(9%) Problem 10: A merry-go round is a playground ride that consists of a large
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(9%) Problem 10: 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 _ 1.4 meters, and a mass M#301 kg. A small boy of mass m- 42 kg runs tangentially to the merry-go-round at a speed of v 2.2 m/s, and umps on R-14 meters M= 301 kg m 42 kg 2.2 m/s Ctheexperita.com ? 17% hrt (a) Calculate the moment of inertia of the merry-go-round, in kg-m2. Grade Summary Potenti 100% sin coso tan789 atan0 acotan0 sinh0 htanh0 cotanho cotanoasin0acos0 Atempts nemaining ( per attemp 12 3 Degrees Radians Suhnit lint 1% deduction per hint, Hits remaining Feedlack per foodback Submission History Date Tiew Answer Hints Feidback 17% Part )Immediately before the boy jumps on the merry go round,calculate his angular speed (in radians/second) about the central xis of the merry-go-round ? 7 % Part c, Immediately after the boy umps on the merry go round. calculate the angular speed in radians second of the merry-go-round nd boy ·17% Part d The boy then crawls towards the center of the merry-go round along radius. What is the angular speed in radians second or he merry-go-round when the boy is half way between the edge and the center of the merry go round? e) The boy then crawls to the center of the merry go round what is the angular speed in radianshcond of the merry go round 17% Part & & 17% Part íf) when the boy is at the center of the merry go round Finally, the boy decides that he has had enough fun He deciges to crawl to the outer edge of the merry-go-round and Jump ft Somchow, he manages to jump in sich a way that he hits the ground with zero velocity with respect to the ground. What is the angular speed n radians/second of the merry go round after the boy jumps offExplanation / Answer
(a) Moment of inertia of merry go round (I1) = (1/2)MR2 = (1/2)*(301)*1.42 = 294.48 kg-m2
(b) Angular speed before the boy jumps (w) = v/R = 2.2/1.4 = 1.571 rad/s
(c)Initial angular momentum of the system = mvR --------(1)
Final angular momentum = I2W
where I2 = (1/2)MR2 + mR2 = 294.48 + (42)*1.42 = 376.8 kg-m2
W is the final angular velocity
Final angular momentum = 376.8*W ------------(2)
On applying conservation of angular momentum
376.8*W = mvR
376.8*W = 42*2.2*1.4
W = 0.3433 rad/s
(d) Angular momentum when the boy is at half of the radius = I3W2
Where I3 = (1/2)MR2 + m(R/2)2 = 294.48 + 42*(1.4/2)2 = 315.06 kg-m2
Now again applying the conservation of angular momentum
315.06 *W2 = 42*2.2*1.4
W2 = 0.41 rad/s
(e) when boy is at center then the angular momentum of the system will be
=I4W3
Where I4 = (1/2)MR2 + m(0)2 = 294.48 kg-m2
Now applying the conservation of angular momentum
294.48 W3 = 42*2.2*1.4
W3 = 0.439 rad/s
(f) Same as e that is 0.439 rad/s
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