Parts e and f confused me please help. Here are my answers for problems A throug
ID: 1787612 • Letter: P
Question
Parts e and f confused me please help. Here are my answers for problems A through D
a)133.7 b) 1.273 c) 0.3444 d) 0.4321
Thank you in advance!
(1796) Problem 6: 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= 221 kg. A small boy of massm41 kg runs tangentially to the merry-go-round at a speed of v = 1.4 m/s, and jumps on. Randomized Variables R= 1.1 meters M= 221 kg m=41kg v= 1.4 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 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 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 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 axis of the merry-go-round and bov 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? Grade Summarv Deductions Potential 0% 100% sin0 cotan0asin acos0 atan0 acotan sinhO cosh0 tanh cotanh0 cos) tan Submissions Attempts ir (390 per attempt) detailed view END Degrees Radians Submit Hint I give up! Hints: 006 deduction per hint. Hints remaining Feedback: 0% deduction per feedback. 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
Using law of conservation of angular momentum
angular momentum when the boy is at outer edge = angular momentum of the system when the boy was at the center of the merry go ground
[(0.5*M*R^2)+(m*R^2)]*w1 =(0.5*M*R^2*w)
(((0.5*221*1.1^2)+(41*1.1^2))*0.3444 = (0.5*221*1.1^2)*w1
w1 = 0.472 rad/s
f) when he jumps off,the angular velocity of the merry go round is w = 0.472 rad/sec