Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.

silas asked in Science & MathematicsPhysics · 5 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
    5 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

    Please remember to select a Favorite Answer from among your results.

Still have questions? Get your answers by asking now.