Ask Question
27 March, 15:50

On the ride "spindletop" at the amusement park six flags over texas, people stood against the inner wall of a hollow vertical cylinder with radius 2.5 m. the cylinder started to rotate, and when it reached a constant rotation rate of 0.60 rev/s, the floor on which people were standing dropped about 0.5 m. the people remained pinned against the wall.

+3
Answers (1)
  1. 27 March, 18:59
    0
    a) In this case the forces are the centrifugal force Fcp, which is directed horizontally toward the wall; the force of static friction Ff with the wall, directed upward; the normal force Fn by the wall, which is directed away the wall; the force of gravity Fg, directed downwards. Then we have that the horizontal forces are all equal in magnitude; similarly the vertical forces are also all equal in magnitude.

    b) The minimum coefficient s occurs when force of gravity is equals the max friction force, that is

    Fg = Ff, max

    m g = s Fn

    Also, the normal force has equal magnitude to the centrifugal force:

    m g = s Fcp

    m g = s m w^2 r

    g = s w^2 r

    s = g / (r w^2)

    With values: g = 9.81 m/s^2; r = 2.5 m; and w = 2pi * 0.60 = 3.77 rad/s; we find

    s = 9.81 / (2.5 * 3.77^2) = 0.276
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “On the ride "spindletop" at the amusement park six flags over texas, people stood against the inner wall of a hollow vertical cylinder with ...” in 📘 Physics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers