Mathematical Immaturity

14.19 Exercises

• A particle moves in a plane perpendicular to the z-axis. The motion takes place along a circle with center on this axis. (a) Show that there is a vector $\pmb{\omega}(t)$ parallel to the $z$-axis such that $$ \mathbf{v}(t) = \pmb{\omega}(t) \times \mathbf{r}(t) $$ where $\mathbf{r}(t)$ and $\mathbf{v}(t)$ denote the position and velocity vectors at times $t.$ The vector $\pmb{\omega}(t)$ is called the angular velocity vector and its magnitude $\omega(t) = \|\pmb{\omega}(t)\|$ is called the angular speed. (b) The vector $\pmb{\alpha}(t) = \pmb{\omega}'(t)$ is called the angular acceleration vector. Show that the acceleration vector $\mathbf{a}(t) = \mathbf{v}'(t)$ is given by the formula \begin{align*} \\ \mathbf{a}(t) = \left[\pmb{\omega}(t) \cdot \mathbf{r}(t)\right]\pmb{\omega}(t) - \omega^2(t)\mathbf{r}(t) + \pmb{\alpha}(t) \times \mathbf{r}(t) \end{align*} (c) If the particle lies in the $xy$-plane and if the angular speed $\omega(t)$ is constant, say $\omega(t) = \omega,$ prove that the acceleration vector $\mathbf{a}(t)$ is centripetal and that, in fact, $\mathbf{a}(t) = -\omega^2\mathbf{r}(t).$
• (a) Recall that for circular motion, the radial component of velocity is zero and the transverse component is $r\frac{d\theta}{dt}$ where $\left|\frac{d\theta}{dt}\right|$ is the angular speed. Refer to the solution of Exercise 18 to derive $\pmb{\omega}(t).$ (b) Recall that for two vector-valued functions $F$ and $G,$ $\left(F \times G\right)' = \left(F' \times G\right) + \left(G' \times F\right).$ (c) If the motion lies in the $xy$-plane, what does this imply about the term $\left[\pmb{\omega}(t) \cdot \mathbf{r}(t)\right]\pmb{\omega}(t)$ in the equation for acceleration? If angular velocity $\omega = \frac{d\theta}{dt}$ is constant, what does this imply about $\pmb{\alpha}(t)?$
• (a) For circular motion with center on the $z$-axis, we have the following relations between the rectangular coordinates $(x, y)$ and the polar coordinates $r$ and $\theta.$ Let $r = \|\mathbf{r}\|,$ then \begin{align*} \\ x &= r\cos\theta, \quad y = r\sin\theta. \end{align*} We now define two perpendicular unit vectors $\mathbf{u}_r$ and $\mathbf{u}_{\theta}:$ \begin{align*} \\ \mathbf{u}_r &= \cos\theta\,\mathbf{i} + \sin\theta\,\mathbf{j} \\ \\ \mathbf{u}_{\theta} &= \frac{d\mathbf{u}_r}{d\theta} \\ \\ &= -\sin\theta\,\mathbf{i} + \cos\theta\,\mathbf{j} \end{align*} Then, we can rewrite position as $\mathbf{r}(t) = r\mathbf{u}_r.