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

[erobz]

I dont agree, to me is more intuitive to think how centrifugal force push bottle along door, then think about why bottle goes out in inertial frame...
In inertial frame at first it seems that something is missing..

Indeed for most people, otherwise people will not ask these questions, I am not first one who ask this question.
In general, I think non inertial frame is more intuitive for human brain.

Confusion start with prejudice that the motion of a body should be along the direction of the total force exerted on the body in all cases.

Yeah, I think its intuitive in a non-inertial frame because that's the frame we tend to experience while accelerating in cars and spinning in amusement park rides, etc..

I'm not sure if the math is generally easier in one or another.

 

[Dale]

I'm not sure if the math is generally easier in one or another.

I think there is no general rule for this. Some problems will be better in an inertial frame and others will be easier in a non-inertial frame.

 

[Ibix]

I think there is no general rule for this. Some problems will be better in an inertial frame and others will be easier in a non-inertial frame.

There's a story in one of Feynman's autobiographies about designing a gun director; apparently it's a fairly straightforward task in polar coordinates when the gunsight is next to the gun. Then the lieutenant in charge of the project asked what if the sight is not next to the gun. That had never occurred to his team. Suddenly the problem was almost insoluble in polar coordinates and the machine had to be redesigned from the ground up using Cartesian coordinates.

Horses for courses, as they say.

 

[sophiecentaur]

Ah, so you want to know why the glass slides along the shelf?

That's still the contact force(s) between the glass and door. The direction "outwards" changes as you open the door. The initial force is outwards, relative to the half open door. Or, at least, has an outwards component.

That can be tricky to visualise until you look at the forces in polar coordinates.

This is definitely a partial help with a problem which pretty well all people have, initially. The cogniscenti can be very smug when using a term like frame of reference and expect the confused to work with that alone. The 'fact is' that things get thrown outwards (increasing distance from the centre of rotation) and a force has caused it. Your "outwards component" is a good way into getting it right in one's head.

 

[Orodruin]

I think there is no general rule for this. Some problems will be better in an inertial frame and others will be easier in a non-inertial frame.

If this was not the case then nobody would teach or study non-inertial frames. Ever.

 

[PeroK]

Confusion start with prejudice that the motion of a body should be along the direction of the total force exerted on the body in all cases.

That prejudice is generally called Newton's second law of motion. Newton's great insight was to see beyond the intuitive prejudices about motion and establish the fundamental laws of motion.

Understanding that it's a centripetal force that keeps an object moving in a circle and that there is not a real force centrifugal force trying to push everything outward.

I thought the whole purpose of this thread was to counter your argument that centrifugal force is real force. The scenario with the fridge was described from the inertial frame of your kitchen. Not from the non-inertial frame of the rotating door. Your whole argument, I assumed, was that centrifugal force was a real force in that inertial frame. If not, then I have no idea what we've been arguing about in these 56 posts.

 

[PeroK]

PS note that in the non-inertial frame of the opening fridge door, there are equally "fictitious" forces on your kitchen and everything in it, causing them to accelerate and rotate, relative to the door. It's not just the glass sliding along the shelf.

 

[A.T.]

right case is correct.

It depends on how you vary the angular velocity of the door.

In the left case, after an initial impulse the door rotates such that it doesn't exert any force on the glass.

In the right case (not exactly that path curvature), after an initial impulse the door rotates too slow, so the force from the door flips direction, and since it is already rotated, this normal force has a component to the right.

 

[sbrothy]

Is it just me or is this is an awfully long discussion for a very simple observation?

 

[vanhees71]

I dont agree, to me is more intuitive to think how centrifugal force push bottle along door, then think about why bottle goes out in inertial frame...
In inertial frame at first it seems that something is missing..

Indeed for most people, otherwise people will not ask these questions, I am not first one who ask this question.
In general, I think non inertial frame is more intuitive for human brain.

Confusion start with prejudice that the motion of a body should be along the direction of the total force exerted on the body in all cases.

Well, yes, that's just a very personal thing about what you find most intuitive. For me the real relieve was when I learnt Lagrangian mechanics rather than having to think about forces ;-)).

 

[user079622]

In the right case (not exactly that path curvature), after an initial impulse the door rotates too slow, so the force from the door flips direction, and since it is already rotated, this normal force has a component to the right.

We said that right case is not possible.
Bottle cant shift to the right from original position, to do that we must have some real outward force..

 

[Orodruin]

We said that right case is not possible.
Bottle cant shift to the right from original position, to do that we must have some real outward force..

No, that is not the case. What makes your right picture unphysical is that it starts with the bottle moving outwards. Were you to simply stop the door’s rotation, the bottle would retain its radial velocity and eventually pass the vertical line.

 

[A.T.]

We said that right case is not possible.
Bottle cant shift to the right from original position, to do that we must have some real outward force..

The final horizontal offset to the right is possible, but the path should start vertically and then deviate to the right. The force to the right comes from the normal force by the door which has already rotated, but is now too slow, to create the force free case on the left.

 

[Orodruin]

The final horizontal offset to the right is possible, but the path should start vertically and then deviate to the right. The force to the right comes from the normal force by the door which has already rotated, but is now too slow, to create the force free case on the left.

Or start down, deviate to the left, then turn to the right. It all depends on exactly how the door opens.

 

[user079622]

I dont understand what are you talking about, what mean deviate?

 

[Orodruin]

I dont understand what are you talking about, what mean deviate?

https://en.m.wiktionary.org/wiki/deviate

1. (intransitive) To go off course from; to changecourse; to change plans.

 

[user079622]

https://en.m.wiktionary.org/wiki/deviate

1. (intransitive) To go off course from; to changecourse; to change plans.

Yes after door pass 270degress ,at some angle bottle will be right from original position.

 

[Orodruin]

View attachment 334969

Yes after door pass 270degress ,at some angle bottle will be right from original position.

This is not what we are talking about. Show me the refrigerator door that opens 270 degrees or more.

 

[user079622]

This is not what we are talking about. Show me the refrigerator door that opens 270 degrees or more.

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,(often happen to me, god damn centrifugal!)

 

[Dale]

Don't put open containers in the door, put those in the main part of the fridge.

 

[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".