The distance from the center of a ferris wheel to a passenger seatis 10 meters.

Question

The distance from the center of a ferris wheel to a passenger seatis 10 meters. The ferris wheel starts from rest and builds up to afinal angular operating speed of 0.80 rad/sec while rotationthrough an angular displacement or 6.0 rad

A. What is the angular acceleration of the ferris wheel from restto it operating speed -assume constant angular acceleration

B. Once the. Ferris wheel has its operating speed what is theapparent weight of a 75 kg passenger at the bottom of the ferriswheel rotation?

Please show final answer and all work done to get that answer

Explanation / Answer

you would use the equation: 2 =o2 + 2() When you plug in the values to solve for :(.8rad/sec)2/2(6rad) = 0.0533rad/sec2 B. The apparent weight at the bottom is equal to thenormal force. Fc = FN + FW FN = FC + FW v = r = .8rad/sec * 10m = FC = mv2/r =75kg(8m/s)2/10m FN= 480N + (75 * 9.8m/s2) = 1215N v = r = .8rad/sec * 10m = FC = mv2/r =75kg(8m/s)2/10m FN= 480N + (75 * 9.8m/s2) = 1215N

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