1
$\begingroup$

Coriolis effect is usually derived in a long twice differentiated equation of motion between a lab frame and rotated frame. Afterwards you group together all the terms and you get all the fictitious forces. However no intuition is gained in it for me.

For the coriolis effect, it seems it is the force that appropriately changes the angular momentum of an object, or the friction force in the rotating frame that is needed to keep an object on a straight path w.r.t. the rotating frame.

So for an object moving radially outward from the center of the rotating disk, the corrective friction force will be (ignoring the other effects): enter image description here

\begin{equation} \text{Angular momentum of the object is:} \\ L = mr^2 \omega= m(vt)^2 \omega, \ \ \ \ \ [r=vt] \\ \end{equation}

\begin{equation} \text{differentiating w.r.t. time} \\ \frac{dL}{dt} = \frac{d}{dt} (mv^2t^2 \omega) = mv^2 \omega \frac{d (t^2)}{dt} \\ \end{equation}

\begin{equation} \text{expressing it in terms of torque} \\ \frac{dL}{dt} = \tau = F_{\text{cor}}r = 2 mv^2 \omega t \\ F_{\text{cor}} = 2 mv \omega \frac{(vt)}{r} = \boxed{2mv\omega} \end{equation}

There is a related derivation for centrifugal force as well using conservation of linear momentum.

$$ \frac{dp}{dt} = \frac{d (p\theta)}{dt} = p\omega = mv \frac{v}{r} = \frac{mv^2}{r}$$

Why aren't these derivations used more often?

$\endgroup$
3
  • 3
    $\begingroup$ your case is just for one dimensional in three dimensional the Coriolis force is $~\vec F_c=2\,m\,(\vec\omega\times \vec v)~$ can you obtain this result ? $\endgroup$
    – Eli
    May 7, 2022 at 15:31
  • $\begingroup$ Isn't that an arbitrary generalization? I mean if you have an object with a random three dimensional velocity in a rotating frame you could decompose the velocity in radial/tangential velocity and apply conservation of (angular momentum) to get it. But the gist is exactly the same, it just gets a bit messier in 3D I believe. Or I could be mistaken $\endgroup$ May 7, 2022 at 19:36
  • $\begingroup$ Related : Velocity in a turning reference frame. $\endgroup$
    – Frobenius
    May 7, 2022 at 21:27

3 Answers 3

3
$\begingroup$

Let me first give a general derivation for the case of an object that is physically going around a center of attraction, with the provided centripetal force dialed in to be equal in magnitude and opposite in direction to the centrifugal effect, at every distance to the center of attraction.

That case is circumnavigating motion with a force according to Hooke's law.

With a centripetal force according to Hooke's law the solution of the equation of motion can be described with the following parametric equation.

$x = a \cos(\Omega t)$
$y = b \sin(\Omega t)$

$a$ half the length of the major axis
$b$ half the length of the minor axis
$\Omega$ 360 degrees divided by the duration of one revolution

So the motion is a superpositon of two harmonic oscillations, perpendicular to each other.

enter image description here

The uploaded animated GIF represents a rotating disk. On the left the rotating disk as viewed from a stationary point of view, on the right the rotating disk as viewed from a co-rotating point of view. The arrow represents the centripetal force. The magnitude of the force increases in proportion to the distance to the center of circumnavigation. Along the rim there is a division in 4 quadrants, the quadrants serve as a reference.

The angular velocity is not constant; the angular velocity increases and decreases, cycling twice for every full circumnavigation. Likewise the distance to the center of attraction cycles twice for every full circumnavigation.

The parametric equation can be rearranged as follows:

$$ x = \frac{a+b}{2} \cos(\Omega t) + \frac{a-b}{2} \cos(\Omega t) $$ $$ y = \frac{a+b}{2} \sin(\Omega t) - \frac{a-b}{2} \sin(\Omega t) $$

After transformation of the motion to the rotating coordinate system the motion relative to the co-rotating coordinate system is as follows:

$$ x = \frac{a-b}{2} \cos(2 \Omega t) $$ $$ y = - \frac{a-b}{2} \sin(2 \Omega t) $$

Notice that the motion along the small circle ( the motion relative to the co-rotating coordinate system) cycles with a frequency of $2\Omega$.

