homework and exercises - Circular motion - vectors

We are asked to show that, if $$\frac{{\rm d} \mathbf{r}}{{\rm d}t} = \mathbf{c}\times\mathbf{r},$$ where $\mathbf c$ is a constant vector, then the body moves in uniform circular motion.

I have done this by considering the time derivative of the dot product and showing that this is zero: $$\frac{{\rm d}\mathbf r}{{\rm d}t}\cdot\frac{{\rm d}\mathbf r}{{\rm d}t}=0,$$ and hence that the speed of the body is constant.

We are then asked to show that if $\hat θ$ and $\hat n$ are tangential and unit normal vectors at a point along the path, and $s$ is the scalar length along the path, that $$\frac{{\rm d}\theta}{{\rm d}s}=\frac{1}{r}\hat{n},$$ where $r$ is the radius of curvature at the point.

Answer

You can decompose the normal and tangential unit vectors $\hat {\mathbf n}$ and $\hat {\boldsymbol {\theta}}$ in Cartesian coordinates. From the diagram below we have: $$\hat {\boldsymbol {\theta}} = -\sin\theta \hat{\mathbf i} + \cos\theta \hat{\mathbf j}$$ $$\hat {\mathbf {n}} = \cos\theta \hat{\mathbf i} + \sin\theta \hat{\mathbf j}$$ Now differentiate both sides of the first equation with respect to $\theta$ to get: $$\frac {d}{d\theta}\hat {\boldsymbol {\theta}}= -\cos\theta \hat{\mathbf i} - \sin\theta \hat{\mathbf j}$$ Which by the second equation means: $$\frac {d}{d\theta}\hat {\boldsymbol {\theta}}= -\hat {\mathbf {n}}$$ Now since $ds = rd\theta$ (arc length = radius $\times$ angle), we finally get: $$\frac {d}{ds}\hat {\boldsymbol {\theta}}= -\frac {1}{r}\hat {\mathbf {n}}$$ The differing minus sign with your equation is probably because of different conventions in choosing the direction of $\hat {\boldsymbol {\theta}}$. I chose the direction of increasing $\theta$ as my convention, as you can see from the diagram below. One can also choose the opposite direction.