An amusement park ride consist of a large verticle cylinder that spins about its axis first fast enough that any person inside is held up againest the wall when the is `mu_(s)` and the radius of the cylider is R

a. Shown that maximum period of revolation neccessary to keep the person from falling is `T = (4 pi^(2) mu_(s) R //g)^(1//2)`
b. Obtain a numerical value for `T` taking `R = 4.00m` and `mu_(s) = 0.400` . How many revolations per minute does the cylinder make?
c. If the rate of revolation of the cylinder is make to be some what larger what happens to the magnitude of each one of the forces acting on the person ? What happens to the motion of the person ?
d. If instead the cylinder rate revolution is made to be somewhat smaller what happens to the magnitude and what happens to the motion of the person ?

Explanation

a. `N = (mv^(2))/(R ), f - mg = 0 , f= mu_(s)N, v = (2 xx pi R)/(T)`
Solving we get `T = sqrt((4 pi ^(2) R mu_(s))/(g))`

b. `T = 0.8 pi s`
`(rev)/(min) = (60)/(0.8 pi)= (75)/(pi) rev min^(-1)`
c. The gravitational and the friction force remain constant . The normal force increases .The person remains in motion with the wall .
d. The gravitational force remain constant . The normal and the frictional forces decrease . THe person slides relative to the wall and downward into the pit