Question

In an amusement park ride passengers stand inside a hollow cylinder of 2.12 meter radius. The cylinder is then spun and the floor of the cylinder is dropped away as the passengers stick to the wall. If the coefficient of static friction between the passengers and the wall is 0.551, with what angular velocity (rpm) does the cylinder have to be rotating so the passengers don’t slide down the wall?

Answer 1

let the mass of the passanger is m and cylinder is rotating at ω.

now for horizontal direction force balance

Normal force ( N ) = Centrifugal force ( m r ω^2 )

N =  m r ω^2 ...... Equation 1

now for vertical direction force balance

Friction force ( f ) = weight of the passanger ( mg )

f = mu * N = mg

from equation 1

mu * m r ω^2 = mg

0.551 * 2.12 *  ω^2 = 9.8

ω = 2.896 rad/s

so rotational speed in RPM = ω * ( 60 / 2pi )

= 2.896 * ( 60 / 2pi )

= 27.655 RPM