physics help please ?
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
The acceleration of gravity is 9.8 m/s
Given g = 9.8 m/s
, 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
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.
- WhomeLv 72 months agoFavorite 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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