$ As such, the equation for velocity becomes: \begin{align*} \\ \mathbf{v}(t) &= \frac{d\left(r\mathbf{u}_r\right)}{dt} \\ \\ &= \frac{dr}{dt}\mathbf{u}_r + r\frac{d\mathbf{u}_r}{dt} \end{align*} Applying the chain rule, with $\frac{d\mathbf{u}_r}{dt} = \frac{d\mathbf{u}_r}{d\theta}\frac{d\theta}{dt},$ we can express velocity as a linear combination of the component vectors $\mathbf{u}_r$ and $\mathbf{u}_{\theta}:$ \begin{align*} \\ \mathbf{v}(t) &= \frac{dr}{dt}\mathbf{u}_r + r\frac{d\theta}{dt}\mathbf{u}_{\theta} \end{align*} where the scalar factors $\frac{dr}{dt}$ and $r\frac{d\theta}{dt},$ respectively, are called the radial and transverse components of velocity. For a particle moving along a circle, we know that the radius remains constant, thus the radial component of velocity is zero. Rewriting $\mathbf{v}(t)$ in terms of the component vectors $\mathbf{i}$ and $\mathbf{j},$ we get: \begin{align*} \\ \mathbf{v}(t) &= r\frac{d\theta}{dt}\mathbf{u}_{\theta} \\ \\ &= \frac{d\theta}{dt}\left(-r\sin\theta\,\mathbf{i} + r\cos\theta\,\mathbf{j}\right) \end{align*} But as we saw in Exercise 18, this linear combination is equal to the cross product $\omega\mathbf{k} \times \mathbf{r},$ where $\omega = \frac{d\theta}{dt}.$ Thus, we have shown that there exists a vector $\pmb{\omega}(t)$ parallel to the $z$-axis satisfying $\mathbf{v}(t) = \pmb{\omega}(t) \times \mathbf{r}(t)$ for circular motion with center on the $z\text{-axis.}\quad \blacksquare$ (b) Calculating $\mathbf{a}(t) = \mathbf{v}'(t),$ we get: \begin{align*} \\ \mathbf{a}(t) &= \left[\pmb{\omega}(t) \times \mathbf{r}(t)\right]' \\ \\ &= \left[\pmb{\alpha}(t) \times \mathbf{r}(t)\right] + \left[\pmb{\omega}(t) \times \mathbf{v}(t)\right] \end{align*} Expanding the rightmost term, we get: \begin{align*} \\ \pmb{\omega}(t) \times \mathbf{v}(t) &= -\omega^2r\cos\theta\,\mathbf{i} - \omega^2r\sin\theta\,\mathbf{j} \\ \\ &= -\omega^2\mathbf{r}(t) \end{align*} where $\omega = \frac{d\theta}{dt}$ is the angular velocity. Then, we note that $\pmb{\omega}(t)$ is parallel to $\mathbf{k}$ and thus perpendicular to $\mathbf{r},$ making the term $\left[\pmb{\omega}(t) \cdot \mathbf{r}(t)\right]\pmb{\omega}(t) = 0.$ As such, the equation for acceleration becomes \begin{align*} \\ \mathbf{a}(t) &= \left[\pmb{\alpha}(t) \times \mathbf{r}(t)\right] + \left[\pmb{\omega}(t) \times \mathbf{v}(t)\right] \\ \\ &= -\omega^2\mathbf{r}(t) + \left[\pmb{\alpha}(t) \times \mathbf{r}(t)\right] \\ \\ &= \left[\pmb{\omega}(t) \cdot \mathbf{r}(t)\right]\pmb{\omega}(t) -\omega^2\mathbf{r}(t) + \pmb{\alpha}(t) \times \mathbf{r}(t) \quad \blacksquare \end{align*} (c) If the particle lies in the $xy$-plane, then the position vector is perpendicular to the $z$-axis, and thus $\left[\pmb{\omega}(t) \cdot \mathbf{r}(t)\right]\pmb{\omega}(t) = O.$ Additionally, if the angular velocity $\omega = \frac{d\theta}{dt}$ is constant for all $t,$ then $\frac{d^2\theta}{dt^2} = 0.$ But $\pmb{\alpha}(t) = \pmb{\omega}'(t) = \frac{d^2\theta}{dt^2}\mathbf{k},$ thus $\pmb{\alpha}(t) = O,$ making $\pmb{\alpha}(t) \times \mathbf{r}(t) = O.$ This leaves us with \begin{align*} \\ \mathbf{a}(t) &= \left[\pmb{\omega}(t) \cdot \mathbf{r}(t)\right]\pmb{\omega}(t) -\omega^2\mathbf{r}(t) + \pmb{\alpha}(t) \times \mathbf{r}(t) \\ \\ &= -\omega^2\mathbf{r}(t) \quad \blacksquare \end{align*}