Why Does a Glass Move Outward When Opening the Fridge Door?

[A.T.]

I dont understand how your glass is right from original position,.Only when you stop a door, and glass continue along shelf and water splash at you,

That is what I meant. It is possible if you stop the door or move too slowly, after an initial jerk.

(often happen to me, god damn centrifugal!)

Nothing to do with centrifugal force. The acceleration to the right comes from the normal force by the rotated door.

 

[user079622]

Nothing to do with centrifugal force. The acceleration to the right comes from the normal force by the rotated door.

How would you explain this to regular people who dont even know what is normal force?

 

[A.T.]

How would you explain this to regular people who dont even know what is normal force?

https://en.wikipedia.org/wiki/Normal_force

 

[kuruman]

Here is my two bits of wisdom while summarizing much of what has already been said.
First the theory. For an object following a curved trajectory the radial component of the acceleration is $$a_r=\ddot r-r\dot{\theta}^2.$$Here we assume that (a) there is no force in the radial direction and (b) $\dot{\theta}=\rm{const.}=\omega.$ Then the radial acceleration $a_r$ is zero which means that $$\frac{d^2r}{dt^2}=\omega^2r.$$One might interpret this as saying that the rate of change of the radial velocity is equal to the centrifugal acceleration. Regardless of interpretation, the solution of this equation is $$r(t)=r_0\cosh(\omega t)\implies r(\theta)=r_0\cosh(\theta).$$ Shown below is a parametric plot of $y=r(\theta)\sin(\theta)$ vs. $x=r(\theta)\cos(\theta)$ in equal 10° increments of $\theta.$ The hinge of the door is at the origin and the door opens counterclockwise towards the y-axis. The length units are arbitrary but have an aspect ratio of 1 on the screen.

Note that from 0° to about 30° there is practically no displacement in the x-direction which is the initial orientation of the door. This is because there is practically no force in the x-direction until the normal force rotates enough to provide a significant component in the x-direction through the normal force. A Taylor series expansion for the horizontal position of the mass yields $$x=r_0\cosh(\theta)\cos(\theta)=r_0\left[1-\frac{\theta^4}{6}+\mathcal{O}(\theta^6)\right].$$ This small angle approximation is what I attempted to illustrate in post #13 albeit more crudely. Finally, it should be clear that the shape of this curve is independent of the constant $\omega$.

 

[Vanadium 50]

Show me the refrigerator door that opens 270 degrees or more.

You mean after the first time?

 

[PeterO]

If centrifugal force dont exist why glass of water moves outward/toward me when I open fridge door?
Glass of water moves out in relation to door and in relation to earth, so what force push glass out?

You are asking the wrong question!!
When you open a fridge drawer, each part of the door moves in a circular path (due to the combined effect of the hinges and your action on the handle)
Why doesn't a glass of water on a shelf also follow the circular path that will make it retain its position on that shelf?
ie:
what force, including direction, does the friction force between the shelf and the glass have to be if the glass is not to slide along the shelf?
In the absence of that force, in which direction, along the shelf, will the glass slide.

 

[Drakkith]

The acceleration to the right comes from the normal force by the rotated door.

Normal force? Shouldn't that be the frictional force of the door on the container?

 

[A.T.]

Normal force? Shouldn't that be the frictional force of the door on the container?

No, I'm assuming no friction so the glass is free to move along the door. The normal forces, which keep the glass within the fridge door shelf, can act form either side (railing and door itself). Once the door is rotated, and the normal force comes force from the door (because it opens too slowly for the already moving glass), it has a component to the right (in the diagram of post #15).

 

[Drakkith]

My mistake. I was confused about what the glass was doing.

 

[user079622]

Once the door is rotated, and the normal force comes force from the door (because it opens too slowly for the already moving glass), it has a component to the right (in the diagram of post #15).

Normal force dont have component to the right when door opens constantly or accelerate(increase door rpm).
It has component to the right only when door decelerate(rpm slows down)

Why doesn't a glass of water on a shelf also follow the circular path that will make it retain its position on that shelf?
ie:

Because of inerita, glass resist to change position in space, centripetal force(friciton between plastic and glass bottom/back) is not enough,so glass slide.

 

[PeterO]

Normal force dont have component to the right when door opens constantly or accelerate(increase door rpm).
It has component to the right only when door decelerate(rpm slows down)Because of inerita, glass resist to change position in space, centripetal force(friciton between plastic and glass bottom/back) is not enough,so glass slide.

Well done, you have successfully explained why the glass slides without reference to, nor need for a centrifugal force - so the original question which began "If centrifugal Force doesn't exist ..." becomes redundant.

