Help Solve Amusement Park Wall of Death Problem

[bulldog23]

The wall of death in an amusement park is comprised of a vertical cylinder that can spin around the vertical axis. Riders stand against the wall of the spinning cylinder and the floor falls away leaving the riders held up by friction. The radius of the cylinder is 3.9 m and the coefficient of static friction between the wall and the rider is .33. Find the tangential velocity so that the riders do not slip from the wall.

I know that tangential velocity equals V^2/r, but I am unsure how to get this only given the static friction and the radius. Can anyone help me?

 

[twotaileddemon]

Well~ this is my outlook on things
1) tangental velocity is not equal to v^2/r, that's the centripetal acceleration
2) Draw a free body diagram
3) When you do, notice that mg is down, normal force is directed to the middle of the circle, and the frictional force must be directed up to prevent everyone from falling.
4) In order to prevent the riders from falling, the frictional force must support the weight of everyone
5) The change of F in the y direction is F_y = (u_s)(F_n) - mg = 0
the change of F in the x direction is F_x = F_n = mv^2/r
We know that force of friction is (u_s)(F_n) and that it equals mg from the above equation
6) (u_s)(mv^2/r) = mg
(u_s)(v^2/r) = g
v^2 = gr/u_s
We see that mass doesn't matter, which it shouldn't - think about it, everyone has different masses anyway
So just find the square root gr/u_s where u_s is .33 and r = 3.9 m and g = 9.8 m/s^2
:) Hope I helped, feel free to reply if you are still confused

 

[bulldog23]

alright, thanks. The part I couldn't figure out was the v^2=gr/u_s.