If centrifugal force doesn't exist, why does a glass move outward in the fridge door?

[user079622]

Depends on a lot of details of the experiment. In the idealised case shown in your original left hand diagram, no it'll go straight down. If the bottle is small compared to the width of the door and it's sliding along the door rather than skating along frictionlessly, then it'll be tangential, yes.

What is difference between sliding and skating?

The left hand diagram is possible if there is zero friction between the door tray and the bottle and the door tray is very wide. The bottle gets an initial downward impulse from the door and then skates frictionlessly downwards.
.

I think it is impossilble to move in straight line , in case of zero friction bottle trajectory will looks like this?
Because of normal force

 

[Ibix]

What is difference between sliding and skating?

Nothing, in this context. I was referring to the difference between the bottle getting an initial bump and then sliding along without interacting with anything, versus the more realistic case of the bottle sliding along a shelf with a railing.

I think this impossible

It's an idealisation, as we've said. In the case that the shelf floor is frictionless, just bump the bottle a little and then rotate the door at the appropriate rate so that it does not touch either the door or the rail. Your diagram is showing the case where the bottle is being constantly pushed by the door. I think it's conflating the direction of motion with the direction of applied force, and is missing the radially inwards frictional force, but qualitatively it's about right.

 

[user079622]

It's an idealisation, as we've said. In the case that the shelf floor is frictionless, just bump the bottle a little and then rotate the door at the appropriate rate so that it does not touch either the door or the rail. Your diagram is showing the case where the bottle is being constantly pushed by the door. I think it's conflating the direction of motion with the direction of applied force, and is missing the radially inwards frictional force, but qualitatively it's about right.

But bump force is not case when you open the fridge, door is constantly push bottle normal to wall.
I never mentioned that I just hit/pull door once.

 

[user079622]

It is weird how in frictionless case path is still curved, even we don't have centripetal force!
Make a turn without centripetal force!

 

[Ibix]

But bump force is not case when you open the fridge, door is constantly push bottle normal to wall.

That's why @Orodruin said your left hand diagram wasn't correct except "in very particular cases", and why I've referred to it as an idealisation. It's possible to engineer it if you have a frictionless surface, but it's not the usual behaviour, no. Your final diagram is much closer to what happens - the tangential force from the door initially knocks the bottle into straight line motion (as in your left hand diagram) but subsequent contact with the door and its changing orientation curves the motion into an outward spiral.

 

[vanhees71]

Ok, let's do the calculation for an idealized model: A bead is gliding along a rigid rod, which is rotating at constant angular speed. Let ##(r,\varphi)## be the usual polar coordinates. Then the position vector of the bead is
$$\vec{r}=\begin{pmatrix} r \cos(\omega t) \\ r \sin(\omega t) \end{pmatrix}.$$
The velocity is
$$\dot{\vec{r}} = \dot{r} \begin{pmatrix} \cos(\omega t) \\ \sin(\omega t) \end{pmatrix} + \omega r \begin{pmatrix} -\sin(\omega t) \\ \cos(\omega t) \end{pmatrix}.$$
Lagrangian is
$$L=T=\frac{m}{2} \dot{\vec{r}}^2 = \frac{m}{2} (\dot{r}^2 + \omega^2 r^2).$$
The equation of motion is given by the Euler-Lagrange equation,
$$\frac{\mathrm{d}}{\mathrm{d} t} \frac{\partial L}{\partial \dot{r}}=\frac{\partial L}{\partial r},$$
i.e.,
$$m \ddot{r} = m \omega^2 r \; \Rightarrow \; r(t)=A \cosh(\omega t) + B \sinh(\omega t),$$
i.e., the bead is flying radially out.

Of course, the equation of motion, here derived entirely in the inertial frame, can be reinterpreted as the equation of motion from the point of view of the non-inertial frame, co-rotating with the rod, leading to the centri-fugal force on the right-hand side.

 

[Orodruin]

Ok, let's do the calculation for an idealized model: A bead is gliding along a rigid rod, which is rotating at constant angular speed. Let ##(r,\varphi)## be the usual polar coordinates. Then the position vector of the bead is
$$\vec{r}=\begin{pmatrix} r \cos(\omega t) \\ r \sin(\omega t) \end{pmatrix}.$$
The velocity is
$$\dot{\vec{r}} = \dot{r} \begin{pmatrix} \cos(\omega t) \\ \sin(\omega t) \end{pmatrix} + \omega r \begin{pmatrix} -\sin(\omega t) \\ \cos(\omega t) \end{pmatrix}.$$
Lagrangian is
$$L=T=\frac{m}{2} \dot{\vec{r}}^2 = \frac{m}{2} (\dot{r}^2 + \omega^2 r^2).$$
The equation of motion is given by the Euler-Lagrange equation,
$$\frac{\mathrm{d}}{\mathrm{d} t} \frac{\partial L}{\partial \dot{r}}=\frac{\partial L}{\partial r},$$
i.e.,
$$m \ddot{r} = m \omega^2 r \; \Rightarrow \; r(t)=A \cosh(\omega t) + B \sinh(\omega t),$$
i.e., the bead is flying radially out.

Of course, the equation of motion, here derived entirely in the inertial frame, can be reinterpreted as the equation of motion from the point of view of the non-inertial frame, co-rotating with the rod, leading to the centri-fugal force on the right-hand side.

Of note is that the x-coordinate of this solution (assuming ##\dot r(0)=0## is given by
$$
x(t) = x_0 \cos(\omega t) \cosh(\omega t).
$$
Here is a plot with ##\omega t## on the horizontal and ##x(t)/x_0## on the vertical axis:

Note how the value of ##x(t)## never exceeds ##x_0##.

