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There is something I don't quite undestand about the role the centrifugal term plays when describing motion in a non-inertial reference frame. In most Classical Mechanics books, you can find similar discussions on non-inertial reference frames. The "effective acceleration" felt by a body in one of those reference frames would be:

$$\vec{a}_{ef} = \vec{a}_{real} - \ddot{\vec{R}} - \dot{\vec\omega}\times\vec{a}-\vec\omega \times (\vec\omega \times \vec r) - 2\vec\omega\times v_{r}$$

Where $\vec{a}_{real}$ are the real forces acting upon the body, like grativational forces, and the rest of the terms correspond to fictitious forces. My doubt is the following:

For a mass moving in circles, like a child in a merry-go-round, there has to be a centripetal force that we have to account for. It is a real force, after all; making it move in circles. But, in that case, wouldn't the centripetal term and the centrifugal term cancel each other out every time we have a situation like this?

I think I am confused about something and would be thankful if someone could clarify this for me.

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    $\begingroup$ When studying physics, there is a rule-of-thumb which usually works: Don't trust the words, trust the math. So, which one of your terms denotes what? For constant motion, i.e. $\dot{\vec{\omega}}$ you'll only find the centrifugal and Coriolis forces, which are expressed through the last two terms in your equation. The centripetal force is an invention for high-school kids and does not correspond to any term in any equation. $\endgroup$ Jan 10, 2020 at 2:39
  • $\begingroup$ @AtmosphericPrisonEscape The centripetal acceleration in this case will be in the $a_\text{real}$ term. $\endgroup$ Jan 10, 2020 at 2:51
  • $\begingroup$ @AaronStevens: $a_{\rm real}$ will contain all accelerations due to fundamental forces. But a centripetal one doesn't exist. I challenge you to prove otherwise by writing it down. $\endgroup$ Jan 10, 2020 at 3:01
  • $\begingroup$ @AtmosphericPrisonEscape An object undergoing uniform circular motiion undergoes centripetal acceleration. This can come from a variety of interactions, but that doesn't make it fake. $\endgroup$ Jan 10, 2020 at 4:24
  • $\begingroup$ $\dfrac{d\omega }{dt}\times a\rightarrow \dfrac{d\omega }{dt}\times r$ $\endgroup$
    – Eli
    Jan 10, 2020 at 9:07

3 Answers 3

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For a mass moving in circles, like a child in a merry-go-round, there has to be a centripetal force that we (I suppose) have to write. It is a real force, after all, making it move in circles. But, in that case, wouldn't the centripetal term and the centrifugal term cancel each other out every time we have a situation like this?

If you are in a frame rotating with the child, then yes, you would want the centripetal force to cancel out the centrifugal force. In your rotating frame you see a child at rest, and this child has a net force of $0$ acting on them. Therefore, you have $F=ma\to0=0$, and everything is consistent.

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  • $\begingroup$ No! Newton's first law only says there is no net force on the child at rest in an inertial frame. The child feels a real force acting on him/her while moving in a circle, and it is the same real force whatever reference system you choose to analyse the problem. Physics doesn't know or care about what coordinate systems you use. $\endgroup$
    – alephzero
    Jan 9, 2020 at 19:48
  • $\begingroup$ @alephzero Fictitious forces allow Newton's second law to be valid in non-inertial frames (while then the 3rd law becomes invalid). In the rotating frame if you choose to include your centrifugal force then you correctly observe no acceleration. $\endgroup$ Jan 9, 2020 at 20:53
  • $\begingroup$ @AaronStevens correct me if I'm wrong: The 3rd law doesn't become invalid, but it is actively used in a non legitimate context, in order to get to correct answers in what then looks like an inertial frame. So there is no wonder it does not hold there anymore. ? $\endgroup$
    – user192234
    Jan 10, 2020 at 3:04
  • $\begingroup$ @user192234 The fictitious forces have no "equal but opposite reactions", so they do not follow Newton's third law. Another way to think about it: fictitious forces do not come from interactions with something else also experiencing fictitious forces. $\endgroup$ Jan 10, 2020 at 4:27
  • $\begingroup$ @AaronStevens are these forces or are these accelerations? Do they depend on the mass of what is accelerated? They seem to be forces because without applying forces to deal with them we would not survive merry go rounds :). The reaction force to, for example, resisting the acceleration when checking out the Coriolis effect of oscillating your arm, is applied by the alarmed system that ultimately wants to keep your head upright and steady (Avoiding the third law, so to speak). There are three elements in sequence. two of them are action and reaction by the child nervous system alone. $\endgroup$
    – user192234
    Jan 10, 2020 at 20:28
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You put considerable effort on merry go rounds (Even when seated) in order to subconsciously show there is no such thing/s. Or you are just wired up to face the rotating system you are in and your brain is taking charge. and $a=0$. You can't fool the Coriolis part is a lot more sneaky. Just look at that term with the angular velcoity vector crossed with linear velocity. Move your hand back and forth, facing the center, perpendicular to the axis of rotation. Then we'll see some $ma$. :)

