Understanding the Coriolis force

[brotherbobby]

TL;DR

I like to understand how the Coriolis acceleration ($-2\vec\Omega\times\vec v_{rot}$)arises, as explained in Kleppner and Kolenkkow's book An Introduction to Mechanics (2014, 2E). It is evident that there are two terms behind the acceleration; a radial and a tangential component. Only, I am not able to understand either of them.

I copy and paste the relavant paragraphs and images from the text. I might have to do it several times, each time underlining in red what I failed to follow. Here's the first.

Statement : "Here the acceleration is in a plane perpendicular to $\mathbf\Omega$".

Sure, the acceleration itself is perpendicular to $\vec\Omega$. But plane?

The Coriolis acceleration is given by $-2\vec\Omega\times\vec v_{rot}$. The author(s) are trying to see it from an inertial frame in which the body has both a rotation and a velocity. In the rotating frame, the body only has a velocity $\vec v_{rot}$. Fair enough. So we have to understand how does the acceleration $+2\vec\Omega\times\vec v_{rot}$ come about. For simplicity, let's assume that the body is moving uniformly in the rotating frame along the $x$ axis, meaning $\vec v_{rot} = v_0 \hat i$. Also, let's assume that the rotating frame is rotating uniformly about the $z$ axis : $\vec \Omega= \Omega_0 \hat k$. In that case, the coriolis acceleration would be $= 2\Omega_0 v_0\hat j$.

The acceleration is in the $y-z$ plane, along the $y$ axis. So is the angular velocity vector $\vec\Omega$, but along the $z$ axis. How is the acceleration in a plane perpendicular to the rotation velocity?

 

[A.T.]

Sure, the acceleration itself is perpendicular to $\vec\Omega$. But plane?

Saying: acceleration is perpendicular to $\vec\Omega$, is the same as saying: acceleration lies within a plane that is perpendicular to $\vec\Omega$. There might be more appropriate terms for perpendicular in 3D, like normal or orthogonal.

The acceleration is in the y−z plane, along the y axis.

There are infinitely many planes that contain the y axis, including the xy-plane, which is perpendicular to $\vec\Omega$. So there is no contradiction to the previous statement.

The reason for the emphasis on the plane perpendicular to $\vec\Omega$, is that usually $\vec\Omega$ is constant so that plane is fixed. Thus, no matter what the direction of velocity in the rotating frame is, the Coriolis acceleration is always contained within that fixed plane.

 

[kuruman]

Imagine a plane rotating like a turntable about an axis perpendicular to it. A line is drawn on this plane from the axis of rotation radially out. The plane is frictionless and you are in a car that has four thrusters, front, rear, right and left. Firing the front thruster moves you backwards, left thruster to the right and so on.

You start at the axis of rotation and your task is to move out keeping the drawn line directly under you at all times. How do you program your thrusters to accomplish this task? Clearly, you need to fire the rear thruster to move away from the axis, but as you do that, you also need to fire the side thruster(s) to keep the line from rotating away from you. Firing the side thrusters to stay over the straight line in the rotating frame is another way of saying that you are providing the necessary force to "cancel the Coriolis force."

For the mathematical details, you may wish to look at this article that deals with motion inside a straight tunnel that rotates about its middle. A spaceship enters through one end, passes through the center and exits out the other end without touching the walls.

 

[Simon F]

TL;DR: I like to understand how the Coriolis acceleration ($-2\vec\Omega\times\vec v_{rot}$)arises, as explained in Kleppner and Kolenkkow's book An Introduction to Mechanics (2014, 2E). It is evident that there are two terms behind the acceleration; a radial and a tangential component. Only, I am not able to understand either of them.

View attachment 368688
I copy and paste the relavant paragraphs and images from the text. I might have to do it several times, each time underlining in red what I failed to follow. Here's the first.

Statement : "Here the acceleration is in a plane perpendicular to $\mathbf\Omega$".

Sure, the acceleration itself is perpendicular to $\vec\Omega$. But plane?

The Coriolis acceleration is given by $-2\vec\Omega\times\vec v_{rot}$. The author(s) are trying to see it from an inertial frame in which the body has both a rotation and a velocity. In the rotating frame, the body only has a velocity $\vec v_{rot}$. Fair enough. So we have to understand how does the acceleration $+2\vec\Omega\times\vec v_{rot}$ come about. For simplicity, let's assume that the body is moving uniformly in the rotating frame along the $x$ axis, meaning $\vec v_{rot} = v_0 \hat i$. Also, let's assume that the rotating frame is rotating uniformly about the $z$ axis : $\vec \Omega= \Omega_0 \hat k$. In that case, the coriolis acceleration would be $= 2\Omega_0 v_0\hat j$.

The acceleration is in the $y-z$ plane, along the $y$ axis. So is the angular velocity vector $\vec\Omega$, but along the $z$ axis. How is the acceleration in a plane perpendicular to the rotation velocity?

The discourse about seeing the opposite of the Coriolis force as the effective acceleration is existing but unfortunately wrong.

The Newtonian acceleration a=F/m derived from Newtonian force F is either disconnected from the rotating frame or connected to the rotation frame by the inertial oscillation process. In both cases, the opposite of the Coriolis force is useless for a physical understanding.

In the latter case, very important in physics, acceleration is centripetal and the Newtonian force as well but trajectory is elliptic or epicyclic. It is an analogue to a keplerian otbit. The difference between the circle and the ellipse is the relative trajectory. Despite a radial-only work of the Newtonian force you can obtain a full oscillation impacting both radial and tangential speeds, shown as a stationary "inertia circle" in the rotating frame. The link between the direction of the Newtonian centripetal force in the inertial frame and the Coriolis force in the rotating frame is very complex because it depends on the timing of the oscillation.

 

[A.T.]

The discourse about seeing the opposite of the Coriolis force

Where did the OP write anything about "seeing the opposite of the Coriolis force"?

 

[kuruman]

The Newtonian acceleration a=F/m derived from Newtonian force F is either disconnected from the rotating frame or connected to the rotation frame by the inertial oscillation process.

What oscillation process?

 

[Simon F]

The above reference introduced: "the Coriolis force is the fictitious force required to balance the real force that provides the Coriolis acceleration". This discourse resembles some of weird Anders Persson's statements (among many other great statements) being a poor attempt to magnify the positive Coriolis acceleration, i.e. the opposite of the Coriolis force.
Without the inertial oscillation process, both positive and negative Coriolis terms have no physical meaning, no real force provides the positive Coriolis acceleration. The best example is to show that a case without any Newtonian force is conducive to both positive and negative Coriolis acceleration.
The inertial oscillation in coherent rotating systems is the key to understand why the negative Coriolis term get a physical meaning in many physical studies. In the meantime, the positive Coriolis term is always deprived of a physical meaning.

 

[renormalize]

(Something or other deleted.)

Sorry to mix metaphors, but this reads like a word salad that's clear as mud.