The motion of the particle satisfies $\textbf{v} = \textbf{c}\times \textbf{r}$

Question

Why does a particle with position $\textbf r(t)$ that satisfies $\textbf{v}(t) = \textbf{c}\times \textbf{r}(t)$ follow a circle that lies in a plane perpendicular to $\textbf{c}$ with centre on a line through the origin in the direction of $\textbf{c}$?

Answer 1

The other two answers provided so far are both very good, but are perhaps a bit too advanced for what I suspect is an introductory calculus-based physics question. So, how to understand this from a an introductory calculus-based physics level as opposed to an upper-undergraduate / graduate physics level? The most advanced concept I'll use is the vector triple product.

An easy way to understand this is to look at acceleration. We have $\dot{\vec r}(t) = \vec c \times \vec r(t)$, where $\vec c$ is a constant vector. Differentiating yields $\ddot{\vec r}(t) = \vec c \times \dot{\vec r}(t) = \vec c \times (\vec c \times \vec r(t))$. Applying the triple vector product identity, this becomes $$\ddot{\vec r}(t) = \vec c (\vec r(t) \cdot \vec c) - \vec r(t)(\vec c \cdot \vec c) = (\vec r(t) \cdot \vec c)\,\vec c - c^2\vec r(t)$$ where $c\equiv ||\vec c||$. The first term on the right hand side involves a factor of $\vec r(t) \cdot \vec c$. Differentiating this with respect to time yields $\frac d{dt}(\vec r(t) \cdot \vec c) = \dot{\vec r}(t) \cdot \vec c = (\vec c \times \vec r) \cdot \vec c$. This is the inner product of two orthogonal vectors, so it is identically zero. That first term on the right hand side is thus a constant. Defining $\vec r_c = \frac {(\vec r(0)\cdot \vec c)\,\vec c}{c^2}$, the expression for $\ddot{\vec r}(t)$ can be rewritten as $$\ddot{\vec r(t)} = -c^2(\vec r(t) - \vec r_c)$$ Now define $\vec u(t) \equiv \vec r(t) - \vec r_c$. Since $\vec r_c$ is a constant, $\dot {\vec u}(t) = \dot {\vec r}(t)$ and $\ddot {\vec u}(t) = \ddot {\vec r}(t)$. Thus $$\ddot {\vec u}(t) = -c^2 \vec u$$ This is uniform circular motion about the origin. Since the original vector $\vec r(t)$ is a fixed displacement from $\vec u(t)$, the vector $\vec r(t)$ undergoes uniform circular motion about a center offset from the origin by length $r_c$ in the direction of $\vec c$. Finally, since $\dot{\vec r}(t) \cdot \vec c$ is identically zero, the plane of the rotation is normal to $\vec c$.

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To loading...: You eventually will learn about Lie groups if you keep loading up on physics. Lie groups are extremely useful in many aspects of physics. One of them is understanding rotation.

Answer 2

Write the cross product as a matrix equation: instead of ${\rm d}_t \vec{r} = \vec{\omega}\times \vec{r}$ we can write:

$${\rm d}_t \vec{r} = \left(\begin{array}{ccc}0&-\omega_z&\omega_y\\\omega_z&0&-\omega_x\\-\omega_y&\omega_x&0\end{array}\right)\,\left(\begin{array}{c}x\\y\\z\end{array}\right)\tag{1}$$

(check that this is the same as the cross product!) and so our solution is :

$$ \vec{r} = \exp\left(t\,\left(\begin{array}{ccc}0&-\omega_z&\omega_y\\\omega_z&0&-\omega_x\\-\omega_y&\omega_x&0\end{array}\right)\right) \vec{r}(0)\tag{2}$$

Arguing as in this answer here, you can show that the matrix above is a rotation matrix, so that your equation must describe a circular path. You can intuitively see this also by taking heed that the eigenvalues of the $\vec{\omega}$ matrix are 0 (corresponding to the rotation axis) and $\pm i\,|\vec{\omega}|$, so we get terms like $e^{\pm i |\vec{\omega}| t}$ in the solution.

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Your Question Backwards

ACuriousMind makes the comment "Because that's how the cross product works". This is not as flippant as it may seem: you can use my reasoning above to define the cross product and get geometric insight into its meaning.

Arguing as in my other answer, the group of $N\times N$ rotation matrices are precisely those matrices of the form $e^\omega$, where $\omega = -\omega^T$ is a real, skew symmetric (its transpose is its negative) $N\times N$ matrix. All rotational motions are those which can be described by $X(t) = e^{\omega \,v(t)}\,X(0)$, where $v:\mathbb{R}\to\mathbb{R}$ accounts for a varying angular speed with time. The particle's instantaneous velocity is:

$$\dot{X} = \dot{v}\,\omega\,X$$

where $\dot{v}\in\mathbb{R}$ and we can absorb it into the skew-symmetric $\omega$ matrix and still have a skew-symmetric real matrix. So, given a position vector $X$, we see that the set of possible circular motion velocities of that vector is precisely the set

$$\{\omega\,X|\,\omega=-\omega^T;\;\omega\text{ an }N\times N\text{ real matrix}\}\tag{3}$$

The subspace of $\mathbb{R}^N$ left invariant by the rotation is that spanned by the eigenvectors of $e^\omega$ corresponding to unit eigenvectors; otherwise put: the space left invariant is the nullspace of $\omega$ (spanned by eigenvectors corresponding to eigenvalue of nought). When we talk about three dimensions, it so happens, by co-incidence, that this nullspace is one dimensional i.e. we can define a three dimensional rotation by a single axis. The vector woth components $(\omega_x,\,\omega_y,\,\omega_z)$ in (1) and (2) is directed along the null vector of the matrix $\omega$. Axes are only meaningful in three dimensions: in four or greater, the $\omega$ matrix has several, linearly independent null vectors and one cannot define a general rotation simply by an axis and an angular speed.

So we define the cross product of the 3-vector $\omega$ with $X$ as instantaneous velocity of $X$ when undergoing the uniform rotation $e^{\omega\,t}\,X$ (where we replace $\omega$ by the $3\times3$ skew-symmetric matrix $\omega$ in (1)). So the cross product is the operation on the RHS of (1), which you can check to be the cross product's wonted definition.

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See also Example 1.3 on the page "Some examples of connected Lie groups" on my website for a fuller discussion of the the above.

Answer 3

Assuming that $\bf{c}$ is the angular velocity vector the motion of any particle that is part of rigid body in general is

$$ {\bf v } = h {\bf c} + {\bf c} \times {\bf r}$$

where $h$ is the motion pitch and ${\bf r}$ is the location of the particle. See this post on details of how the screw motion of a rigid body defined.

The first part is not position dependent and represents a velocity parallel to the rotation axis defined by ${\bf c}$. The second part is position dependent and only the perpendicular distance to axis of rotation matters, and hence the plane definition which is normal to motion axis.

Linear velocity is a vector (screw) field defined by the line of rotation and a pitch (the amount of parallel translation per rotation). If the point of interest moves parallel to the rotation axis it does not change the observed velocity because the distance to the line hasn't changed.

In the end, the motion of rigid body is defined by an infinate line in space with 6 Pluecker Coordinates $({\bf v},\,{\bf c})$. This means that any translation parallel to the line does not change the velocity.