When you are moving in a planar circle then by construction there is an acceleration vector perpendicular to the linear velocity at each instant, and it keeps rotating the velocity vector while keeping it fixed in size. Otherwise, you would not be moving in a circle.

If you are part of the rotating system then mathematically:

  1. You modify with an acceleration vector in the other direction. That's what is happening.

Now if you take your make believe coordinates and formally differentiate twice, nothing happens. Clearly not what you should expect.

On paper it is often much cleaner and easier to restrict yourself to one particular non-inertial system and get a correct answer. In reality you would trip and fall most probably if you walk around as if it is not a restricted non-inertial system you are in.

But, if you recall Newton's postulates are meaningful for your prior coordinates, then you do the correct thing and substitute back to inertial coordinates into the expression, and only then do you satisfy Newtonian postulates. What you have written is a nicely packed vector expression of correct Newtonian physics in an inertial frame. The accelerations are all real. You apply force in order to nullify them.

Again try to toy around really with the Coriolis expression. It's real. Like bold letters! And like proper books on Newtonian Mechanics.

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  • $\begingroup$ Non inertial coordinates are well understood and perfectly legitimate. You are free to use whatever coordinates you like to simplify a given scenario $\endgroup$
    – Dale
    Jan 10, 2020 at 2:43
  • $\begingroup$ But it is not oh so simple a scenario. You have to work with these expressions, work with legit coordinate system, if you want to have the slightest chance of accomplishing anything on the ground from miles high and mach speed, for instance. $\endgroup$
    – user192234
    Jan 10, 2020 at 2:53
  • $\begingroup$ If down voters spell out a reason I'll be much obliged! $\endgroup$
    – user192234
    Jan 10, 2020 at 3:06
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The confusion is not realizing that real forces are real, and fictitious forces are not real. Therefore, they can not "cancel each other out" in real life.

Using a rotating frame of reference and introducing fictitious forces is just adding and equal quantities (a force, and a mass $\times$ acceleration) to both sides of the math equations. It has nothing to do with the real forces acting on the system. It is no different from solving $x + 3 = 7$ by saying $x + 3 - 3 = 7 - 3$, and therefore $x + 0 = 7 - 3$ and $x = 4$.

The physics of the situation (i.e. the interaction between the various parts of the system which we call "force", "stress", etc) does not depend on what coordinate system you choose to model it. You get the same values of the physical quantities in any coordinate system.

The only reason for choosing one coordinate system or a different one is because the math is easier in one than in the other, not because it changes the physics.

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    $\begingroup$ Though if you do choose to work in the rotating frame and use fictitious forces, you will find that they do indeed cancel out. The OP doesn't seem to be confused about which forces are real, the use of different frames, etc. The title specifies work is being done in the non-inertial frame. $\endgroup$ Jan 9, 2020 at 21:23
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    $\begingroup$ Fictitious forces certainly can and do cancel out real forces in the rotating reference frame. They can add to real force and cause acceleration and even do work in non inertial frames where energy is conserved. $\endgroup$
    – Dale
    Jan 10, 2020 at 1:43
  • $\begingroup$ The rotating coordinate system is not Newton legit. That's the whole grand idea here. $\endgroup$
    – user192234
    Jan 10, 2020 at 2:40

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