(8%) Problem 7: A merry-go-round is a playground ride that consists of a large d
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Question
(8%) Problem 7: 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= 251 kg. A small boy of mass m 42 kg runs tangentially to the merry-go-round at a speed of v 1.9 m/s, and jumps on. Randomized Variables R= 1.1 meters M = 251 kg m=42 kg v=1,9 m/s ©theexpertta.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 jumps on the merry go round, calculate the angular speed in radians second of the merry-go-round an Grade Summary Potential Submissions Attempts remaining: 1 2 18% 82% Sin cotan asi asin) acos % per attempt) detailed view atan0acotan) si coshOtanhO) cotanh0 4% 4% 4% 4% 0 END Degrees O Radians 4 Submit Hint give up Hints: 1 for a 2% deduction. Hints remaining: 4 Feedback: 2% deduction per feedback. Taking the boy and the merry-go-round as a system, there is a quantity that is conserved in this interaction. 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? là 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) moment of inertia of disc = M R^2 /2 = 251 x 1.1^2 / 2 =151.85 kg m^2
b) initially it was at rest , so w = 0
c) using angular momentum conservation,
m v R = ( I + mR^2)w
42 x 1.9 x 1.1 = ( 151.85 + (42 x 1.1^2)) w
w = 0.433 rad/s
d) using angular momentum again
42 x 1.9 x 1.1 = ( 151.85 + ( 42 x (1.1/2)^2) w
w = 0.533 rad/s
e) using angular momentum again
42 x 1.9 x 1.1 = ( 151.85 + ( 42 x0) w
w = 0.578 rad/s
f) using angular momentum again
42 x 1.9 x 1.1 = ( 151.85 ) w
w = 0.578 rad/s
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