\[\mathbf{\omega}\]
about the origin. Where should a unit square be placed in the \[z\]
plane so that the velocity flux across it is minimized?We can write
\[\mathbf{\omega}=\omega \mathbf{i}, \mathbf{r}= x \mathbf{i} + \mathbf{j} + z \mathbf{k}\]
then\[\begin{equation} \begin{aligned} \mathbf{v} &= \mathbf{\omega} \times \mathbf{r} \\ &= \omega \mathbf{i} \times x \mathbf{i} + \mathbf{j} + z \mathbf{k} \\ &= - \omega y \mathbf{i} + \omega x {j} \end{aligned} \end{equation}\]
Let
\[M\]
be the mass of flux passing through the unit square per second. then \[M= \int_S \rho \mathbf{v} \cdot \mathbf{n} dS\]
The rotation is uniform around the \[z\]
axis so we can pick a point to place the unit square to make the calculations simple. Pick a point on the y axis with coordinates \[(0,a/2,1/2)\]
to be the centre of the square, then \[\mathbf{n} = \mathbf{i}\]
. Then\[\begin{equation} \begin{aligned} M= \int_S \rho \mathbf{v} \cdot \mathbf{n} dS &= \rho \omega \int^1_0 \int^{a+1}_a y dy dz \\ &= \rho \omega \int^1_0 [\frac{y^2}{2}]^{a+1}_a dz \\ &= \rho \omega \int^1_0 [\frac{2a=1}{2}]^{a+1}_a dz \\ & =\rho \omega \frac{2A+1}{2} \end{aligned} \end{equation}\]
\[m\]
will be zero at \[a=- \frac{1}{2}\]
.