| Grade Summary Submissions (11%) Problem 9: A merry-go-round is a playground ri
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(11%) Problem 9: 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 = 1.5 meters, and a mass M = 226 kg. A small boy of mass m = 42 kg runs tangentially to the merry-go-round at a speed of v = 1.3 m/s, and jumps on.Randomized Variables R = 1.5 meters
M = 226 kg
m = 42 kg
v = 1.3 m/s 17% Part (a) Calculate the moment of inertia of the merry-go-round, in kg m2. I =
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Attempts remaining: 15 (1% per attempt) detailed view Hints: 1% deduction per hint. Hints remaining: 3 Feedback: 1% deduction per feedback. 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 boy.
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?
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)
For the merry-go-round, I = 0.5 * m * r^2 = 0.5 * 226 * 1.5^2 = 254.25 kg m^2
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.
Angular velocity = tangential velocity / radius
Initial angular velocity of the boy = 1.3 / 1.5 = 0.867 rad/s
c)
Momentum is always conserved
Angular momentum = I * omega
For the boy, I = m * r^2 = 42 * 1.5^2 = 94.5
Initial angular momentum of the boy = 94.5 * 0.867 = 81.9
For the merry-go-round, I = 0.5 * m * r^2 = 0.5 * 226 * 1.5^2 = 254.25
Initial angular velocity of disk = 0
Initial angular momentum of disk = 0
Total initial angular momentum = 81.9
Since angular momentum is conserved, the total angular momentum of boy and merry-go-round will always be 81.9
When the boy is on the merry-go- round, he is rotating at the angular velocity as the disk.
Let omega * f = final angular velocity as the disk and boy
Final angular momentum of boy = 94.5 * omega * f
Final angular momentum of merry-go- round = 254.25 * omega * f
94.5 * oemga* f + 254.25 * omega *f = 348.75 * omega*f
This is the angular momentum of the boy and merry-go- round, after the boy has jumped onto the merry-go- round.
Total final angular momentum = Total initial momentum
348.75 * omega*f = 81.9
omega *f = 81.9 / 348.75 = 0.235 rad/s
This is the angular velocity of the boy and merry-go- round immediately after the boy jumps onto the merry-go- round.
d)
As the crawls towards the halfway to the center of the merry-go-round along a radius, his moment of inertia decreases. The moment of inertia of the merry-go-round is constant.
New moment of inertia of boy = 42 * 0.75^2 = 23.625
New angular momentum of boy = 23.625 * omega
New angular momentum of merry-go-round = 254.25 * omega
23.625 * omega + 254.25 * omega = 277.87 * omega
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
277.87 * omega = 81.9
omega = 81.9 / 277.87 = 0.294rad/s
This is the angular velocity of boy and merry-go- round, when the boy is halfway to the center.
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