physics help please ?

Question

An amusement park ride consists of a large
vertical cylinder that spins about its axis fast
enough that a person inside is stuck to the
wall and does not slide down when the floor
drops away.
The acceleration of gravity is 9.8 m/s
2
.
Given g = 9.8 m/s
2
, the coefficient µ =
0.614 of static friction between a person and
the wall, and the radius of the cylinder R =
5.2 m. For simplicity, neglect the person’s
depth and assume he or she is just a physical
point on the wall. The person’s speed is
v =
2πR
T
where T is the rotation period of the cylinder
(the time to complete a full circle).
Find the maximum rotation period T of the
cylinder which would prevent a 73 kg person
from falling down.
Answer in units of s.

Whome · Accepted Answer

To keep from sliding down, the minimum available friction force must be equal to the weight of the rider.

mg = μN 
       mg = μ(mω²R)
          g = μ(ω²R)
    g/μR = ω²
√(g/μR) = ω
√(g/μR) = 2π/T
          T = 2π/√(g/μR)

T = 2π/√(9.8/(0.614(5.2))
          T = 3.586350...  s

The question numerals are limited to two significant digits by the values of g and R.

T ≈ 3.6 seconds

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physics help please ?

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