Past exam of the mathematics course of the University of Cambridge / 2008 / ib / Paper 3 / 18B (Fluid Dynamics)

The rigid-rotation velocity field is $\mathbf{v}=(-\Omega y,\Omega x,0)$, with spatially uniform $\Omega (t)$. Its divergence is zero, since ${\partial }_x(-\Omega y)={\partial }_y(\Omega x)=0$. It therefore satisfies the incompressible continuity equation. On the cylindrical wall, $\mathbf{v}\cdot (x,y,0)=0$, so it has zero normal velocity; on either end cap its vertical component is zero. These are the impermeability boundary conditions for an inviscid flow. Tangential slip is permitted; a no-slip condition is not part of this inviscid model.

At height $z$, $\mathbf{r}\times \mathbf{v}=(-\Omega zx,-\Omega zy,\Omega (x^2+y^2))$. Integrating over a circular cross-section cancels the horizontal components by symmetry. Thus the angular momentum per unit length isThe coefficient is the polar moment of inertia per unit length of the liquid column.

The Euler equations for an inviscid fluid are ${\partial }_t\mathbf{v}+(\mathbf{v}\cdot \nabla )\mathbf{v}=-\nabla p/\rho +\mathbf{f}$. Here the vorticity is $2\Omega {\mathbf{e}}_z$, and direct differentiation gives $(\mathbf{v}\cdot \nabla )\mathbf{v}=-{\Omega }^2(x,y,0)$, whose curl vanishes. Taking the curl therefore givessoFor constant force coefficients and the initial rest condition, $\Omega (t)=(\gamma -\beta )t/2$. Only the antisymmetric part of the horizontal force gradient generates rotation; its symmetric part is balanced by pressure.

Substitute the rotating velocity field into the Euler equations for an inviscid fluid. Its acceleration is $(-\dot{\Omega }y-{\Omega }^{2}x,\dot{\Omega }x-{\Omega }^{2}y,0)$, soThe preceding angular acceleration makes both mixed coefficients $(\beta +\gamma )/2$, ensuring integrability. ThusPressure is determined up to the spatially uniform function $p_0(t)$. The specified force has no vertical component; if gravity were additionally included in the body force, the usual hydrostatic term $-\rho gz$ would also appear. It is not implied by the displayed force here.