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

[user079622]

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?

 

[PeroK]

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?

Presumably the glass of water is held by the fridge door? What force opens the fridge door?

 

[user079622]

Presumably the glass of water is held by the fridge door? What force opens the fridge door?

yes, glass of water is in door slot, when I open door fast ,glass accelerate toward me...

 

[PeroK]

yes, glass of water is in door slot, when I open door fast ,glass accelerate toward me...

Is that a surprise?

 

[user079622]

Is that a surprise?

If centrifugal force dont exist then yes it is surprise..If centrifugal force exist then it is not surprise.

 

[Lnewqban]

... 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, ...

Does the glass of water slide in direction that is parallel or perpendicular to the door?

 

[PeroK]

If centrifugal force dont exist then yes it is surprise..If centrifugal force exist then it is not surprise.

I fail to see the relevance of centrifugal force. The glass is essentially part of the door and moves with the door. It has no choice, unless it breaks away from the door.

Centrifugal force usually relates to the apparent outward force when an object is not constrained to move in a circle.

What you have in your fridge door is a real outward force on the door handle. I can see no sense in assuming that force does not exist.

 

[user079622]

Does the glass of water slide in direction that is parallel or perpendicular to the door?

View attachment 334805

what force push glass of water out from center of rotation if centrifugal force dont exist?

 

[PeroK]

what force push glass of water out from center of rotation if centrifugal force dont exist?

The contact force with the fridge shelf, which holds the fridge contents in place.

 

[user079622]

Does the glass of water slide in direction that is parallel or perpendicular to the door?

parallel, if you put glass of water at inner part of shelf ,when open door too fast ,glass will slide out

 

[PeroK]

parallel, if you put glass of water at inner part of shelf ,when open door too fast ,glass will slide out

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.

 

[user079622]

I fail to see the relevance of centrifugal force. The glass is essentially part of the door and moves with the door. It has no choice, unless it breaks away from the door.

glass is sliding along door shelf, from inner part toward outward part if you open door too fast..

The contact force with the fridge shelf, which holds the fridge contents in place.

But why this force has outward direction if we say that centirufgal force dont exist?

 

[kuruman]

Consider a bead with a rod through it (see figure below). When rod rotates as shown, the bead will move away from the axis of rotation.

If you are a non-believer, you will say that the bead will tend to stay in place while the rod rotates through it.
If you are a believer, you will say that opening the door creates a centrifugal force the pushes the bead away from the axis.

The glass on the refrigerator door is a similar case.

Non-believers are often called observers in an inertial frame.
Believers are often called observers in a rotating frame.
Both kinds are good people.

 

[PeroK]

glass is sliding along door shelf, from inner part toward outward part if you open door to fast..

But why this force has outward direction if we say that centirufgal force dont exist?

See my latest post.

 

[user079622]

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.

Outward force in inertial frame?

View attachment 334806

your glass has zero right movement(left case), that is not correct
right case is correct.

 

[PeroK]

Outward force in inertial frame?

Yes.

 

[user079622]

Yes.

In inertial frame centrifugal force dont exist, so how can we have outward force?

 

[Dale]

If centrifugal force dont exist

The centrifugal force does exist in a rotating reference frame.

In inertial frame centrifugal force dont exist, so how can we have outward force?

It doesn’t accelerate outward in an inertial frame. Why would there need to be an outward force when it doesn’t accelerate outward?

 

[Orodruin]

In inertial frame centrifugal force dont exist, so how can we have outward force?

You don’t. Any motion in the outwards direction is due to tangential acceleration followed by the door rotation resulting in that tangential velocity becoming radial due to the rotation.

 

[user079622]

It doesn’t accelerate outward in an inertial frame. Why would there need to be an outward force when it doesn’t accelerate outward?

Where is accelerate?
You want to say that glass dont have right acceleration as I show in right case?

 

[PeroK]

In inertial frame centrifugal force dont exist, so how can we have outward force?

The initial force is "outward". If we say that the door is initially in the x direction and the initial force is in the y direction. Then that force can generate motion in the y direction. Once the door has rotated to an open position, then the initial y direction is now partly along the length of the door - partly in the "outward" direction.

In other words, the direction you call "outwards" changes with time. And the motion of the glass from the forces on the door becomes "outward" as the direction you call "outward" changes.

See also the example of bead on the rod.

In either case, the "outward" motion is caused by a lack of centripetal force to ensure the object moves in a circle. Friction with the door can provide some centripetal force, but not enough if you open the door quickly.

 

[user079622]

You don’t. Any motion in the outwards direction is due to tangential acceleration followed by the door rotation resulting in that tangential velocity becoming radial due to the rotation.

any graph of this forces ?

 

[russ_watters]

If centrifugal force dont exist...

If you're commenting on it being a "fictitious force", that's not the same as "does not exist".

https://en.m.wikipedia.org/wiki/Fictitious_force

 

[Dale]

Where is accelerate?
You want to say that glass dont have right acceleration as I show in right case?

View attachment 334816

That is correct. In that drawing there is clearly no outward acceleration at any point. Outward acceleration in an inertial frame would mean that the path would be concave outward. But the path is concave inward at each point. Therefore there is clearly no outward acceleration at any point do no outward force is needed at any point.

