A fluid with viscosity `eta` fills the space between two long co-axial cylinders of radii `R_1 `and `R_2`, with `R_1 lt R_2`. The inner cylinder is stationary while the outer one is rotated with a constant angular velocity `omega_2`. The fluid flow is laminar. Taking into account that the friction force acting on a unit area of a cylindrical surface of radius r is defined by the formula `sigma=etar(delomega//delr)`, find:
(a) the angular velocity of the rotating fluid is as a function of radius r,
(b) the moment of the friction forces acting on a unit length of the outer cylinder.

Explanation

(a) Let us consider an elemental cylinder of radius r and thickness `dr` then from Newton's formula
`F=2pi r l eta r(domega)/(dr)=2pi l eta r^2(domega)/(dr)`
and moment of this force acting on the element,
`N=2pir^2l eta(domega)/(dr)r=2pir^3l eta(domega)/(dr)`
or, `2pi l eta d omega=N(dr)/(r^3)` (2)
As in the previous problem N is constant when conditions are steady
Integrating, `2pi l eta underset(0)overset(omega)intdomega=Nunderset(R_1)overset(r)int(dr)/(r^3)`
or, `2pi l eta omega=N/2[(1)/(R_1^2)-1/r^2]` (3)
Putting `r=R_2, omega=omega_2`, we get
`2pi l eta omega_2=N/2[(1)/(R_1^2)-(1)/(R_2^2)]` (4)
From (3) and (4),
`omega=omega_2(R_1^2R_2^2)/(R_2^2-R_1^2)[(1)/(R_1^2)-(1)/(r^2)]`
(b) From Eq. (4),
`N_1=N/l=4pi eta omega_2(R_1^2R_2^2)/(R_2^2-R_1^2)`