question, no idea wat to do plz help?

Question

In a “Rotor-ride” at a carnival, people are rotated in a cylindrically walled “room”. The room radius is 4.6 m, and the rotation frequency is 0.50 revolutions per second when the floor drops out. What is the minimum coefficient of static fric-tion so that the people will not slip down?

Answer 1

The force of static friction is the product of the force normal to the surface and the coefficient of friction. This force must be greater than or equal to the passenger's weight. First, we find the acceleration:

a = v²/r = ω²r, where ω is the angular velocity in rad/s
ω=.5 rev/s = π rad/s
a=4.6π² m/s² ≈ 45.40018 m/s²

The normal force producing this acceleration is:
F = ma = m*4.6π² m/s²
So the maximum force of static friction is:
μ*m*4.6π² m/s²

(Note: those units are correct - I'm not writing the kg because the units of mass are contained in the term for mass - if I were to substitute a particular mass, I would get something like μ*(60 kg) * 4.6π² m/s², so writing kg a second time would be redundant).

Now, the weight of a passenger is:
F=mg ≈ m*9.81 m/s²

So we have:
μ*m*4.6π² m/s² ≥ m*9.81 m/s²

dividing by m*4.6π² m/s:

μ ≥ 0.216079

However, since our initial values were only given to two significant digits:

μ ≥ 0.22