What is the minimum speed of a roller coaster at the top of a 39.0 meter tall vertical loop if the passengers are "weightless" at that point?
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Expert's answer
2011-05-12T11:22:38-0400
In the top point of the motion the centripetal acceleration should be equal to the gravitation acceletation: g = V2/R Thus V = √(gR) = √(10*39) = 19.75 m/s
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