Question

In a classical carnival ride, patrons stand against the wall in a cylindrically shaped room. Once the room gets spinning fast enough, the floor drop from the bottom of the room! Friction between the walls of the room and the people on the ride make them the "stick" to the wall so they do not slide down. In one ride, the radius of the cylindrical room is R=6.4m and the room spin with a frequency of 22.1 revolutions perminute. What is the speed of a person "stuck" to the wall and what is the normal force of the wall on a rider of m=54kg,what is the minimum coefficient of friction needed between the wall and the person.?

Answer 1

Answer:

- the speed of a person "stuck" to the wall is 14.8 m/s

- the normal force of the wall on a rider of m=54kg is 1851 N

- the minimum coefficient of friction needed between the wall and the person is 0.29

Explanation:

Given information:

the radius of the cylindrical room, R = 6.4 m

the room spin with frequency, ω =  22.1 rev/minutes = 22.1 $\frac{2\pi }{60}$ = 2.31 rad/s

mass of rider, m = 54 kg

the speed of a person "stuck" to the wall

v = ω R

  = 2.31 x 6.4

  = 14.8 m/s

the normal force of the wall on a rider

F = m a

a  = ω^2 R

   =  $\frac{v^{2} }{R^{2} }$ R

   = $\frac{v^{2} }{R}$

F = $\frac{mv^{2} }{R}$

  = $\frac{(54)(14.8)^{2} }{6.4}$

  = 1851 N

the minimum coefficient of friction needed between the wall and the person

F(friction) = μ N

W =  μ N

m g =  μ $\frac{mv^{2} }{R}$

g = μ $\frac{v^{2} }{R}$

μ = $\frac{gR}{v^{2} }$

  = $\frac{(9.8) (6.4)}{14.8^{2} }$

  = 0.29