Posted by **Bob** on Wednesday, November 19, 2008 at 11:59am.

In a popular amusement-park ride, a cylinder of radius 3.00 m is set in rotation at an angular speed of 5.00 rad/s (counter-clockwise). The floor then drops away, leaving the riders suspended against the wall in a vertical position. What minimum coefficient of friction between a rider's clothing and the wall of the cylinder is needed to keep the rider from slipping? (Hint: Recall that Fs = [Mu symbol]s*Fn. where the normal force is the force that maintains circular motion.)

- Physics -
**Bob**, Wednesday, November 19, 2008 at 12:18pm
( And the answer's supposed to be 0.131 )

- Physics -
**bobpursley**, Wednesday, November 19, 2008 at 12:45pm
Goodness. The hint takes all the fun out of it. Thinking is fun.

Fs=mu*Fn

but Fs has to equal weight to keep it from sliding

Weight=mu(mw^2*r)

mg=mu*m*w^2 r

m divides out

solve for mu. YOu are given w, r, and you know g.

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