A hollow cylinder of radius 10 cm rotates about its axis which is vertical. A small body remains in contact with the inner wall if the frequency of rotation is 200 per minute but falls at lower frequencies. Find the co-efficient of friction between the body and the cylinder.
Let us find centrifugal acceleration for that small body
a= v^2/r = nu^2*r
where r is radius and nu is angular velocity
nu=2*pi/T = 6.28/(200/60) = 1.884 s^(-1) the acceleration is a = 1.884^2*0.1 = 0.3549456& m/s^2
coefficient of friction can be find from equilibrium condition:
mg=mu*ma mu=g/a = 9.8/0.3549456 = 27.6 friction coefficient& between the body and the cylinder is approximately 27.6