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

[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.