Fyneman exercise II ch2-5

Q.

The velocity of a solid object rotationg about an axis is a field $\vec{v}(x,y,z)$. Show that

a)

$\vec{\nabla} \cdot \vec{v}=0$

b)

$\vec{\nabla}\times \vec{v} = 2\vec{\omega}$ where $\vec{\omega}$ = angular velocity.

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Trial.

a)

If velocity has radial component or magnitude of angular velocity is not constant, solid object should be broken, so velocity has no radial component and magnitude of angular velocity is constant.

Let’s take coordinate such that solid object is rotating in $z=0$ plane, axis is $x=0,\ y=0$ and use $\omega$ for magnitude of angular velocity.

In cylinderical coordinate, we can write $x=r\cos{\theta}$, $y=r\sin{\theta}$.

Then
$dx=\cos{\theta}dr-r\sin{\theta}d\theta\tag{1}$
$dy=\sin{\theta}dr+r\cos{\theta}d\theta\tag{2}$

Divide by $dt$
$v_x = \cos{\theta}{dr \over dt}-r\sin{\theta}{d\theta \over dt}\tag{3}$
$v_y = \sin{\theta}{dr \over dt}+r\cos{\theta}{d\theta \over dt}\tag{4}$

Radial componet of velocity is zero and ${d\theta \over dt}=\omega$, so from eq(3), (4) we get

$v_x=-r\sin{\theta}\omega=-y\omega\tag{5}$
$v_y=r\cos{\theta}\omega=x\omega\tag{6}$
$\omega=const.\tag{7}$

$\vec{\nabla}\cdot\vec{v}={\partial v_x \over \partial x}+{\partial v_y \over \partial y}\tag{8}$

Using eq(5),(6) and (7), eq(8) can be wrriten

$\vec{\nabla}\cdot\vec{v}=0\cdot \omega+0\cdot \omega= 0\tag{9}$

b)

Suppose rotation is counter clock wise.

Then we can write
$\vec{\omega}=\omega\hat{e}_z\tag{10}$

Curl $\vec{v}$ is
$\vec{\nabla}\times\vec{v}=({\partial v_y \over \partial x}-{\partial v_x \over \partial y})\hat{e}_z\tag{11}$

Using eq(5) & (6) and (10), eq(11) can be written

$\vec{\nabla}\times\vec{v}=(\omega-(-\omega))\hat{e}_z=2\omega\hat{e}_z=2\vec{\omega}\tag{12}$