 

[A.T.]

Normal force dont have component to the right when door opens constantly or accelerate(increase door rpm). It has component to the right only when door decelerate(rpm slows down)

Yes, but it's not easy to open it at constant RPM in practice, and you cannot accelerate it forever.

 

[vanhees71]

Normal force dont have component to the right when door opens constantly or accelerate(increase door rpm).
It has component to the right only when door decelerate(rpm slows down)Because of inerita, glass resist to change position in space, centripetal force(friciton between plastic and glass bottom/back) is not enough,so glass slide.

I don't know, why we still discuss this. There are at least two postings, solving the (idealized) problem. Just once again:

We assume a mass that can glide without friction along a rod, which is rotating with constant angular velocity in a horizontal plane. There's no net external force, i.e., the Lagrangian is
$$L=T=\frac{m}{2} \dot{\vec{r}}^2.$$
The position vector can be described as
$$\vec{r}(t)=r(t) \begin{pmatrix} \cos (\omega t) \\ \sin(\omega t) \\ 0 \end{pmatrix}.$$
The derivative is
$$\dot{\vec{r}} = \dot{r} \begin{pmatrix} \cos (\omega t) \\ \sin(\omega t) \\ 0 \end{pmatrix} + r \omega \begin{pmatrix}-\sin(\omega t) \\ cos(\omega t) \\0 \end{pmatrix}.$$
Thus
$$L=\frac{m}{2}(\dot{r}^2+r^2 \omega^2).$$
The equation of motion is
$$\mathrm{d}_t \frac{\partial L}{\partial \dot{r}}=m \ddot{r}=\frac{\partial L}{\partial r}=m r \omega^2.$$
The solution is
$$r(t)=r_0 \cosh(\omega t) + \frac{v_0}{\omega} \sinh(\omega t).$$
The force on the mass is
$$\vec{F}=m \ddot{\vec{r}}=m (\ddot{r}-\omega^2 r) \begin{pmatrix} \cos (\omega t) \\ \sin(\omega t) \\ 0 \end{pmatrix} + 2m \omega \dot{r} \begin{pmatrix}-\sin(\omega t) \\ cos(\omega t) \\0 \end{pmatrix} = 2 m \omega^2 [A \sinh(\omega t) + B \cosh(\omega t)]\begin{pmatrix}-\sin(\omega t) \\ \cos(\omega t) \\0 \end{pmatrix}.$$
This force is the force imposed by the rod to keep the point on the rod.

 

[weirdoguy]

I don't know, why we still discuss this

Because OP does not know Lagrangian mechanics? I think it would be clear if you read other posts...

 

[vanhees71]

Then it gets complicated, because you have to calculate the said reaction force in a somehow different way. The only alternative is to use d'Alembert's principle, which however is more complicated than Lagrange :-(.

An alternative is to transform to the co-rotating frame. Then the motion is described as one-dimensional motion along the $x$-axis, but you get the additional inertial forces. From this point of view there's only the centrifugal force. The Coriolis force is compensated by the constraint force from the rod (the force, calculated above in the inertial frame). This of course gives the same equation for $x$ as for $r$ in the Lagrange formalism.

 

[weirdoguy]

Well, since others didn't have to use d'Alembert's principle to discuss OP, I guess there are other ways.

 

[erobz]

Newtons laws work just fine, it's just a longer derivation.

 

[erobz]

$$m \ddot x = -N \sin( \omega t) \tag{1} $$
$$ m \ddot y = N \cos( \omega t ) \tag{1} $$

Then with $x = r \cos( \omega t ), y = r \sin( \omega t)$ ( contstant $\omega$ )

$$ \ddot x = \ddot r \cos( \omega t ) - 2 \dot r \omega \sin( \omega t) - r \omega^2 \cos( \omega t ) $$

$$ \ddot y = \ddot r \sin( \omega t ) + 2 \dot r \omega \cos( \omega t) - r \omega^2 \sin( \omega t ) $$

Eliminating $N$ between (1) and (2), canceling mass:

$$ \cos( \omega t) \ddot x = - \sin( \omega t) \ddot y $$

After you rearrange (applying trig identity) you are left with:

$$\ddot r - \omega^2 r = 0 $$

ect...

 

[vanhees71]

Well, since others didn't have to use d'Alembert's principle to discuss OP, I guess there are other ways.

Where has the concrete equation of motion been described in this thread? I must admit, I've no clue, how else than with at least using D'Alembert's principle you could derive it. After all it's a problem with a holonomic, rheonomic constraint.

