Velocity viscous flow around rotating cylinder I am struggling to find an equation of flow velocity at distance $r$ around rotating cylinder with radius $R$, angular velocity $w$ in stationary viscous fluid with some density $ρ$ and viscosity $\mu$.


I found "Hagen–Poiseuille equation"
$$
U = \frac{(P_{2} - P_{1}) * (R^2-r^2)}{4\mu L}
$$
But that equation is for pipe with radius $R$ and flow radius $r$ and require to know pressure difference, but rotating cylinder rotate flow due to viscosity, but not pressure explicitly.
 A: We can obtain a closed-form solution for steady unidirectional flow in a cylindrical coordinate system driven only by the rotation of the cylinder.  All components of velocity vanish except the azimuthal component $u_\theta$ which is a function only of the radial coordinate $r$.  For incompressible, viscous flow the Navier-Stokes equations in cylindrical coordinates reduce to
$$\mu\left[\frac{1}{r} \frac{\partial}{\partial r}\left(r \frac{\partial u_\theta}{\partial r} \right)- \frac{u_\theta}{r^2} \right]= 0,$$
Replacing partial with ordinary derivatives (since $u_\theta$ is a function of $r$ alone), we obtain
$$\frac{1}{r} \frac{d}{d r}\left(r \frac{d u_\theta}{d r} \right)- \frac{u_\theta}{r^2} = \frac{d^2 u_\theta}{dr^2}+\frac{1}{r} \frac{d u_\theta}{dr} - \frac{u_\theta}{r^2}=0$$
Multiplying both sides by $r^2$ we have the usual form of a second-order, homogeneous  Euler-type differential equation
$$r^2\frac{d^2 u_\theta}{dr^2}+r\frac{d u_\theta}{dr} - u_\theta=0$$
This type of ODE can be solved by assuming solutions of the for $r^n$.  Upon substitution of that form we get $n^2 = 1$ and a general solution
$$u_\theta = Ar + Br^{-1}$$
The boundary conditions for an infinite domain where the fluid is quiescent far from the cylinder are $u_\theta(R) = \omega R$ and $u_\theta(r) \to 0$ as $r \to \infty$.  Applying these conditions and solving for the constants $A,B$ we get $A= 0$ and $B= \omega R^2$.
Hence, the velocity is
$$u_\theta = \frac{\omega R^2}{r}$$