 

[user079622]

That is correct. In that drawing there is clearly no outward acceleration at any point. Outward acceleration in an inertial frame would mean that the path would be concave outward. But the path is concave inward at each point. Therefore there is clearly no outward acceleration at any point do no outward force is needed at any point.

But what is this outward force in inertial frame that other members talk about?
I dont understand where it comes from..

 

[Dale]

But what is this outward force in inertial frame that other members talk about?

There is no outward force in the inertial frame. If there were then path would be concave outward.

 

[user079622]

There is no outward force in the inertial frame. If there were then path would be concave outward.

"Any motion in the outwards direction is due to tangential acceleration followed by the door rotation resulting in that tangential velocity becoming radial due to the rotation."

Can you explain this ?

 

[Vanadium 50]

OP, I see that you are responding in seconds - sp fast you don't even have time to use capital letters, I suggest that things will go faster for you if you took a few minutes to read and think about the messages people are sending you before responding.

 

[Orodruin]

right case is correct.

No, it isn’t. For several reasons. Nor is the left picture except in very particular cases.

"Any motion in the outwards direction is due to tangential acceleration followed by the door rotation resulting in that tangential velocity becoming radial due to the rotation."

Can you explain this ?

Consider the closed door starting to open. This gives a force perpendicular to the door, which starts putting the glass into motion downwards in your picture. At the insrant of opening downwards will be tangential and not radial. The glass will continue the induced downward motion unless acted upon by another force. However, as the door swings open by even a minor angle. This means that downwards is no longer purely tangential, it has a radial component. Hence, the glass now has a radial component of velocity without ever being acted upon by a radial force.

 

[Lnewqban]

what force push glass of water out from center of rotation if centrifugal force dont exist?

The door would move linearly toward you (assuming the door handle is located by its CG), together with the glass of water, as your force is simultaneously applied on both bodies and directed toward you.

That theoretical linear movement of the door is restricted by the hinges, and it instead rotates about the vertical line of the hinges.

That theoretical linear movement of the glass of water is not restricted by the door's hinges, therefore, it tends to move linearly toward you.
At the same time, the points of the door in non-solid contact with the glass describe a circular motion, which deviates the sliding glass from that theoretical linear movement.

Hence, the glass describes a trajectory that is between circular (hinged door) and linear (non-hinged door).
Please, see animation in post #6 above.

 

[user079622]

No, it isn’t. For several reasons. Nor is the left picture except in very particular cases.Consider the closed door starting to open. This gives a force perpendicular to the door, which starts putting the glass into motion downwards in your picture. At the insrant of opening downwards will be tangential and not radial. The glass will continue the induced downward motion unless acted upon by another force. However, as the door swings open by even a minor angle. This means that downwards is no longer purely tangential, it has a radial component. Hence, the glass now has a radial component of velocity without ever being acted upon by a radial force.

Why right case is not correct and is left case possible in reality?

 

[Ibix]

Why right case is not correct and is left case possible in reality?

You drew that image using the inertial reference frame where centrifugal force is zero. So the bottle cannot move to the right; there are no forces in that direction. Thus the right hand diagram is wrong.

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.

In practice that's not possible, and the bottle path will curve slightly to the left due to friction and bumping into the door.

 

[vanhees71]

I fail to see the relevance of centrifugal force. The glass is essentially part of the door and moves with the door. It has no choice, unless it breaks away from the door.

Centrifugal force usually relates to the apparent outward force when an object is not constrained to move in a circle.

The centrifugal force occurs only in rotating non-inertial frames. It's a inertial force. Some people call it "fictitious force", because it's not a force due to an interaction but just due to the reinterpretation of parts of the expression for the acceleration wrt. an inertial frame in terms of coordinates referring to the non-inertial rotating frame.

What you have in your fridge door is a real outward force on the door handle. I can see no sense in assuming that force does not exist.

Of course, it's always simpler to analyze a problem in an inertial frame. The force on the glass when opening the door is of course true to the interaction of the glass with the door (i.e., electromagnetic interactions).

 

[user079622]

You drew that image using the inertial reference frame where centrifugal force is zero. So the bottle cannot move to the right; there are no forces in that direction. Thus the right hand diagram is wrong.

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.

In practice that's not possible, and the bottle path will curve slightly to the left due to friction and bumping into the door.

Bottle will curve to the left like this, so acceleration is inward?

1.If I stop door rotation in time t2 and if door end is open, then bottle will leave door in radial direction vr?

2.If bottle leave door(open end) when door is still rotating, it will leave door with tangential direction?

 

[Ibix]

Bottle will curve to the left like this, so acceleration is inward?

The acceleration is to the left. I wouldn't characterise it as radial.

1.If I stop door rotation in time t2 and if door end is open, then bottle will leave door in radial direction vr?

It'll bump into the door or the railing of the shelf it's on and bounce off, or it may already be sliding along the railing. The result of the force from the railing will be to push the bottle outwards, yes.

2.If bottle leave door(open end) when door is still rotating, it will leave door with tangential direction?

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.

 

[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.