If centrifugal force is a pseudo-force then why we can observe its effects outside of rotating frame?

Question

I know that centrifugal force only "exists" when the reference frame is rotating. In this case, if there is centripetal force, there must also be centrifugal force so that bodies on the rotating frame will be motionless.

But if centrifugal force is a pseudo-force then why we can observe its effects outside of rotating frame? I mean if we start to spin a table with objects on it very fast, we can see the objects flung to different directions. Hence there must be acceleration outwards and we can observe it when our reference frame is outside of rotation. The centrifuge device is another example.

Answer 1

When objects fly off a spinning table, from an external reference frame they simply stop travelling in circles. At the point of "unsticking" they do not accelerate radially outwards from the centre of rotation but merely stop accelerating inwards; they continue in a straight line tangential to the circle and at right angles to the radial direction. Thus, no centrifugal force is present - indeed, it is the very absence of any force subsequent to breakaway which explains their behaviour.

Answer 2

When you look at the rotating table or the centrifuge from outside you are not seeing the centrifugal force, but only Newton's first law, that inertial objects continue to move in a straight line.

Answer 3

Suppose that using an inertial frame of reference you observe a body undergoing a centripetal acceleration because of a force which is acting on it.
Using the inertial frame you can apply Newton's laws of motion to predict what is happening to the body and that force acting on the body has a Newton third law twin (action/reaction).

Now choose a frame of reference which is not moving relative to the body which is undergoing a centripetal acceleration when observed in the inertial frame.
What do you observe?
The body is not accelerating (moving) relative to the non-inertial frame and yet it has a force acting on it, the force that caused the centripetal acceleration in the inertial frame.
This immediately means that Newton's second law does not work $F\ne m\,0$.
So that Newton's laws can be used in the non-inertial frame a fictitious/pseudo force is added so that the net force on the body is zero and now Newton's second law does predict the correct acceleration of the bod, zero, in the non-inertial frame $F-F_{\rm pseudo} = m\,0$.
This fictitious force does not have a Newton third law twin but enables Newton's second law to be used in the non-inertial frame.

In a sense you have looked at the situation the wrong way round.
The pseudo force is a way of making Newton's law work in a non-inertial frame but there is a real force which is there in both the inertial and non-inertial frame which causes the centrpetal acceleration of the body when observed in the inertial frame.

Answer 4

The situation you describe is the following : a mass $m$ in rotational motion on the edge of a turntable of radius $R$. The mass has no other forces acting on it, and is held at rest relative to the turning table by static friction $F_{static}$. The assumption of the mass being on the edge is to simplify the analysis (more on this later)

In the "lab" frame, we observe the rotational motion, say at angular velocity $ω$. The fact that a mass $m$ is in rotational motion in a circle of radius $R$ with angular velocity $ω$ implies that there must be an radially inward force acting on the mass of magnitude $mω^2R$. This is a simple geometrical fact. This inward force is called the "centripetal force". The only source of this force is the static friction, hence $F_{static} = mω^2R$. It goes without saying that this is a "real" or "inertial" force.

Now, let us analyze the same situation in the turntable frame. In this frame, there is an inward "inertial force" of $F_{static} = mω^2R$, and the mass is at rest. So, there must be an equal and opposite pseudo force, which we call the "centrifugal force".

As we can see, higher the $\omega$, higher the $F_{static}$ required to hold the mass at rest relative to the turntable. Note that between any two materials, there is a maximum possible $F_{static}$ ; $F_{static} \leqslant \mu_s mg$, where $\mu_s$ is the "static coefficient of friction". If we imagine steadily increasing the angular velocity, the mass will rotate with the turntable up to $\omega \leqslant \sqrt{\frac{\mu_s g}{R}}$.

