A 39.0 kg child takes a rideon a Ferris wheel that rotates four times each minut

Question

(a) What is the centripetal acceleration of the childat the lowest point of the ride?

(b) What force (magnitude and direction) does the seat exert on thechild at the lowest point of the ride?

magnitude=

direction=

(c) What force does the seat exert on the child at thehighest point of the ride?
magnitude=

direction=

(d) What force does the seat exert on the child whenthe child is halfway between the top and bottom?

magnitude=

direction=

Explanation / Answer

centrifugal acceleration is given at v2/r and we have v=r thus the acceleration can be described as 2r = 4revs/min = .4189rad/s r = 16m a = 2.81m/s2 at the bottom the centrifugal acceleration and acceleration due togravity will add together so at the bottom the total acceleration is 12.61m/s2 so the force is 39kg * 12.61m/s2 at the top the will subtract so the total acceleration is9.8m/s2-2.81m/s2 = 6.99m/s2 so the force will be 39kg * 6.99m/s2 we have to use the pythagorean theorem for the acceleration at themid point you will have a force equal to 9.8m/s2 * 39kg pushingup and a force equal to 2.81m/s2 *39kg pushing the childinward let me know if you need any more help from here

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