Does it make sense to take the grad of this? I have been asked to calculate the grad of a function $V(r)$, where $r$ is a position vector and $\omega$ is the angular velocity. But surely, $V(r)$ is a scalar? So how is this possible? Context: I want to show that a force is conservative with this potential, the force is $F=-m\omega\land(\omega\land r)$
$$V(r)=-\frac12 m |\omega \land r|^2 $$

The gradient operator transforms a scalar field $\Phi(\vec t)$ into the vector field $\nabla \Phi(\vec r)$.  It is written in Cartesian Coordinates as
$$\nabla \Phi(\vec r)=\hat x\frac{\partial \Phi(\vec r)}{\partial x}+\hat y\frac{\partial \Phi(\vec r)}{\partial y}+\hat z\frac{\partial \Phi(\vec r)}{\partial z}$$
Let $V(\vec r)=-\frac12m|\vec \omega \times\vec r|^2$.  Then, we can write $V(\vec r)$ as 
$$\begin{align}
V(\vec r)&=-\frac m2 (\vec \omega \times \vec r)\cdot (\vec \omega \times \vec r)\\\\
&=-\frac m2 \vec \omega \cdot (\vec r\times (\vec \omega \times \vec r))\\\\
&=-\frac m2\omega \cdot(r^2\vec \omega-(\vec r\cdot \vec\omega)\vec r)\\\\
&=\frac m2((\vec \omega\cdot \vec r)^2-\omega^2r^2)
\end{align}$$  
Then, the gradient of $V$ is given by 
$$\begin{align}
\nabla V(\vec r)&=\frac m2 \left(2(\vec \omega \cdot \vec r)\nabla (\vec \omega \cdot \vec r)-2r\omega^2\nabla (r)\right)\\\\
&m\left((\vec \omega \cdot \vec r)\vec \omega-\omega^2\vec r\right)\\\\
&=m(\vec\omega\times(\vec \omega\times\vec r))
\end{align}$$
