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I do not know whether this problem is even correct or not (the problem constitutes a 'what if' thought), but nonetheless providing a description

Suppose the viscous flow of a liquid in circular path. The density of the liquid is $\rho$ and coefficient of viscosity is $\eta$. Find the angular velocity as a function of radial distance from from the centre $(\omega(r))$

enter image description here First taking n infinitesimal segment, it could be seen that $\eta A \frac{dv}{dr}=2\pi \eta r^3h\frac{d\omega}{dr}$ will be constant as the velocity of fluid should not be a function of time and for the segment, viscosity on one side (inner or outer) would be directed such that it may increment angular velocity and the other so that it may decrement angular velocity.This gives

\begin{align}2\pi\eta r^3h\frac{d\omega}{dr}=k\implies \omega(r_2)-\omega(r_1)=\frac{k}{2r_1^2}-\frac{k}{2r_2^2}\end{align}

But this does not seems to give a definite value at any point. I further tried using $AdP=m\omega^2r$ but that too does not help. It should be assumed that the distribution is infinite for obvious reasons, I believe.

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"Irrotational" and "viscous" are contradictory here. If the fluid is initially flowing irrotationally, any viscoscity will replidy make the flow to have a rotational component.

You will have to give up "irrotational".

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  • $\begingroup$ edited @mikestone $\endgroup$
    – Shimura Variety
    Commented Jul 9, 2022 at 13:29
  • $\begingroup$ At steady state, the fluid will be rotating with the same velocity distribution as a rigid body. $\endgroup$
    – Chet Miller
    Commented Jul 12, 2022 at 10:02

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