Hi Jessie.

You are watching: In a rotor ride at a carnival

There is no pressure pressing them against the wall. The forces affiliated are:

1. Heaviness (acting downwards that course), 2. Friction (counteracting gravity) 3. Centripetal force of rotation (acting native the center of the cylinder right into the wall surface of the cylinder) 4. Normal pressure (counteracting the centripetal force)

The force of heaviness is merely (*mg*). In order because that the friction to be adequate to store a person from falling, the frictional force must it is in (-*mg*).

Frictional pressure for things at remainder (static friction) is *F*=μ*N* whereby μ is the co-efficient of revolution friction and *N* is the normal force from the wall surface towards the person. Therefore, μ=*F/N*. However *F=mg*, so μ=*mg/N*.

The size of the normal force is merely the exact same as the centripetal force because you don\"t loss through or lift off the side of the cylinder. Centripetal force is calculated as *F=ma* wherein *a* is the centripetal acceleration. therefore μ = *mg / N = mg / ma = g / a*.

However, centripetal acceleration is *a*=*v*2/*r* where *v* is the tangential velocity and *r* is the radius that the cylinder. The tangential velocity is *v*=ω*r*, wherein ω is the angular velocity in radians every unit time. hence μ= *g* / *a* = *g* / (*r*ω2).

Now you have actually an equation for the coefficient of static friction (μ) in regards to the radius (*r*) and also the angular velocity (ω), which you know.

You\"ll require to transform your angular velocity from transformations per 2nd to radians per 2nd before substituting for *g*, *r* and ω.

Hope this helps, Stephen La Rocque.

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