On the ride "spindletop" at the amusement park six flags over texas, people stood against the inner wall of a hollow vertical cylinder with radius 2.5 m. the cylinder started to rotate, and when it reached a constant rotation rate of 0.60 rev/s, the floor on which people were standing dropped about 0.5 m. the people remained pinned against the wall.

Question

Elizabeth Wheeler

Physics

16 October, 18:15

On the ride "spindletop" at the amusement park six flags over texas, people stood against the inner wall of a hollow vertical cylinder with radius 2.5 m. the cylinder started to rotate, and when it reached a constant rotation rate of 0.60 rev/s, the floor on which people were standing dropped about 0.5 m. the people remained pinned against the wall.

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Bobbie · Answer 1

a) In this case the forces are the centrifugal force Fcp, which is directed horizontally toward the wall; the force of static friction Ff with the wall, directed upward; the normal force Fn by the wall, which is directed away the wall; the force of gravity Fg, directed downwards. Then we have that the horizontal forces are all equal in magnitude; similarly the vertical forces are also all equal in magnitude.

b) The minimum coefficient s occurs when force of gravity is equals the max friction force, that is

Fg = Ff, max

m g = s Fn

Also, the normal force has equal magnitude to the centrifugal force:

m g = s Fcp

m g = s m w^2 r

g = s w^2 r

s = g / (r w^2)

With values: g = 9.81 m/s^2; r = 2.5 m; and w = 2pi * 0.60 = 3.77 rad/s; we find

s = 9.81 / (2.5 * 3.77^2) = 0.276