Asked by andres
An amusement park ride consists of a large vertical cylinder that spins about its axis fast enough that a person inside is stuck to the wall and does not slide down when the floor drops away.
The acceleration of gravity is 9.8 m/s2 .Given g = 9.8 m/s2, the coefficient μ = 0.2 of static friction between a person and the wall, and the radius of the cylinder R = 6.5 m. For simplicity, neglect the person’s depth and assume he or she is just a physical point on the wall. The person’s speed is v=((2*pi)*radius)/T.
where T is the rotation period of the cylinder (the time to complete a full circle).
Find the maximum rotation period T of the cylinder which would prevent a 48 kg person from falling down.
Answer in units of s
thanks
The acceleration of gravity is 9.8 m/s2 .Given g = 9.8 m/s2, the coefficient μ = 0.2 of static friction between a person and the wall, and the radius of the cylinder R = 6.5 m. For simplicity, neglect the person’s depth and assume he or she is just a physical point on the wall. The person’s speed is v=((2*pi)*radius)/T.
where T is the rotation period of the cylinder (the time to complete a full circle).
Find the maximum rotation period T of the cylinder which would prevent a 48 kg person from falling down.
Answer in units of s
thanks
Answers
Answered by
samuel
dont get it
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