Question

The Spinny-Spin ride at the fair is a giant rotating cylinder with people on the inside. Once it spins up to speed, the floor drops out from under the people, leaving them pinned against the wall. The inside surface of the cylinder is coated with rubber which gives it a coefficient of static friction with clothing of 0.85. If the cylinder has a diameter of 3.5-m, what is the lowest number of rotations per minute that will keep the riders from sliding down once the floor drops out?

Notes

 The spinny-spin ride must rotate at least 24.5 rpm.(This is the answer)

 The riders are undergoing uniform circular motion. Aka, they are moving in a circle.

Answer 1

given
mue_s = 0.85
d = 3.5 m

r = d/2 = 3.5/2 = 1.75 m

let F is the centripetal force needed and w is the angular speed.

let m is the mass of a person.

F = m*a_centripetal

F = m*r*w^2

As the person are in euilibrium in y-direction net force in y-direction must be zero.
now apply, Fnety = 0

fs_max - m*g = 0

F*mue_s - m*g = 0

m*r*w^2*mue_s - m*g = 0

m*r*w^2*mue_s = m*g

w^2 = g/(r*mue_s)

w = sqrt(g/(r*mue_s) )

= sqrt(9.8/(1.75*0.85) )

= 2.567 rad/s

= 2.567*60/(2*pi) rev/min

= 24.51 rev/min <<<<<<<<<<---------------Answer