Curl Operator/Examples/Rotation of Rigid Body

Example of Curl Operator
Consider a rigid body $B$ in rotary motion with angular velocity $\omega$ about an axis $OA$, where $O$ is some fixed point inside $B$.

Let $P$ be an arbitrary point inside $B$.

Let $B$ also be subject to a linear velocity $\mathbf v_0$ in an arbitrary direction.

Let the instantaneous velocity of $P$ be $\mathbf V$

Then:
 * $\bsomega = \dfrac 1 2 \curl \mathbf V$

where $\bsomega$ is the angular velocity (axial) vector along the axis $OA$ in the sense according to the right-hand rule.

Proof
$P$ moves in a circular path $C$ around $OA$ with an instantaneous velocity $v$, tangent to $C$ where:
 * $r$ is the distance of $P$ from $O$
 * $\theta$ is the angle between $OR$ and $OP$.


 * Curl-and-angular-velocity.png

As noted, the angular velocity is a vector (axial) $\bsomega$ along the axis $OA$ in the sense according to the right-hand rule.

Let the position vector of $P$ be given by $\mathbf r$.

Then the tangential velocity $v$ of $P$ is $\bsomega \times \mathbf r$, which has a magnitude $\omega r \sin \theta$ and a direction which is perpendicular to the plane containing $OA$ and $OP$, that is, the plane containing $\bsomega$ and $\mathbf r$.

We also have that in addition to the angular velocity undergone by $B$, let it also be subject to a linear velocity $\mathbf v_0$ in an arbitrary direction.

Then the total velocity $V$ of $P$ is given by:
 * $\mathbf V = \mathbf v_0 + \bsomega \times \mathbf r$

and the curl of $V$ is given by:
 * $\curl V = \nabla \times \mathbf V = \nabla \times v_0 + \nabla \times \bsomega \times \mathbf r$

Because $v_0$ is constant for all points in $B$, and so is independent of any arbitrary Cartesian coordinates $x$, $y$ and $x$ that may be imposed upon it, $\nabla \times \mathbf v_0$ is zero.

The angular velocity $\bsomega$ is also constant for all points in $B$ at any one time, as $B$ is rigid.

It can be written:
 * $\bsomega = \omega_x \mathbf i + \omega_y \mathbf j + \omega_z \mathbf k$

By definition of vector cross product:
 * $\bsomega \times \mathbf r = \paren {\omega_y z - \omega_z y} \mathbf i + \paren {\omega_z x - \omega_x z} \mathbf j + \paren {\omega_x y - \omega_y x} \mathbf k$

Because $\omega_x$, $\omega_y$ and $\omega_z$ are not functions of $x$, $y$ and $z$, their partial derivatives $x$, $y$ and $z$ are all zero.

Hence: