A thin horizontal disc of radius R=10cm is located with in a cylindrical cavity filled with oil whose viscosity `eta=0.08`P (figure) The distance between the disc and the horizontal planes of the cavity is equal to `h=1.0` mm find the power developed by the viscous forces acting ont he disc when it rotates with the angular velocity `omega=60rad//s`. The end effect are to be neglected.

Explanation

When the disc rotates the fluid in contact, with, carotates but the fluid in contact with the walls of the cavity does not rotate. A velocity gradient is then set up leading to viscous forces. At a distance r from the axis the linear velocity is `omegar` so there is a velocity gradient `(omegar)/(h)` both in the upper and lower clearance. The corresponding force on the element whose radiul width is `dr` is
`eta2pidr(omegar)/(h)` (from the formula `F=etaA(dv)/(dx)`)
The torque due to this force is
`eta2pirdr(omegar)/(h)r`
and the net torque considering both the upper and lower clearance is
`2underset(0)overset(R)inteta2pir^3dr(omega)/(h)`
`=piR^4omegaeta//h`
So power developed is
`P=piR^4omega^2eta//h=9*05W` (on putting the values).
(As instructed end effects i.e. rotation of fluid in the clearance `rgtR` has been neglected.)