Physics...HELP!!!?

Question

In an amusement park ride, a cylinder of radius 2.50 m is set in rotation at an angular speed of 4.10 rad/s. 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?

I have found linear speed of 10.25 m/s but now I am stuck. Any help??

Answer 1

I presume it is the centripetal force that keeps the poor rider from sliding down and out of this contraption from hell?

I guess Lem has answered your question.

Answer 2

Determine the "centrifugal force" which is the mass of the rider times the centripetal acceleration. (a = v^2/r), (f = ma)

The coefficient of friction has to be high enough so that the centrifugal force times the coefficient is greater than the force of gravity. (kma = mg)

so k= mg/ma = g/a = g/(v^2/r) = gr/v^2

Apply calculator.

Answer 3

Momentum is a vector quantity (course concerns). even as drawing near the aspect of the table, its area of momentum perpendicular to the aspect is +0.2x15xsin(60) (in course of the aspect is defined as +) After hitting the aspect of the table, its area of momentum perpendicular to the aspect is -0.2x15xsin(60) faraway from the aspect (this course is -) So replace (very last - preliminary) is -0.2x15xsin(60) - (+0.2x15xsin(60) ) = -2x0.2x15xsin(60) = -3xsqrt(3). Momentum parallel to the aspect is unchanged. earlier, momentum is 0.2x15xcos(60) in a given course After, momentum is 0.2x15xcos(60) interior a similar course So replace is 0.2x15xcos(60) - 0.2x15xcos(60) = 0. subsequently entire replace in momentum is -3xsqrt(3) the position - signal exhibits "faraway from the aspect".

Answer 4

V^2 / r = centripetal acceleration.

the ratio of the centripetal acceleration to gravtitational acceleration is your required frictional coefficient.