Spin Out! An interesting amusement park activity involves a cylindrical room tha

Question

Spin Out! An interesting amusement park activity involves a cylindrical room that spins about a vertical axis (figure below). Participants in the "ride" are in contact with the wall of the room, and the circular motion of the room results in a normal force from the wall on the riders. When the room spins sufficiently fast, the floor is retracted and the frictional force from the wall keeps the people "stuck" to the wall. Assume the room has a radius of 1.4 m, and the coefficient of static friction between the people and the wall is s = 0.51.

D) Apply Newton's second law along both the vertical and the radial directions. Find the minimum rotation rate for which the riders do not slip down the wall.

Explanation / Answer

here,

radius of the cyclinder , r = 1.4 m

coefficient of static friction , us = 0.51

let the velocity of the cyclinder be v

let the normal reaction of the wall be N

equating the forces horizontally

m*v^2/r = N

equating the forces vertically

for the rider not to slip

us*N - m*g = 0

us*m*v^2/r - m*g = 0

0.51 * v^2/1.4 - 9.8 = 0

v = 5.19 m/s

the rotation rate , w = v/r

w = 3.7 rad/s

the rotation rate of the cyclinder is 3.7 rad/s

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