Math CentralQuandaries & Queries


Question from Jessie, a student:
In a "Rotor-ride" at a carnival, people are rotated in a cylindrically walled "room". The room radius is 4.6m and the rotation frequency is 0.550 revolutions per second when the floor drops out. What is the minimum coefficient of static friction so that people will not slip down? People on this ride say they were "pressed against the wall". Is there really an outward force pressing them against the wall? If so, what is its source? If not, what is the proper description of their situation (besides scary)?

Hi Jessie.

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 = μN where μ is the co-efficient of static friction and N is the normal force from the wall towards the person. Therefore, μ = F/N. But F = mg, so μ = 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 μ = 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 = ωr, where ω is the angular velocity in radians per unit time.
Thus μ = g / a = g / (rω2).

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

Hope this helps,
Stephen La Rocque.

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