circular-motion addict of mass 60 kgrides a Ferris wheel around in a vertical ci

Question

circular-motion addict of mass 60 kgrides a Ferris wheel around in a vertical circle of radius 10 m ata constant speed of 6.1 m/s. (a) What is the period of the motion?
1 s

(b) What is the magnitude of the normal force on the addict fromthe seat when both go through the highest point of the circularpath?
2 N

(c) What is it when both go through the lowest point?
3 N (a) What is the period of the motion?
1 s

(b) What is the magnitude of the normal force on the addict fromthe seat when both go through the highest point of the circularpath?
2 N

(c) What is it when both go through the lowest point?
3 N

Explanation / Answer

1) period = 2r / velocity = 2(10 m)/ 6.1m/s = 10.3 s 2) Fc=mv2/r = 60 kg * (6.1m/s)2 / 10m = 223.26 N      Fg=mg = 60 kg * 9.8m/s2 = 588 N so at the highest point, gravity is pulling down at 588 N, androtational force is pushing up at 223.26 N, so 588 N - 223.26 N =364.74 N. Normal force is equal in magnitude but opposite indirection to this force. They only ask for magnitude, so364.74 N 3) at the low point, gravity and rotational force arepulling in the same direction, so 588 N - 223.26  N=811.26 N. Normal force is equal in magnitude but opposite indirection to this force. They only ask for magnitude, so811.26 N

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