A cylindrical drum made of steel and of diameter 20 cm is rotating about its own vertical axis. A small body made of steel remains stuck inside the cylinder when the drum rotates at the rate of 200 rpm. When the velocity of rotation decreases, the body falls down. What is the coefficient of friction between the body and the surface of the drum?

Explanation

Let the point where the body remains stuck be P
For the equilibrium of the body,
weight of the body
= limiting frictional force
or, mg = F = `mu`R
[ R = normal force on the body by the wall of the durm]
Again, R supplies the necessary centripetal force to the body for its revolution.
Hence, R = `(mv^(2))/(r ) = m omega^(2) r `
`therefore " " mg = mu m omega^(2) r `
or, `" " mu = (g)/(omega^(2) r) = (980 xx 9)/((20 pi)^(2) xx 10) [ because omega = (200xx2pi)/(60) = (20pi)/(3) ] `
= 0.2234