Consider the following equation (10.10) in Morin's book on classical mechanics, it is meant to provide a correct prediction of the measured acceleration of a particle for someone taking measurements inside an accelerating frame (granted they know that they are in one and know some of the details of its motion):

$$m\vec{a}=\vec{F}\overbrace{-m\dfrac{d^2\vec{R}}{dt^2}}^{\vec{F}_{trans}}\overbrace{-m\vec{\omega}\times(\vec{\omega}\times\vec{r})}^{\vec{F}_{cent}}\overbrace{-2m\vec{\omega}\times\vec{v}}^{\vec{F}_{cor}}\overbrace{-m\dfrac{d\vec{\omega}}{dt}\times\vec{r}}^{\vec{F}_{azim}}$$

The coordinates in this expression are measured in two different frames and is meant to be interpreted in the following way: $\vec{r}$, $\vec{v}$ and $\vec{a}$ are the position, velocity and acceleration of a particle of mass $m$ as measured in the $\textbf{accelerating frame}$, while $\vec{\omega}$ and $\vec{R}$ are the angular velocity and position vector $\textbf{of the accelerating frame}$ as measured from the inertial frame (lab frame). Finally, $\vec{F}$ are all the $\textbf{real forces}$ as measured on $\textbf{the particle}$ from the inertial frame (lab frame).

Now consider the following simple example to get a feel for the use of this equation: consider a particle stuck to the floor on the inside of a linearly accelerating train. An observer on the train does not see this particle accelerating, it sees it stationary. In order for $\vec{a}=0$ on the left hand side, there would need to be a real force acting on the particle to be included in the $\vec{F}$ term in order to cancel out the fictitious acceleration it would feel (from $\vec{F}_{trans}$) if it were not bound to the floor by friction. Of course the real force is friction, and the right hand side cancels out nicely to obtain zero on the left hand side as desired.

Now consider a spinning disk rotating with constant angular velocity (as measured in the lab frame). Now picture a particle stuck to the surface of the spinning disk. An observer on the disk does not see the particle accelerating, it sees it stationary, so in order to obtain zero on the left hand side, we need a real force to cancel out the fictitious centrifugal force (from $\vec{F}_{cent}$) on the right hand side. Of course, this real force is known as the centripetal force, and including it into $\vec{F}$ on the right hand side cancels out the centrifugal force and we obtain zero on the left hand side as desired.

If we now consider a particle on a rotating disk moving at constant velocity (as measured by an observer in the accelerating disk frame) along the diameter of the disk, once again an observer on the disk does not see the particle accelerating, it is merely moving at constant velocity. Therefore once again we desire to obtain zero on the left hand side of the equation above. In order to do so, we must include two real forces into $\vec{F}$: one to cancel out the fictitious centrifugal force, and one to cancel out the fictitious coriolis force (from $\vec{F}_{cor}$) . Ofcourse the first is the centripetal force and the second I know the quantitative details of (its exact magnitude is that of the coriolis force, and its direction is opposite to that of the coriolis force) but....

$\textbf{My question is the following:}$ does this real force analogue of the coriolis force have a $\textit{name}$? (like the centrifugal/centripetal pair)

I have the same question for the true force analogue of the azimuthal force.

Bonus: If it doesn't have a name, why is this so? Is it just because we deal with these true forces less often than the centripetal force and thus just never coined a name?

  • The real force counterpart to the centrifugal force is just called centripetal force, it is due to other forces and need not be friction, spin a ball on a string and you can now replace friction with tension. Spin a ball in a vertical circle and now tension and gravity supply the centripetal force. A ball rolling inside a vertical loop has both gravity and the normal force for centripetal force. A satellite in orbit has gravity to thank. – Triatticus Dec 6 at 19:36

Now consider a spinning disk rotating with constant angular velocity (as measured in the lab frame). Now picture a particle stuck to the surface of the spinning disk. An observer on the disk does not see the particle accelerating, it sees it stationary, so in order to obtain zero on the left hand side, we need a real force to cancel out the fictitious centrifugal force (from F⃗ cent) on the right hand side. Of course, this real force is known as the centripetal force

No, that real force is friction. That friction has a centripetal direction, but that's just your choice of coordinates. One could also call the acceleration in the first case "longitudinal force", but that doesn't add anything.

If we now consider a particle on a rotating disk moving at constant velocity (as measured by an observer in the accelerating disk frame) along the diameter of the disk, once again an observer on the disk does not see the particle accelerating, it is merely moving at constant velocity.

So just to be sure I understand the setup, like an ant was walking around the outer edge of a record that was being played?

My question is the following: does this real force analogue of the coriolis force have a name? (like the centrifugal/centripetal pair)

Yes, "friction".

If you were to do this in "real life", make a huge record and walk along the edge, you would notice it immediately. As you walked you would feel both the "outward force" and the "twisting force" on your shoes. If you wore curling shoes you'd go flying off.

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