Posted by shelby on .
The Wall of Death in an amusement park is comprised of a vertical cylinder that can spin around the vertical axis. Riders stand against the wall of the spinning cylinder and the floor falls away leaving the riders held up by friction. The radius of the cylinder is 4.2 m and the coefficent of static friction between the rider and the wall is 0.47. Find the minimum number of revolutions per minute necessary so that the riders do not slip down the wall (note: to specify revolutions per minute as the units in your answer, enter rev/min).
Centripetal acceleration = w^2 R
where w is angular velocity in radians per second
then the force holding the rider against the wall is m a = m w^2 R
then the maximum friction force preventing a slide to oblivion = 0.47 m w^2 R
when m g is bigger than that we slide
9.8 = 0.47 w^2 (4.2)
solve for w
then convert w
w radians/s * (1 rev/2 pi radians) *(60 s/min) = answer in revs/min