A merry-go-round is a playground ride that consists of a large disk mounted to t

Question

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 = 201 kg. A small boy of mass m = 44 kg runs tangentially to the merry-go-round at a speed of v = 1.1 m/s, and jumps on.

Moment of Inertia calculated: 196.98 kgm^2

Angular Speed about the central axis of the merry-go-round calculated: 0.786 m/s^2

1.Immediately after the boy jumps on the merry go round, calculate the angular speed in radians/second of the merry-go-round and boy.

2.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?

3.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?

4.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

radius R = 1.4m mass of merrygoround M = 201kg

mass of the boy m= 44kg veocity v = 1.1m/s

1.Angular speed of the boy and the merry go round w = v/R = 1.1 / 1.4 = 0.786 rad/s

moment of inertia of the boy I = mr2 = 44 x 1.4 x 1.4 = 86.24 kg m2

angular momentum of the boy Lb = Iw = 86.24 x 0.786 = 67.78 kgm2/s

moment of inertia of the merry go round I = 0.5mr2  = 0.5 x 201 x 1.4 x 1.4 = 196.98 kgm2

angular momentum of the merry go round L = I w = 0 since initiallu it is at rest

total initial angular momentum of the system = 67.78 + 0 = 67.78 kgm2/s

let the final angular speed be wf

from conservation of angular momentum

initial angular momentum = final angular momentum

67.78 = (86.24 +196.98) wf

angular speed wf = 67.78/ 283.22 = 0.2393 rad/s

2. for boy r = 1.4/2 = 0.7 m

angular momentum of the boy L = Iw = 44 x 0.7x 0.7 w = 21.56 w

angular momentum of the merry go round L = 196.98 w

toal angular momentum = 21.56w + 196.98 w = 218.54 w

from conservation of angular momentum it should be equal to 67.78

hence 218.54w = 67.78

angular speed w = 67.78/218.54 = 0.31 rad/s

3.at the centre moment of inertia of the boy is zero hence angular momentum will be zero

hence the toal angular momentum is the angular momentum of the merry go round

ie. 196.98 w

  from conservation of angular momentum it should be equal to 67.78

hence 196.98w = 67.78

angular speed w = 67.78/196.98 = 0.344 rad/s

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