Angular velocity

Question

The angular velocity $\omega$ of rotation of a rigid body has the direction of the rotaion axis and magintude equal to the rotation rate in rad per second. The orientation of $\omega$ is determined by the right-hand rule.

1. Let $\overrightarrow{r}$ a vector of the axis to a point $P$ of the rigid body. Show that $\overrightarrow{v}=\overrightarrow{\omega} \times \overrightarrow{r}$ gives the velocity of $P$.
2. Interpret the result at the case of the rotation of a cylinder around its axis, with $P$ a point in the region.

Answer 1

Let the axis of rotation coincide with the $z$ axis of a $x,y,z$ coordinated system, and let $P\in xy$ plane. Then

$$\boldsymbol \omega \times \bf{r} = \det\begin{bmatrix}\bf{\hat i} &\bf{\hat j}&\bf{\hat k} \\0&0&\omega\\r\cos\theta& r \sin\theta & 0\end{bmatrix}=\omega r \cos\theta \bf{\hat j} -\omega r \sin\theta \bf{\hat i}.$$

Where $\theta$ is the angle between $\bf r$ and the positive $x$ semi-axis. Clearly $|\boldsymbol \omega \times \bf{r} |=\omega r$. Moreover its direction is given by a $\pi/2$ rotation of $\bf r$, which corresponds to the description of $\bf v$. The rotation of $P$ about the axis describes a circumference. That's why actually $\bf v$ is tangent to the path of $P$, hence perpendicular to $\bf r$.

Answer 2

One minor problem with this question is that there may be several (ultimately equivalent) definitions of angular velocity and it is not clear from which you would like an answer to start from. So I'll pick a definition that seems the simplest to me and go from there.

Place your spatial origin at the rigid body's center of mass. Let $R(t)$ denote the time-dependent rotation matrix corresponding to your (constant) angular velocity vector $\omega$. Note that the matrix $\hat{\Omega}(t)=\dot{R}(t)R(t)^{-1}$ is antisymmetric because $R(t)^{-1}=R(t)^T$. Therefore there is a unique vector $\Omega(t)$ defined by the formula

$$\hat{\Omega}(t)v=\Omega(t)\times v,$$ where $v$ is an arbitrary vector. The definition of the angular velocity is $\omega=\Omega(t)$.

Given a point $\mathbf{r}$ within the rigid body at $t=0$, the subsequent evolution of that point as it is dragged along with the rigid body is given by $\mathbf{r}(t)=R(t)\mathbf{r}$. Therefore the velocity of the moving point is given by $$\dot{\mathbf{r}}(t)=\dot{R}(t)\mathbf{r}=\dot{R}(t)R(t)^{-1}\mathbf{r}(t).$$ But by the definition of angular velocity given above, $\dot{R}(t)R(t)^{-1}\mathbf{r}(t)=\omega\times\mathbf{r}(t)$, which proves (1). I can't say I understand (2) entirely.