Physics?!?! Please help!?

Question

An amusement park ride consists of a large vertical cylinder that spins about its axis sufficiently fast that any person inside is held up against the wall when the floor drops away. The coefficient of static friction between person and wall is Mu(s) and the radius of the cylinder is R.

a) show that the maximum period of revolution necessary to keep the person form falling is T = (4pi^2RMu(s) / g)^1/2
b) Obtain a numerical value for T assuming that R = 4.00m and Mu(s) = 0.400. How many revolutions per minute does the cylinder make?

Buh? Help!

Answer 1

Ok, here's how you do it

First of all don't forget to make a drawing of the situation.

When you see the drawing, you will realize that the weight of the person has to equal the friction the person makes with the cylinder, so

Weight = Friction

Because of Newton's law, you get that

Weight = mg

where m is the mass of the person and g is the gravity acceleration.

you know that

friction = Mu*N

N = m*ac, where ac is the centripetal acceleration

ac = W^2R

where W is the angular velocity

W = 2*pi/T

Now that you have all the necesary equations you do as follows

f=mg

Mu*N=mg

Mu*m*ac=mg

Mu*ac=g

Mu*W^2*R=g

Mu*4*pi^2*R/T^2=g

and solving for T

T = (Mu*4*pi^2*R/g)^1/2

There you have the answer for a

now for b you just substitute the values you have in the equation you got previously

T = (0.4*4*3.1416^2*4m/9.81m/s^2)^1/2

T = 2.537s

since the frecuency = 1/T

F = 0.394 rev/s*60s/1min = 23.6 Rev/min

Hope this helps.