(Obviously this is assuming an infinitely long door. In the case of a finite door this plot will stop being valid when the glass reaches the end of the door.)

 

[erobz]

Ok, let's do the calculation for an idealized model: A bead is gliding along a rigid rod, which is rotating at constant angular speed. Let ##(r,\varphi)## be the usual polar coordinates. Then the position vector of the bead is
$$\vec{r}=\begin{pmatrix} r \cos(\omega t) \\ r \sin(\omega t) \end{pmatrix}.$$
The velocity is
$$\dot{\vec{r}} = \dot{r} \begin{pmatrix} \cos(\omega t) \\ \sin(\omega t) \end{pmatrix} + \omega r \begin{pmatrix} -\sin(\omega t) \\ \cos(\omega t) \end{pmatrix}.$$
Lagrangian is
$$L=T=\frac{m}{2} \dot{\vec{r}}^2 = \frac{m}{2} (\dot{r}^2 + \omega^2 r^2).$$
The equation of motion is given by the Euler-Lagrange equation,
$$\frac{\mathrm{d}}{\mathrm{d} t} \frac{\partial L}{\partial \dot{r}}=\frac{\partial L}{\partial r},$$
i.e.,
$$m \ddot{r} = m \omega^2 r \; \Rightarrow \; r(t)=A \cosh(\omega t) + B \sinh(\omega t),$$
i.e., the bead is flying radially out.

Of course, the equation of motion, here derived entirely in the inertial frame, can be reinterpreted as the equation of motion from the point of view of the non-inertial frame, co-rotating with the rod, leading to the centri-fugal force on the right-hand side.

My Linear Algebra is rusty. How do we multiply the matrices to square ##{ \vec{\dot{r} } }^2##, because it seems like we have column matrices. We need to transpose one of them to multiply them by themselves (or each other). What is the justification for that?

Basically I'm seeing we have:

$$ \begin{bmatrix} a \\ b \end{bmatrix} \cdot \begin{bmatrix} a \\ b \end{bmatrix}$$

But I know that we need for valid multiplication:

$$ \begin{bmatrix} a & b \end{bmatrix} \cdot \begin{bmatrix} a \\ b \end{bmatrix}$$

I don't remember what is going on here.

 

[Orodruin]

My Linear Algebra is rusty. How do we multiply the matrices to square ##{ \vec{\dot{r} } }^2##, because it seems like we have column matrices. We need to transpose one of them to multiply them by themselves (or each other). What is the justification for that?

Basically I'm seeing we have:

$$ \begin{bmatrix} a \\ b \end{bmatrix} \cdot \begin{bmatrix} a \\ b \end{bmatrix}$$

But I know that we need for valid multiplication:

$$ \begin{bmatrix} a & b \end{bmatrix} \cdot \begin{bmatrix} a \\ b \end{bmatrix}$$

I don't remember what is going on here.

It is not a matrix multiplication. It is a vector inner product.

 

[erobz]

It is not a matrix multiplication. It is a vector inner product.

That makes sense now...

 

[vanhees71]

My Linear Algebra is rusty. How do we multiply the matrices to square ##{ \vec{\dot{r} } }^2##, because it seems like we have column matrices. We need to transpose one of them to multiply them by themselves (or each other). What is the justification for that?

Basically I'm seeing we have:

$$ \begin{bmatrix} a \\ b \end{bmatrix} \cdot \begin{bmatrix} a \\ b \end{bmatrix}$$

But I know that we need for valid multiplication:

$$ \begin{bmatrix} a & b \end{bmatrix} \cdot \begin{bmatrix} a \\ b \end{bmatrix}$$

I don't remember what is going on here.

The usual definition of the "vector squared" is that it's the dot-product of the vector with itself, i.e.,
$$\dot{\vec{r}}^2=\dot{\vec{r}} \cdot \dot{\vec{r}}=\dot{x}_1^2 + \dot{x}_2^2 + \dot{x}_3^2,$$
where ##x_j## (##j \in \{1,2,3 \}##) are components of ##\vec{r}## with respect to a Cartesian basis.

 

[erobz]

The usual definition of the "vector squared" is that it's the dot-product of the vector with itself, i.e.,
$$\dot{\vec{r}}^2=\dot{\vec{r}} \cdot \dot{\vec{r}}=\dot{x}_1^2 + \dot{x}_2^2 + \dot{x}_3^2,$$
where ##x_j## (##j \in \{1,2,3 \}##) are components of ##\vec{r}## with respect to a Cartesian basis.

For some reason the vertical notation and absence of unit vectors caused me to completely forget you were working with vectors.

 

[Orodruin]

For some reason the vertical notation and absence of unit vectors caused me to completely forget you were working with vectors.

This is why I don’t like representing vectors as column vectors. I much prefer a notation using the basis vectors. Students also tend to mess up the component notation for anything other than Cartesian coordinates …

 

[erobz]

This is why I don’t like representing vectors as column vectors. I much prefer a notation using the basis vectors. Students also tend to mess up the component notation for anything other than Cartesian coordinates …

I'm sure your reasons are justified. My own are that I never did enough physics/engineering that required any significant level of competence with any vectoral notation in practice. I think 1 or 2 coordinates describe most of our problems!

I was trying to solve ## \sum F_x, \sum F_y ## by taking the derivatives of ##x = r\cos(\omega t ), y = r\sin(\omega t )## and substituting for ##N##, etc... And I saw how slick what @vanhees71 did appears to be in comparison... So I finally decided I can no longer pretend there isn't a better way!

 

[user079622]

Of course, it's always simpler to analyze a problem in an inertial frame.

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.

 

[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.