Assume that the particle has the velocity of $v(t)$, and $v = (wk) \times r, w>0$, where $v(t) = r'(t)$. Apparently, $k$ is the unit direction vector of the vertical axis in the 3d-space.
I need to prove that the particle moves along a circle, with a constant angular speed $w$. I tried proving that the path has constant curvature, because if it is a circle, than the curvature should be constant everywhere, but:
$$\kappa(t) = \frac{\lVert r''(t) \times r'(t) \rVert}{\lVert r'(t) \rVert^3} = \frac{\lVert wk \rVert}{\lVert (wk) \times r \rVert^3} = \frac{w}{w^3\big(\lVert r(t)\rVert^2 – (k \cdot r(t))^2\big)^{3/2}}$$
I cannot simplify it further to arrive at a constant. How can I prove that the movement is along a circle?
Best Answer
You can dot multiply your condition by $\mathbf{r}(t)$. The triple product on the left will be zero and you obtain $$ \mathbf{r}'(t)\cdot \mathbf{r}(t)=\frac{1}{2}\frac{d}{dt}\|\mathbf{r}(t)\|^2=0 $$ and therefore $r=\|\mathbf{r}(t)\|$ is a constant. So the point is within a constant distance from the origin.
Next, you dot multiply by $w\mathbf{k}$. Again the left hand side is zero and you get $$ \mathbf{r}'(t)\cdot w\mathbf{k}=0 $$ that is $\mathbf{v}(t)$ is parallel to the $XY$ plane. Therefore your motion is circular on a plane parallel to the $XY$-plane. Finally, taking modules and taking into account that $\mathbf{r}(t)$ forms a constant angle $\alpha$ with $\mathbf{k}$ you have $$ v=\|\mathbf{v}(t)\|=wr\sin\alpha=wd $$ where $d$ is the distance from the particle to the $z$-axis which means exactly that your angular velocity is $w$.