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Dmitry_Shevchenko [17]
3 years ago
12

A girl and a boy are riding on a merry-go-round that is turning at a constant rate. The girl is near the outer edge, and the boy

is closer to the center. For a given elapsed time interval, which rider has greater angular displacement?
A. Both the girl and the boy have the same nonzero angular displacement.
B. Both the girl and the boy have zero angular displacement.
C. The boy has greater angular displacement.
D. The girl has greater angular displacement.
Physics
1 answer:
masha68 [24]3 years ago
4 0

Answer:

A. Both the girl and the boy have the same nonzero angular displacement.

Explanation:

Since girl and boy both are standing on the same merry go round so here we can say that boy and girl both will revolve with same angular speed as the speed of the merry go round

so the angular displacement is given as

\theta = \omega t

so here since both has same angular speed that is the speed of the disc so they both will turn by same angle in the same interval of time

So we have correct answer as

A. Both the girl and the boy have the same nonzero angular displacement.

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To solve the problem it is necessary to take into account the concepts of kinematic equations of motion and the work done by a body.

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The distance in this case is a composition between number of steps and the height. Then,

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On the other hand we have the speed changes, depending on the displacement and acceleration (omitting time)

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W = 50*9.8*(0.3*30)

W = 4.41kJ

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PART B) To find the acceleration (from which we can later find the time) we start from the previously given equation,

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Here,

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Replacing,

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a = -\frac{0.9^2}{2*30*0.3}

a = -45*10^{-3}m/s^2

At this point we can calculate the time, which is,

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