Mathematical Immaturity

14.19 Exercises

• A body is said to undergo a rigid motion if, for every pair of particles $p$ and $q$ in the body, the distance $\|\mathbf{r}_p(t) - \mathbf{r}_q(t)\|$ is independent of $t,$ where $\mathbf{r}_p(t)$ and $\mathbf{r}_q(t)$ denote the position vectors of $p$ and $q$ at time $t.$ Prove that for a rigid motion in which each particle $p$ rotates about the $z$-axis we have $\mathbf{v}_p(t) = \pmb{\omega}(t) \times \mathbf{r}_p(t),$ where $\pmb{\omega}(t)$ is the same for each particle, and $\mathbf{v}_p(t)$ is the velocity of particle $p.$
• Suppose particle $q$ sits at the origin, what happens to $\|\mathbf{r}_p(t) - \mathbf{r}_q(t)\|?$ What does this imply about $\mathbf{v}_p(t)?$
• Since the body rotates about the $z$-axis, we can express the motion of any given particle in terms of polar coordinates. For any given particle $p$ in the body with position $\mathbf{r}_p = (x, y),$ we have the transformation: \begin{align*} \\ x &= r_p\cos\theta, \quad y = r_p\sin\theta \end{align*} where $r_p = \|\mathbf{r}_p\|$ and $\theta$ is the polar angle at time $t.$ Then, we 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*} This means that we can express the position and velocity of $p$ in terms the component vectors $\mathbf{u}_{r}$ and $\mathbf{u}_{\theta}:$ \begin{align*} \\ \mathbf{r}_p(t) &= r_p\mathbf{u}_r \\ \\ \mathbf{v}_p(t) &= \frac{dr}{dt}\mathbf{u}_r + r_p\frac{d\theta}{dt}\mathbf{u}_{\theta} \end{align*} where the scalar factors $\frac{dr}{dt}$ and $r_p\frac{d\theta}{dt}$ are called the radial and transverse components of velocity, respectively. But since the body is rigid, we know that for any pair of particles $p$ and $q,$ the distance $\|\mathbf{r}_p(t) - \mathbf{r}_q(t)\|$ is constant for all $t.$ Suppose $q$ sits at the origin. Then, every particle $p$ inside the body undergoes circular motion about the origin, which means that the radial component of velocity $\frac{dr}{dt} = 0,$ making $\mathbf{v}_p(t) = r\frac{d\theta}{dt}\mathbf{u}_{\theta}.$ As we showed in Exercise 19, we can express this velocity as the cross product $\pmb{\omega}(t) \times \mathbf{r}_p(t),$ where $\pmb{\omega}(t) = \omega(t)\mathbf{k}$ and $\omega(t) = \frac{d\theta}{dt}:$ \begin{align*} \\ \mathbf{v}_p(t) &= r_p\frac{d\theta}{dt}\mathbf{u}_{\theta} \\ \\ &= \frac{d\theta}{dt}\left(-r_p\sin\theta\,\mathbf{i} + r_p\cos\theta\,\mathbf{j}\right) \\ \\ &= \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 0 & \frac{d\theta}{dt} \\ r_p\cos\theta & r_p\sin\theta & 0 \end{vmatrix} \\ \\ &= \omega(t)\mathbf{k} \times r_p\mathbf{u}_r \\ \\ &= \pmb{\omega}(t) \times \mathbf{r}_p(t) \end{align*} And as we can see, $\pmb{\omega}(t)$ is the same regardless of position. Thus, it must be the same for all particles in the body. This completes the proof.