 Solution

Let $R(t)$ rotate vectors from the rotating frame to the inertial frame. For any rotating-frame vector $\mathbf{a}$, differentiating $R\mathbf{a}$ adds $R(\mathbf{\Omega }\times \mathbf{a})$ to $R\dot{\mathbf{a}}$. Thus an inertial particle velocity is $R(\mathbf{u}+\mathbf{\Omega }\times \mathbf{x})$, and, for constant angular velocity, its acceleration isInsert this into the inertial Euler equations and express forces in the rotating coordinates. This givesThe additional terms are the Coriolis acceleration and the centripetal term associated with frame rotation. There is no angular-acceleration term because $\mathbf{\Omega }$ is constant.

Write the conservative body force as $\mathbf{g}=-\nabla {\Phi }_g$. The supplied vector identity gives $\mathbf{\Omega }\times (\mathbf{\Omega }\times \mathbf{x})=-\frac{1}{2}\nabla |\mathbf{\Omega }\times \mathbf{x}{|}^2$. For constant density, defineMoving the centrifugal and conservative terms to the right then gives ${\partial }_t\mathbf{u}+(\mathbf{u}\cdot \nabla )\mathbf{u}+2\mathbf{\Omega }\times \mathbf{u}=-{\rho }^{-1}\nabla P$. Only the gradient of this reduced pressure matters, so an additive spatial constant is immaterial.