But now, imagine increasing $\omega$ to $ \sqrt{\frac{\mu_s g}{R}} + \epsilon$, where $\epsilon \to 0^{+}$. The mass will now come unstuck, and will experience $kinetic$ friction $F_{kinetic} = \mu_k mg$. Note that $\mu_k$ is understood to be less than $\mu_s$. In order to continue moving in circular motion with angular velocity $\omega$, the inward force required is $\mu_s mg$, however, only $\mu_k mg$ is available. Hence the mass "flies off". In our case, when the mass is at the edge of the turntable, it flies off tangentially. In case the mass was inside (not at the edge), the analysis is a little more involved; its trajectory will resemble a spiral of increasing radius from the point at which $\omega$ exceeded $\sqrt{\frac{\mu_s g}{R}}$ (spiral, not straight line, because there is still a frictional force acting on the mass) till the spiral hits the edge, at which point, the mass just continues in a straight line, and "flies off".

Note that the only forces that are "real" (with reference to the equivalence class of inertial frames) are the ones measured in an inertial frame. The "fictitious" force is a result of analyzing this motion in an accelerating frame. But this may just be a matter of convention. After all, we can think of an equivalence class of frames rotating with $\omega$ just as well as we can think of an equivalence class of inertial frames. However, I still want to argue that it is more appropriate to say that he mass flies off due to there not being enough centripetal force, rather than to attribute the flying off to the effects of centrifugal force. Let me explain why.

The reason the mass flies off is (superficially) different depending on whether we answer this question in the inertial "lab" frame, or the rotating frame. In the lab frame, there is no longer enough inward (centripetal) force to force the mass around a circle of radius $R$ at angular velocity $\omega$. In the rotating frame, there is no longer enough inward (centripetal) force to keep the mass at rest, given that there is a (outward, centrifugal) fictitious force $\mu_s mg$ also acting.

You may call it a matter of taste or convention, but if I had to sum up the above in a catchy phrase "the mass flies off because of there is no longer enough centripetal force" seems a more apt conclusion from both the above statements (if for no other reason, just the fact that this phrase repeats in the reasons given by both frames for why the mass flies off) than "the mass flies off because of the effects of centrifugal force". I rest my case.

Answer 5

It depends on what we call pseudo. Lets name it other way. I suggest you to use the word coordinate force. Here is the thing observed by Newton - his second law $F=ma$ only worked when he was using inertial frame. And in the same way, it will be the definition of inertiality. $F=ma \Leftrightarrow \mathsf{inert},$ which certainly looks like self-dependency and there is no real "easy way" to cut into this issue on school-tier of knowledge.

In higher tier of mechanics you have Lagrangian approach $A=\int L(q,q',t) dt = \mathsf{max}$ leading to Newton laws of Euler form $$ \frac{\partial L}{\partial q} -\frac{d}{dt}\frac{\partial L}{\partial q'} = 0, $$ where $q'$ is the velocity (mathematicians would call it "tangent space" to our "normal coordinate space"). And it turns out that rotational symmetry of lagrangian $L$ is the rotational momentum conservation law. It is shown in textbooks everywhere. And from it follows that only symmetric coordinate systems (in turn to symmetric lagrangians) will be able to be consistent (i.e. validating) with corresponding newton-type equation for system.

For your system it is obvious you have non-symmetry because you have chosen to stick with particular axis or rotation. Any point will be accelerated with rotational additive $a_1=a_0+R\omega^2$. It is actually corresponding with additional force on opposite side of your equation $F_1=F_0+F_{circ}=m(a_0+R\omega^2),$ in which you can see that your system breaks up back into "core system" plus circular additionale.

As for about action of coordinated forces, it is absolutely real. For example, centrifugal force will really change the shape of water. It is not "pseudo force", it is real. Because your basic Newton's law is not working in non-symmetric system.

Is that what you are asking for?

Answer 6

Wikipedia explains:

In the inertial frame of reference (upper part of the picture), the black ball moves in a straight line. However, the observer (brown dot) who is standing in the rotating/non-inertial frame of reference (lower part of the picture) sees the object as following a curved path due to the Coriolis and centrifugal forces present in this frame.

Image by Hubi