[Help] Amusement Park Physics Problem

[helium-4]

Homework Statement

You are designing a new amusement park ride called Barrel o’ Fun. The idea is that people will stand inside a barrel with their backs against the wall. It takes 1.5 seconds for the barrel to complete one revolution. The diameter of the barrel is 15 meters. What coefficient of friction will be necessary for the people to stick to the wall so that when the barrel is spinning the floor can drop away?

Homework Equations

Force of Static Friction = Coefficient of Static Friction × Normal Force = mv^2/r
Fs = μs×N = mv^2/r
μs = mv^2/rN
μs = v^2/rg
μs = 4∏^2r^2/rgt^2

The Attempt at a Solution

μs = 4∏^2r^2/rgt^2
μs = 13.4143


This does not seem right. I would appreciate your help.

 

[frogjg2003]

Draw a force body diagram.

 

[NasuSama]

As one user just said, you need to draw the free body diagram. You need to include the normal force, the weight vector and the corresponding forces that take place during the ride.

You might want to show us the free body diagram, so we can check to see if you are at the right track!

 

[helium-4]

I already solved the problem. I am sorry for the delay in response - I did not have access to internet for much of yesterday.

I misplaced the Force of Static Friction on the FBD. In this particular problem, the Fs points up, and the Fg points down (y direction).

Fs - Fg = ma_y = 0. So μsN - mg = 0.
μs = mg/N.

The N force points towards the center of the barrel (x direction).

N = ma_x = mv^2/R = 4∏^2mR/T^2

So, μs = (mg)/(4∏^2mR/T^2)
μs = ((g)(T^2)) / ((4)(∏^2)(R)) = 0.74547