# Amusement park question

1. Jun 22, 2009

### PeterPumpkin

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

In an amusement park, a person leans against the inner wall of a spinning cylinder. Suppose the minimum angular speed needed so that the person doesn’t slide is w min. At this minimum speed call the friction force, F1. Suppose the cylinder rotates twice as fast. What is the new frictional force?

2. Relevant equations

F1 = mg
Normal force = N = m * v * v /R
friction <= coefficient static friction * N

3. The attempt at a solution

If w is doubled, then v is doubled and therefore N quadruples.
Therefore there is a net upward force (as mg is constant but friction has increased).
Therefore the person moves upwards.
I don't believe this will happen.
Where did I go wrong?

Obviously friction must remain <= mg but that is inconsistent with the equation:
friction <= coefficient static friction * N
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jun 22, 2009

### rl.bhat

Frictional force comes into picture only during the relative motion between the surfaces.
Therefore friction quadruples means even heavier person may stay without slipping.

3. Jun 22, 2009

### PeterPumpkin

"Therefore friction quadruples means even heavier person may stay without slipping." Agreed.

"Frictional force comes into picture only during the relative motion between the surfaces." Agreed for dynamic friction.

Sorry. I still don't see why friction would remain the same if Friction = μ static * N

4. Jun 22, 2009

### rl.bhat

The friction is a self adjusting force, which prevents the relative motion between two surfaces due to irregularities on the surfaces. By adding extra load, the irregularities on the surface will not change. If you add force gradually, the frictional force will also increase gradually up to a certain limit. After this limiting value the body starts moving