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A circular-motion addict of mass 83.0 kg rides a Ferris wheel around in a vertical circle of radius 13.0 m at a constant speed of 6.10 m/s. (a) What is the period of the motion? What is the magnitude of the normal force on the addict from the seat when both go through (b) the highest point of the circular path and (c) the lowest point?

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Answer:

(A) Time period T = 6.28 SEC

(B) At highest point fore is - 575.83 N

(B) At lowest point force is 1050.97 N

Step-by-step explanation:

We have given that mass m = 83 kg

Radius r = 13 m

Speed v = 6.10 m/sec

(A) Time period of the motion is given by
T=(2\pi r)/(v)=(2* 3.14* 13)/(6.10)=6.28sec

(b) Net force is given by
F_(NET)=(mv^2)/(r)=(83* 6.10^2)/(13)=237.571N

Force due to gravity
F_(gravity)=mg=83* 9.8=813.4N

At highest point
F_(NORMAL)=F_(NET)-F_(GRAVITY)=237.57-813.4=-575.83

(B) At lowest point
F_(NORMAL)=F_(NET)+F_(GRAVITY)=237.57+813.4=1050.97N

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