 

[ALBAR]

Centrifugal force is absolutely a very real inertial force in an inertial reference frame. The old argument that this radial force does not exist uses the example of an object whirling around on a string. Cut the string and the object zooms away along a tangential path. Well, cutting the string eliminates the centripetal and centrifugal forces, so of course the object moves off straight forward.

 

[A.T.]

Centrifugal force is absolutely a very real inertial force in an inertial reference frame.

In this context, the term "real" usually refers to interaction forces that obey Newton's 3rd Law, which inertial forces do not. Terms like "real" & "fictitious" just lead to pointless philosophical debates, when that merely technical definition is taken out of context. It's better to to use "interaction" & "inertial" instead here.

ETA: And as @Orodruin points out below, inertial forces exist only in non-inertial frames.

 

[Orodruin]

Centrifugal force is absolutely a very real inertial force in an inertial reference frame.

The centrifugal force is zero in an inertial frame by definition. So no.

 

[jbriggs444]

Cut the string and the object zooms away along a tangential path. Well, cutting the string eliminates the centripetal and centrifugal forces, so of course the object moves off straight forward.

If centripetal and centrifugal forces are equal and opposite and if they both vanish when you cut the string then the object should have been moving in a straight line even before you cut the string.

Indeed, it did have constant velocity in the rotating frame.
However, in the inertial frame, it did not.

 

[ALBAR]

I don't understand why people don't stay in the essentially inertial laboratory reference frame when pondering basic mechanics. I agree that if centrifugal force were applied externally as the centripetal is, the object would go straight (with perhaps some lateral stretching). The centripetal force acts normal to the velocity while the object's matter does what matter always does: it resists any change of its momentum by generating inertial force. Neither force can exist without the other. All forces must have counter force. If you don't believe Newton, try pushing or pulling on nothing. The roles of these forces are cause and effect. Centripetal causes acceleration, simultaneously triggering production of inertial force by Newton's second and third laws. That force is what I boldly tag the word "centrifugal" onto.

 

[jbriggs444]

I don't understand why people don't stay in the essentially inertial laboratory reference frame when pondering basic mechanics. I agree that if centrifugal force were applied externally as the centripetal is, the object would go straight (with perhaps some lateral stretching). The centripetal force acts normal to the velocity while the object's matter does what matter always does: it resists any change of its momentum by generating inertial force. Neither force can exist without the other. All forces must have counter force. If you don't believe Newton, try pushing or pulling on nothing. The roles of these forces are cause and effect. Centripetal causes acceleration, simultaneously triggering production of inertial force by Newton's second and third laws. That force is what I boldly tag the word "centrifugal" onto.

That is the "reactive centrifugal force". Yes, it is a real force. But it is not what is meant by "centrifugal force". The "centrifugal force" arises only in the non-inertial rotating frame.

As you point out, the centrifugal force in the non-inertial sense does not have a corresponding third law reaction force.

 

[Ibix]

The centripetal force acts normal to the velocity

Right.

the object's matter does what matter always does: it resists any change of its momentum by generating inertial force.

That's not what the phrase "inertial force" means, and I don't think there's any concept similar to what you are describing here.

All forces must have counter force.

I assume you're intending to cite Newton's third law here. If so, I don't think it means what you think it means. The centripetal force on the bottle has a third law pair often called the "centrifugal reaction force" which acts on the fridge door. This is not an inertial force and is not the same as the centrifugal force, which is an inertial force and is therefore zero in the inertial frame.

 

[Orodruin]

I don't understand why people don't stay in the essentially inertial laboratory reference frame when pondering basic mechanics. I agree that if centrifugal force were applied externally as the centripetal is, the object would go straight (with perhaps some lateral stretching). The centripetal force acts normal to the velocity while the object's matter does what matter always does: it resists any change of its momentum by generating inertial force. Neither force can exist without the other. All forces must have counter force. If you don't believe Newton, try pushing or pulling on nothing. The roles of these forces are cause and effect. Centripetal causes acceleration, simultaneously triggering production of inertial force by Newton's second and third laws. That force is what I boldly tag the word "centrifugal" onto.

As others have already said, you are confusing "centrifugal force" with "reactive centrifugal force". The actual centrifugal force is defined as the inertial force due to angular velocity and distance from the rotational center in a rotating frame. This force is zero in any inertial frame. You can attach words as you wish, but unless you use standard terminology, you will have a hard time to make yourself understood.

