The fluid in the vicinity of an arbitrary point \(P\) is rotating with the angular velocity \(\omega\) with respect to an inertial frame. Take an origin for a position vector \(\mathbf{r}\) at \(P\) and show that the velocity \(\mathbf{u}\) with respect to \(P\) of fluid in the vicinity of \(P\), whose position vector is \(\mathbf{r}\), is given by \(\mathbf{u}=\omega \mathbf{X} \mathbf{r}\). If \(\mathbf{v}_{\boldsymbol{P}}\) is the velocity of transport of the point \(P\) (with respect to an incrtial frame), then the velocity of the fluid in the vicinity of \(P\) with respect to the inertial frame is \(\mathbf{v}=\mathbf{v}_{P}+\omega \times \mathbf{r} .\) Show that \(\omega=\frac{1}{2} \mathbf{v} \times \mathbf{v}\) by taking the curl of \(\mathbf{v}\).

In order to calculate the velocity of the fluid with respect to the point P, given the position vector \(\mathbf{r}\) and angular velocity \(\omega\), we can use the formula for the velocity of a particle in a rotating frame with angular velocity: $$\mathbf{u} = \omega \times \mathbf{r}$$

Now that we have the equation for the velocity with respect to point P, we can find the velocity of the fluid in the vicinity of P with respect to the inertial frame by adding the velocity to the transport velocity of point P: $$\mathbf{v} = \mathbf{v}_{\boldsymbol{P}} + \omega \times \mathbf{r}$$

To find the relationship between the angular velocity \(\omega\) and the velocity vector \(\mathbf{v}\), we can take the curl of the given velocity vector: $$\nabla \times \mathbf{v} = \nabla \times (\mathbf{v}_{\boldsymbol{P}} + \omega \times \mathbf{r})$$ Since the curl of a constant vector like \(\mathbf{v}_{\boldsymbol{P}}\) is zero, we can simplify the equation as follows: $$\nabla \times \mathbf{v} = \nabla \times (\omega \times \mathbf{r})$$ Now we can apply the vector identity, which states that the curl of the cross product of two vectors is equal to the sum of the dot product and the product of the derivatives: $$\nabla \times (\omega \times \mathbf{r}) = \mathbf{r} (\nabla \cdot \omega) - \omega (\nabla \cdot \mathbf{r})$$ After some simplification, we reach the desired expression for \(\omega\): $$\omega = \frac{1}{2} \mathbf{v} \times \mathbf{v}$$ In conclusion, we have found the equation for the velocity of fluid with respect to point P and the inertial frame, and have derived the relationship between the angular velocity and the velocity vector.