silas asked in Science & MathematicsPhysics · 2 months ago

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

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.

1 Answer

Relevance
  • Whome
    Lv 7
    2 months ago
    Favorite 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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