An amusement park ride called the “Rotor” is a very large cylinder, oriented ver

Question

An amusement park ride called the “Rotor” is a very large cylinder, oriented vertically, and is made to spin up to a speed sufficient to allow people (their weight) to be balanced by friction after the floor is dropped out. Two people are shown in the Figure below, one weighing 100 lbs, the other 200 lbs. The coefficient of friction is 0.48 and the drum diameter is 12 m. What is the minimum speed (tangential) in which both people will not slip? Determine also the angular speed in revolutions per second. Plan: Identify forces on either person and use Newton’s 2nd Law to generate a relationship that includes speed

Explanation / Answer

Normal force form the wall must supply the centripetal force needed to keep them moving in a circle.

N = m . v^2 / R

The friction force from the wall must be equal to their weight force ( otherwise they would slide down ( or slide up ))

F = µ . N = m . g

N = m . g / µ = m . v^2 / R

v^2 = R . g / µ

v^2 = 6*9.81 / 0.48

v = 11.07 m/s

w = v/R = 1.84 rad/s

(or)

since the one person is 200 lbs as compared to the 100, that person is the one that will really make a difference, so you need just consider that person. Basically, if you can hold up the heavier person, the lighter person will be easy. The next thing would be to convert to kg.
180lb = 90.72 kg
if we do not want the person to fall the frictional force has to be equal to the weight

F f = 90.72*9.81 = 889.96 N

and we also know that the frictional force is the normal force, or force perpendicular to the surface, times the coefficient of friction.

Ff = Fn*0.48
so Fn = 1854.1

Also, in this case, the normal force is equal to the centripetal force/rotational force.
Fc = Fn
Fc = mv^2/r
1854.1 = 90.72 v^2 / 12
v= 15.66 m/s

w = v/R = 2.61 rad/s

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