# Circular Motion and Static Friction Problem The Wall of Death ride

1. Mar 2, 2013

### jklops686

Circular Motion and Static Friction Problem.. "The Wall of Death" ride

1. The problem statement, all variables and given/known data

A fairground ride called "The Wall of Death" consists of a cylindrical
container of internal diameter 6.50m, mounted on a cylindrical axis.
The passengers feel as if they are being pushed against the wall as the container
begins to rotate. Eventually, the floor is lowered, leaving the miserable passengers
pinned to the wall, apparently defying gravity. When the ride slows down, the
passengers just begin to slide down the wall when they are rotating at 0.400
revolutions per second. Calculate the coefficient of static friction between the wall
and the passengers' backs.

2. Relevant equations

max static friction=coefficient*normal force?

v= 2pi*r/T

3. The attempt at a solution

I calculated how long it takes to go around in one second (period T) .4x=1 So, 1 rev=2.5 seconds. Now that I have T I also calculated the tangential velocity using the equation above and got 8.17m/s. I think the friction force would be set up as v-f=0 (used free body diagram...tangential velocity and friction cancel out to equal zero?) Which would mean force of friction is 8.17 too. I'm not sure if this is correct, but anyways, i don't know how to find the coefficient of static friction. I would think you would need to know the mass of the person.

2. Mar 2, 2013

### rude man

Balance the centripetal force with the force of static friction. Don't need mass.

3. Mar 2, 2013

### jklops686

how do I do that? Doesn't the equation for static friction need mass to find normal force?

4. Mar 2, 2013

### haruspex

Yes, but it doesn't matter what the mass is for this question. Just write it as 'm'. It will cancel out.

5. Mar 3, 2013

### rude man

Well, yes, but so does the expression for centripetal force. If you're really lucky they might cancel each other out ...