Asked by Bob

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.)

Answers

Answered by Bob
( And the answer's supposed to be 0.131 )
Answered by bobpursley
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.
There are no AI answers yet. The ability to request AI answers is coming soon!

Related Questions