(20%) Problem 3: A merry-go-round is a playground ride that consists of a large
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(20%) Problem 3: 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= 301 kg. A small boy of mass m = 43 kg runs tangentially to the merry-go-round at a speed of v = 1.9 m/s, and jumps on. Randomized Variables R=1.5 meters M= 301 kg m = 43 kg y = 1.9 m/s Ctheexpertta.com 2 / 17% Part (a) Calculate the moment of inertia of the merry-go-round, in kg .m“. C / 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. 2 % 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? Grade Summary 03 = 0.266667| Deductions 0% Potential 100% sin() cos() tan) n ( 7 8 9 HOME Submissions cotan() asin() acos() 4 5 6 Attempts remaining: 4 atan() acotan() sinh) 1 2 3 (0% per attempt) detailed view cosh() | tanh). cotanh) END O Degrees O Radians BACKSPACE DEL CLEAR Submit Hint Feedback I give up! Hints: 0% deduction per hint. Hints remaining: 2 Feedback: 0% deduction per feedback. Submission History Answer Hints Feedback Totals W3 = 0.266667 0 % 0% 0% 03 = 0.2667 Totals 0% 0% 0% 09% 2 * 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? m 4 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_merry = 0.5*M*R^2
= 0.5*301*1.5^2
= 338.6 kg.m^2
b) w = v/R
= 1.9/1.5
= 1.27 rad/s
c) Apply conservation of angular mometum
(I_merry + m*R^2)*wf = m*v*R
wf = m*v*R/(I_merry + m*R^2)
= 43*1.9*1.5/(338.6 + 43*1.5^2)
= 0.281 rad/s
d) (I_merry + m*(R/2)^2)*wf' = (I_merry + m*R^2)*wf
wf' = (I_merry + m*R^2)*wf/(I_merry + m*(R/2)^2)
= (338.6 + 43*1.5^2)*0.281/(338.6 + 43*(1.5/2)^2)
= 0.337 rad/s
e) I_merry*wf'' = (I_merry + m*R^2)*wf
wf'' = (I_merry + m*R^2)*wf/I_merry
= (338.6 + 43*1.5^2)*0.281/338.6
= 0.361 rad/s
e) (I_merry + m*R^2)*wf = I_merry*w
==> w = (I_merry + m*R^2)*wf/I_merry
= (338.6 + 43*1.5^2)*0.281/338.6
= 0.361 rad/s
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