I will refer to the motion depicted on the right hand side of the animated GIF as 'motion along the epi-circle'


The next step is to obtain an expression for the acceleration of the moving object relative to the co-rotating coordinate system.

Let $a_c$ be the acceleration relative to the co-rotating coordinate system.
The magnitude of the acceleration towards the center of the epi-circle is given by the standard expression for the required centripetal acceleration for sustained circular motion:

$$ a_c = \omega^2r \tag{1} $$

(I use the lowercase $\omega$ here for the motion along the epi-circle to distinguish it from the uppercase $\Omega$ that I used for the angular velocity of the system.)

This expression gives the magnitude. To sustain uniform circular motion the acceleration must be perpendicular to the instantaneous velocity. That gives the direction of $a_c$: perpendicular.

As stated earlier, the motion along the epi-circle occurs at twice the frequency of the angular velocity of the system: 2$\Omega$

Inserting that into (1):

$$ a_c = (2\Omega)^2r \tag{2} $$

(2) does not feature the velocity relative to the rotating coordinat system explicitly.

Let $v_c$ be the velocity of the object relative to the co-rotating coordinate sytem. Then we can use this expression:

$$ 2\Omega r = v_c \tag{3} $$

to modify (2).

This modification gives us the following expression for the acceleration of the object with respect to the co-rotating coordinate system:

$$ a_c = 2\Omega v_c \tag{4} $$



Discussion

So we see that if the force of attraction increases linear with distance to the center of attraction, Hooke's law, then the motion with respect to the co-rotating coordinate system is given by the coriolis acceleration $2\Omega v_c$

The required centripetal force to sustain circular motion is $m \omega^2 r$
So: when the provided centripetal force is according to Hooke's law there is no opportunity for centrifugal effect; a Hooke's law centripetal force counteracts centrifugal effect everywhere.

Therefore: when the centripetal force is according to Hooke's law the acceleration with respect to the co-rotating coordinate system is described entirely by the Coriolis term: $2\Omega v_c$


The coriolis effect is the same in all directons.
The magnitude is always $2\Omega v_c$
The direction is always perpendicular to the instantaneous velocity.


In many cases you effectively get Hooke's law automatically. For instance, let's say you are on a merry-go-round, and you have been instructed to hold on. When the angular velocity of the merry-go-round increases you hold on more firmly, in response to the need for more centripetal force. In many cases the setup is designed to produce required centripetal force automatically. Example: a liquid mirror telescope


Planar motion

This derivation is for motion in a plane.
The following property is general: when the centripetal force is always towards a single point of attraction the resulting motion is inherently planar motion.

Because of that planar nature it is sufficient to derive for the case with two spatial degrees of freedom.

$\endgroup$
1
$\begingroup$

indeed you can use it for three dimensional case

start with the angular momentum

$$\vec L=\mathbf I\,\vec\omega$$

with the inertia tensor for a mass point $$\mathbf I=-m\,\mathbf{r}_\times\,\mathbf{r}_\times\quad, \mathbf r_\times=\left[ \begin {array}{ccc} 0&-r_{{z}}&r_{{y}} \\ r_{{z}}&0&-r_{{x}}\\ -r_{{y}}&r_{{x}}&0\end {array} \right] $$

you obtain

$$\vec L=-m\,\mathbf{r}_\times\,\mathbf{r}_\times\,\vec\omega$$

with $~\vec r=\vec v\,t$

$$\vec L=-m\,t^2\,\mathbf{v}_\times\,\mathbf{v}_\times\,\vec\omega\quad\Rightarrow\\ \vec{\dot{ L}}=-2\,m\,t\,(\vec v\times\vec v\times\vec \omega)=\vec r\times\vec F_c =\vec{v}\,t\times \vec F_c\quad\Rightarrow\\ +2\,m\,\vec v\times (\vec \omega\times \vec v)=\vec v\times\vec F_c $$ thus the coriolis force $$\boxed{\vec F_c=2\,m\,(\vec\omega\times\vec v)}$$


