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

[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.

 

[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 ;-)).