Question

Needs to document an amusement park ride. For example: Round rotating cylinder where the floor drops out. The information can come from any online website. Needs to know the radius and speed at which the ride needs to rotate so that you don't fall?

Calculate the coefficient of static friction for this ride also

Answer 1

There is no force pressing them against the wall. The forces involved are:

1. gravity (acting downwards of course),
2. friction (counteracting gravity)
3. centripetal force of rotation (acting from the center of the cylinder into the wall of the cylinder)
4. normal force (counteracting the centripetal force)

The force of gravity is simply (mg). In order for the friction to be sufficient to keep a person from falling, the frictional force must be (-mg).

Frictional force for an object at rest (static friction) is F = ?N where ? is the co-efficient of static friction andN is the normal force from the wall towards the person. Therefore, ? = F/N. But F = mg, so ? = mg/N.

The magnitude of the normal force is simply the same as the centripetal force because you don't fall through or lift off the side of the cylinder. Centripetal force is calculated as F = ma where a is the centripetal acceleration.
So ? = mg / N = mg / ma = g / a.

However, centripetal acceleration is a = v²/r where v is the tangential velocity and r is the radius of the cylinder. The tangential velocity is v = ?r, where ? is the angular velocity in radians per unit time.
Thus ? = g / a = g / (r?²).

Now you have an equation for the coefficient of static friction (?) in terms of the radius (r) and the angular velocity (?), which you know.

You'll need to convert your angular velocity from revolutions per second to radians per second before substituting for g, r and ?.