c,d,e,f (5 %) Problem 20: A merry-go-round is a playground ride that consists of
ID: 2038607 • Letter: C
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c,d,e,f
(5 %) Problem 20: 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 = 1.1 meters, and a mass M= 206 kg. A small boy of mass m - 41 kg runs tangentially to the merry-go-round at a speed of v -2.5 m/s, and jumps on Randomized Variables R= 1.1 meters M=206 ke m=41k v 2.5 m/s Otheexpertta.com 17% Part (a) Calculate the moment of inertia of the merry-go-round, in kg-m2 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 i umps On the merry go round, calculate the angular speed in radians/second of the merry-go-round and boy Grade Summary Deductions Potential Late Work % 80% Late Potential 80% 2(2.273*41 * (1.1)2)M 0% 400% 7 89 HOME cosO cotan)asinO Sin ? acoso Submissions Attempts remaining: 10 1 2 3 atan)acotan(O coshO tanh0 cotanhO % per attempt) detailed view 0 END Degrees Radians VO BACKSPACECLEAR Submit Hint Hints: 1 % deduction per hint. Hints remaining: 5 Feedback: 0% deduction per feedback. the merry-go-round when the boy is half way between the edge and the center of the merry go round? when the boy is at the center of the merry go round? 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 I7% 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 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 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 in radians/second of the merry-go-round after the boy jumps off?Explanation / Answer
a) I = 0.5*M*R^2
= 0.5*206*1.1^2
= 124.6 kg.m^2
b) w = v/R
= 2.5/1.1
= 2.27 rad/s
c) Apply conservation of angular momentum
final angular momentum = initial angulau momentum
(0.5*M*R^2 + m*R^2)*w2 = m*v*R
==> wf = m*v*R/(0.5*M*R^2 + m*R^2)
= 41*2.5*1.1/(0.5*206*1.1^2 + 41*1.1^2)
= 0.647 rad/s
d)
Again apply conservation of angular momentum
(0.5*M*R^2 + m*(R/2)^2)*w3 = (0.5*M*R^2 + m*R^2)*w2
w3 = (0.5*M*R^2 + m*R^2)*w2/(0.5*M*R^2 + m*(R/2)^2)
= (0.5*206*1.1^2 + 41*1.1^2)*0.647/(0.5*206*1.1^2 + 41*(1.1/2)^2)
= 0.823 rad/s
e)
Again apply conservation of angular momentum
0.5*M*R^2*w4 = (0.5*M*R^2 + m*R^2)*w2
w4 = (0.5*M*R^2 + m*R^2)*w2/(0.5*M*R^2)
= (0.5*206*1.1^2 + 41*1.1^2)*0.647/(0.5*206*1.1^2)
= 0.904 rad/s
e) 0.5*m*R^2*wf = (0.5*M*R^2 + m*R^2)*w2
wf = (0.5*M*R^2 + m*R^2)*w2/(0.5*m*R^2)
= (0.5*206*1.1^2 + 41*1.1^2)*0.647/(0.5*206*1.1^2)
= 0.904 rad/s
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