An amusement park ride consists of a large veritcal cylinder that spins about it

Question

An amusement park ride consists of a large veritcal cylinder that spins about its axis fast enough that a person inside is stuck on the wall and does not slide down when the floor drops away. The acceleration of gravity is 9.8 m/s^2. Given g= 9.8 m/s^2, the coefficient 0.521 of static friction between the person and the wall, and the radius of the cylinder R= 4.4 m. For simplicity neglect the person's depth and assume he/ she is just a physical point on the wall. The person's speed is v= 2(pi)R/T. Where T is the rotational period of the cylinder. (the time to complete a full circle) Find the maximum rotation period T of the cylinder which would prevent a 40kg person from falling down. Answer in units of s.

Explanation / Answer

T = (4*pi^2*R*Us/g)^(1/2) T=(4*pi*pi*4.4*.521/9.8)^(1/2)=3.04

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