How can you show that the centrifugal force is conservative? (From the definition of being the grad of some potential)

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My lecturer has derived, from considering a rotating inertial frame, that the centrifugal force is given as:

$\mathbf{F_{cen}} = -m\boldsymbol{\omega} \times (\boldsymbol{\omega} \times \mathbf{x})$

By the definition that $\mathbf{F_{cen}}$ is conservative iff it is the grad of some potential, i.e. we have $\mathbf{F_{cen}} = - \nabla\mathbf{V_{cen}}$, for some $\mathbf{V_{cen}}$.

He then gave that the required $\mathbf{V_{cen}}$ is:

$\mathbf{V_{cen}} = -\frac{m}{2} |\boldsymbol{\omega} \times \mathbf{x}|^{2}$

I don't know how to take the grad of such a potential and get the required result.

My best attempt was as follows:

$\mathbf{F_{cen}} = -\nabla (- \frac{m}{2} |\boldsymbol{\omega} \times \mathbf{x}|^{2})\\ = \frac{m}{2} 2 |\boldsymbol{\omega} \times \mathbf{x}| \frac{(\boldsymbol\omega \times \mathbf{x})}{|\boldsymbol{\omega} \times \mathbf{x}|} \times \nabla(\boldsymbol\omega \times \mathbf{x})\\ = m (\boldsymbol\omega \times \mathbf{x}) \times (\nabla\boldsymbol\omega \times \mathbf{x} + \boldsymbol\omega \times \nabla\mathbf{x}))\\ =m (\boldsymbol\omega \times \mathbf{x}) \times (\mathbf{0} \times \mathbf{x} + \boldsymbol\omega)\\ =m (\boldsymbol\omega \times \mathbf{x}) \times \boldsymbol\omega$

But this doesn't yield the result and I fear I have misunderstood/misapplied some of the properties of $\nabla$ in my working.

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Such calculations are most easily carried out by using index notation and summation convention. As $$(\vec{\omega} \times \vec{x})_k= \varepsilon_{k \ell m} \,\omega_\ell\, x_m, \tag{1} \label{1}$$ you have $$\nabla_i(\vec{\omega} \times \vec{x})_k= \varepsilon_{k \ell m} \, \omega_\ell\, \delta_{im} = \varepsilon_{k \ell i}\, \omega_\ell =-\varepsilon_{i\ell k} \omega_\ell,$$ where the total antisymmetry of the Levi-Civita symbol was used in the last step. Thus we obtain $$\nabla_i(\vec{\omega}\times \vec{x})^2=\nabla_i\left[(\vec{\omega} \times \vec{x})_k(\vec{\omega}\times \vec{x})_k \right]=-2 \, \varepsilon_{i\ell k} \, \omega_\ell (\vec{\omega}\times \vec{x})_k, $$ equivalent to $$\vec{\nabla} (\vec{\omega}\times \vec{x})^2=-2 \,\vec{\omega}\times(\vec{\omega} \times \vec{x}) $$ in vector notation.