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.)
2 answers
( And the answer's supposed to be 0.131 )
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.
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.
1 answer