because $~\vec{\dot{r}}=\vec{v}~$ you should use this derivation

$$\vec L=-m\,\mathbf{r}_\times\,\mathbf{r}_\times\,\vec\omega= -m\,\left(\vec{r}\times\vec{r}\times\vec\omega\right)$$ $$\vec{\dot{L}}=-m\,\left(\vec{\dot{r}}\times\vec{r}\times\vec\omega \right)-m\,\left(\vec{r}\times\vec{\dot{r}}\times\vec\omega\right)\\ 2\,m\,\vec r\times\vec\omega\times\vec v=\vec r\times\vec F_c\quad\Rightarrow\\ 2\,m\,(\vec\omega\times\vec v)=\vec F_c$$

remarks $~\vec\omega~$ is constant and not depending on time

$\endgroup$
2
  • $\begingroup$ Thank you. Is there a similar one for linear momentum? $\endgroup$ May 7, 2022 at 21:32
  • 1
    $\begingroup$ @bananenheld The centrifugal force from Newton second law $\begin{aligned}m\dfrac{d\overrightarrow{v}}{dt}=\overrightarrow{F}\\ \overrightarrow{v}=\overrightarrow{w}\times \overrightarrow{r}\\ m\left( \overrightarrow{w}\times \overrightarrow{v}\right) =\overrightarrow{F}\\ m\left( \overrightarrow{w}\times \overrightarrow{w}\times \overrightarrow{r}\right) =\overrightarrow{F}\end{aligned}$ $\endgroup$
    – Eli
    May 8, 2022 at 11:46
1
$\begingroup$

In my earlier answer I derived the coriolis effect for the general case. The general case is that the circumnavigating object can have any direction of velocity with respect to the rotating coordinate system.

In this answer I will addres the more limited case that you raised in your question: if a object is moving radially with respect to a rotating disk, what is the magnitude of the tendendy of that object to change angular velocity?

In your derivation you used a substitution $r=vt$ to introduce a factor $t$ for time, but that substitution isn't necessary.

The expression for angular momentum:

$$ m\omega r^2 $$

The mass $m$ is constant anyway, so the statement of conservation can without loss of validity be narrowed down to:

$$ \omega r^2 = \text{constant} \tag{1} $$

Hence:

$$ \frac{d(\omega r^2)}{dt} = 0 \tag{2} $$

Differentiating:

$$ r^2 \frac{d\omega}{dt} + \omega \frac{d(r^2)}{dt} = 0 \tag{3} $$

Using the chain rule to convert $\tfrac{d(r^2)}{dt}$ into $2r\tfrac{dr}{dt}$:

$$ r^2 \frac{d\omega}{dt} + 2 r \omega \frac{dr}{dt} = 0 \tag{4} $$

Divide by $r$, and then rearrange:

$$ r \frac{d\omega}{dt} = - 2 \omega \frac{dr}{dt} \tag{5} $$

I will use $a_t$ for acceleration in the direction tangential to a concentric circle (hence perpendicular to the radial direction), and $v_r$ for velocity in radial direction.

Then $r\tfrac{dw}{dt} = a_t$ and $\tfrac{dr}{dt} = v_r$

The result:

$$ a_t = -2\omega v_r \tag{6} $$

Multiplying both sides with $m$ for mass gives the amount of force required to keep the object at the same angular velocity when it has a radial velocity $v_r$

$$ F_t = -2m\omega v_r \tag{7} $$



Not the general coriolis effect

This derivation addresses only the case of radial velocity. However, the general coriolis effect is the same for every direction of motion with respect to the rotating coordinate system, so this derivation isn't general enough.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.