| m v2 |
 |
| r |
There is an amusement park ride called The Gravitron or The Rotor, which is essentially a big cylinder. You get on, lean against the wall, and it starts to spin. When it's spinning sufficiently quickly, the floor drops away. You don't fall down because you are pinned to the wall by some mysterious force. How does this work?
As usual, construct a free-body diagram. The interface is vertical, so the normal force is horizontal, perpendicular to the interface. The force keeping you from sliding down the wall is the static force of friction. It must point up, balancing the gravitational force that points down.
The simulation shows what happens as the angular velocity of the system gradually slows. The red line on the graph represents the maximum value of the force of friction. The black line is the value of mg. Note that when the object has fallen to the floor, the upward force is a normal force, not a frictional force.
An appropriate coordinate system has +y up and +x toward the center. Summing forces in the y-direction, where there is no acceleration, gives:
fs = mg
The only force there is in the x-direction is the normal. Applying Newton's second law tells us:
| ΣFx | = | max | = |
|
| N | = |
|
= | mω2r |
Friction must balance the force of gravity. The maximum possible value of the force of static friction, however, is:
fs max = μs N
In this situation we have:
fs max = μs mω2r
If the rotation rate is fast, there's nothing to worry about. If the speed decreases, however, the maximum force of static friction decreases. If it drops below mg, a person on the wall would start to slide down.
This occurs at an angular velocity given by:
| ω2 | = |
|
One final note: the simulation shows an upward frictional force acting on the mass after it has slid down the wall and come to rest at the bottom. There is an upward force, but it's a normal force.