Reactive centrifugal force is not what people generally mean when they say "centrifugal force" as per the above. It is just the third law pair of whatever the centripetal force is and it does not act on the object itself - it is a force from the object on whatever is keeping it in rotational motion.

inertial force by Newton's second and third laws

That's not what an inertial force is. An inertial force is acting on the object itself and has no third-law partner. It is merely due to the effects of using a non-inertial reference frame and is therefore always proportional to the mass of the object.

 

[A.T.]

All forces must have counter force.

Only interaction forces obey Newton's 3 Law. Inertial forces do not, thus momentum is not conserved in non-inertial reference frames.

The roles of these forces are cause and effect.

There is nothing in Newton's 3 Law that allows you to tell which is cause and which effect, so those roles are arbitrary and irrelevant.

production of inertial force by Newton's second and third laws.

Inertial forces are never part of a Newton's 3 Law force pair.

That force is what I boldly tag the word "centrifugal" onto.

You can do that, but by itself it is not sufficient and ambiguous, because there are different types of "centrifugal" froce:

Inertial centrifugal force (exists only in rotating frames, not part of any Newton 3rd force pair):
https://en.wikipedia.org/wiki/Centrifugal_force

Reactive centrifugal force (exists in every frame, forms Newton 3rd pair with centripetal force):
https://en.wikipedia.org/wiki/Reactive_centrifugal_force

See this diagram:

 

[sophiecentaur]

Reactive centrifugal force is not what people generally mean when they say "centrifugal force"

But it's what they feel on fairground rides etc.. That experience trumps the intellectual appreciation of formal Physics. The "doesn't exist" statement that we all got first at school is responsible for a lot of later confusion. Using the term 'frame of reference' is accurate but clouds the whole thing in yet more mystery.

I really sympathise with 'beginners' who read through this thread because they are very likely to leave it none the wiser. It can be a sledge hammer to crack a wallnut.

My friendly Nuthatch has just come onto the feeder outside my window. That has eased my frustration.

 

[A.T.]

Reactive centrifugal force is not what people generally mean when they say "centrifugal force"

But it's what they feel on fairground rides etc.

No, what the rider "feels" is the centripetal force by the ride. The reactive centrifugal force is what the ride "feels".

 

[Orodruin]

But it's what they feel on fairground rides etc

No it isn’t - at least not if you consider them as a whole. That would be the centripetal force. The reactive centrifugal force does not act on them, it is the force they exert on the ride.

If you start making fbds of body parts, then yes, the centripetal force on the head will be the third law pair of the head’s reactive centrifugal force on the neck.

 

[Ibix]

But it's what they feel on fairground rides etc.

Not in general - the reactive centrifugal force is what their body applies to the outer wall of the ride. Unless somebody on the inwards side of you slides into you, you don't feel the reactive centrifugal force.

The only thing you feel, in the strict sense of feel ("something is activating my touch receptors"), is the centripetal force from the floor or walls. In the broader sense of feel ("intuit the presence of"), you feel the centrifugal force (the inertial one, not the reactive one) as an outward g-force in your own rotating frame.

 

[sophiecentaur]

Not in general - the reactive centrifugal force is what their body applies to the outer wall of the ride

It's so easy, when you know the right answer, to assume that the uninitiated interpret things according to accepted Physics. If you ask the average man on the Clapham omnibus what he is 'feeling' he will say that he feels the force of gravity pushing him into his seat. This is the level at which answers to this sort of question should start off - that is if we actually want people to grasp what's what. You can put people right if you start in the right place.
Explaining to me what happens is not helpful to him because you use terms that both of us understand but not he.

 

[Orodruin]

It's so easy, when you know the right answer, to assume that the uninitiated interpret things according to accepted Physics. If you ask the average man on the Clapham omnibus what he is 'feeling' he will say that he feels the force of gravity pushing him into his seat. This is the level at which answers to this sort of question should start off - that is if we actually want people to grasp what's what. You can put people right if you start in the right place.
Explaining to me what happens is not helpful to him because you use terms that both of us understand but not he.

Perpetuating that misconception by saying “indeed, gravity pushes you into your seat” also will not do anything to remedy the understanding.

Furthermore, it is not the equivalent. The equivalent is that centrifugal force pushes you outward, not reactive centrifugal force.

 

[A.T.]

If you ask the average man on the Clapham omnibus what he is 'feeling' he will say that he feels the force of gravity pushing him into his seat.

Nobody in this thread was talking about "feeling" before you started to talk about it. First you claimed that people "feel" the reactive centrifugal force, that doesn't even act on them. Now you switched to "what people believe they feel" and also switched to gravity acting on them.

you use terms that both of us understand

I have my doubts.