Centripetal force question?

Question

In an automatic clothes drier, a hollow cylinder moves the clothes on a vertical circle (radius r = 0.29 m), as the drawing shows. The appliance is designed so that the clothes tumble gently as they dry. This means that when a piece of clothing reaches an angle of above the horizontal, it loses contact with the wall of the cylinder and falls onto the clothes below. How many revolutions per second should the cylinder make in order that the clothes lose contact with the wall when = 67.0°?

ok I used the formula tan(67)=v^2/(0.29(9.8))

so i get V= 2.59 and then I divide it by 2piR, but i get the wrong answer. what am i doing wrong?

Answer 1

Dont use numbers, they can be confusing. Use algebraic expresions So..

Try

m(v^2)/R=mgcos(90-67)

v= sqrt(g R cos(23))
rpm= [v / (2 pi R)]60
rpm)= 60 sqrt(g R cos(23))/(2 pi R)
V(in rppm) = (30/pi) sqrt(g cos(23)/R)