Mathematical Immaturity

14.19 Exercises

• A particle (moving in space) has velocity vector $\mathbf{v} = \omega\,\mathbf{k} \times \mathbf{r},$ where $\omega$ is a positive constant and $\mathbf{r}$ is the position vector. Prove that the particle moves along a circle with constant angular speed $\omega.$ (The angular speed is defined to be $|d\theta/dt|,$ where $\theta$ is the polar angle at time $t.$)
• If the particle moves along a circle with constant angular speed $\omega,$ then the radial component of velocity will be zero (since the radius will be constant) and the transverse component will be a constant $r\omega.$
• It will suffice to show that the radial component of velocity is zero and the transverse component is $r\omega,$ as this describes motion along a polar curve with a radius of fixed length (ie. a circle of radius $r$) with constant angular velocity $\omega.$ Let $\mathbf{r} = r\mathbf{u}_r,$ where $\mathbf{u}_r$ is the unit vector defined by: \begin{align*} \\ \mathbf{u}_r &= \cos \theta\,\mathbf{i} + \sin \theta\,\mathbf{j} \end{align*} where $\theta$ is the polar angle at time $t.$ Then, $\mathbf{v} = \mathbf{r}'$ is given by: \begin{align*} \\ \mathbf{v} &= \left(r\mathbf{u}_r\right)' \\ \\ &= \frac{dr}{dt}\mathbf{u}_r + r\frac{d\mathbf{u}_r}{dt}. \end{align*} We can use the chain rule to find $\frac{d\mathbf{u}_r}{dt}:$ \begin{align*} \\ \frac{d\mathbf{u}_r}{dt} &= \frac{d\mathbf{u}_r}{d\theta}\frac{d\theta}{dt} \\ \\ &= \mathbf{u}_{\theta}\frac{d\theta}{dt} \end{align*} where $\mathbf{u}_{\theta} = -\sin\theta\,\mathbf{i} + \cos\theta\,\mathbf{j}$ is the unit vector perpendicular to $\mathbf{u}_r.$ Thus, we can express velocity in terms of the perpendicular unit vectors $\mathbf{u}_r$ and $\mathbf{u}_{\theta}$ \begin{align*} \\ \mathbf{v} &= \frac{dr}{dt}\mathbf{u}_r + r\frac{d\theta}{dt}\mathbf{u}_{\theta} \end{align*} where the scalar factors $dr/dt$ and $r\,d\theta/dt$ are called the radial and transverse components of velocity, respectively. Now, we use our previous definition of $\mathbf{r}$ to find $\mathbf{v} = \omega\mathbf{k} \times \mathbf{r}$ \begin{align*} \\ \mathbf{v} &= \omega \mathbf{k} \times \mathbf{r} \\ \\ &= -r\omega \sin\theta\,\mathbf{i} + r\omega \cos \theta\,\mathbf{j} \\ \\ &= r\omega \mathbf{u}_{\theta} \end{align*} But this means that the radial component of velocity $dr/dt = 0,$ and the transverse component of velocity $r\,d\theta/dt = r\omega.$ Thus, the particle must be moving on a circle of radius $r$ with angular speed $\omega. \quad \blacksquare$