History All Date times are displayed in Eastern Standard Time Red submission date times indicate late work. Date Answer Hints Feedbaclk ? 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 17% 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 bo 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? -là 17% 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? -? 17% 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?Explanation / Answer
a) I = ½ * m * r^2 = ½ * 271 * 1.3^2 = 229 kg.m^2
b) Angular velocity = tangential velocity ÷ radius
Initial angular velocity of the boy = 1.7 ÷ 1.3 = 1.308 rad/s
c) For the boy, I = m * r^2 = 43 * 1.3^2 = 72.67 kg.m^2
By conservation of angular momentum:
(Iw)initial = (Iw)final
=> (72.67*1.308) = (229+72.67)wf
=> wf = 0.315 rad/s
d) New moment of inertia of boy = 43 * 0.65^2 = 18.17
New angular momentum of boy = 18.17 * w
New angular momentum of merry-go-round = 229 * w
18.17 * w + 229 * w = 247.17 * w
This is the angular momentum of the boy and merry-go- round, when the boy is halfway to the center.
Total new angular momentum = Total initial angular momentum
247.17 * w = 72.67*1.308
w = 0.385 rad/s
This is the angular velocity of boy and merry-go- round, when the boy is halfway to the center.
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