Question

A 25.25 kg child is riding a playground merry-go-round that is rotating at 35.5 rev/min.

(A) What net force is acting on her if she is standing on the merry-go-round 1.4 m from its center in newtons? The net force in this situation is sometimes called "centripetal force".

(B) What net force (in N) is acting on her if she is standing on an amusement park merry-go-round that rotates at 3.2 rpm and she is 7.6 m from its center?

(C) How many times her weight is the force in part (a)?

(D) How many times her weight is the force in part (b)?

Answer 1

(a)

Expression for centripetal force,

F = mω²r

Substitute numerical values

F = ( 25.25 kg ) (35. 5 rev/min * 2πrad/rev * 1min/60s)² * (1.4 m)

F = 488.54 N

If round off the result to 3 significant digits

F = 489 N

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( B )

Circumference of the path = 2 π (7.6 m) = 47.75 m

Find the distance the child moves during 3.2 rotations.

d = 3.2 ( 47.75 m ) = 152.8 m

Find the velocity in m/s by divide this distance by 60. ( since 1 min = 60 second)

v = 152.8 m / 60 s = 2.546 m/s

Therefore, the centipetal force

Fc = m v^2 /r = (25.25 ) (2.546 m/s)^2 / 7.6 = 21.54 N

In three significant digits, Fc = 21.5 N

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(c)

Weight of child, W = mg = 25. 25 kg x 9.8 m/s^2 = 247.45 N

Ratio of force in part (a) to weight ;  488.54 N / 247.45 N = 1.97

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(d)

Ratio of force in part (b) to weight ;  21.54 N / 247.45 N = 0.0871