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While at the county fair, you decide to ride the Ferris wheel. Having eaten too many candy apples and elephant ears, you find the motion somewhat unpleasant. To take your mind off your stomach, you wonder about the motion of the ride. You estimate the radius of the big wheel to be 14 mm , and you use your watch to find that each loop around takes 23 ss .

1 Answer

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Answer:


\omega = 0.273 rad/s


v = 3.825 m/s

Step-by-step explanation:

Assuming that the question is asking for the angular and linear speed of the Ferris wheel, we can solve by converting from period of 1 revolution (23 s) to angular speed knowing that 1 revolution would sweep and angle of 2π.


\omega = (2\pi)/(T) = (2\pi)/(23) = 0.273 rad/s

We can calculate the linear speed of the motion from the angular speed and the radius


v = \omega r = 0.273 * 14 = 3.825 m/s

User Dennis Novac
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