Answer to Question #26244 in Mechanics | Relativity for thakshi

Question

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.

Answer 1

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:

weight=Friction force

mg=mu*ma
mu=g/a = 9.8/0.3549456 = 27.6
friction coefficient& between the body and the cylinder is approximately 27.6