Centrifugal forces don't exist in reality?

[Dale]

isn't that contradictory

No. I am not sure why you think it would be contradictory.

In order for the radial velocity to increase we must have radial acceleration in the same direction as radial velocity, how else does the radial velocity increases?

No, this is definitely wrong. You do not need to have outward radial acceleration. You only need to have inward radial acceleration that is less than the usual circular centripetal acceleration at that radius.

Also, note that the tangential acceleration plays a role. If you have a circular path then your outward velocity is a fixed constant 0. If you also accelerate tangentially at a constant rate then your inward acceleration will be continuously increasing.

I encourage you to work a few of these problems for yourself. Start with some standard spiral equations and simply calculate the outward components of velocity and acceleration.

 

[PeterDonis]

In order for the radial velocity to increase we must have radial acceleration in the same direction as radial velocity, how else does the radial velocity increases?

You can't decompose things this way; "radial" is not always the same direction, because we're working in an inertial frame. If you want to have a workable separation into "radial" and "tangential", you need to work in a non-inertial frame with time-dependent angular velocity relative to an inertial frame. In such a frame, as I said in post #46, there is an outward coordinate acceleration. But not in an inertial frame.

In an inertial frame, the way to think about it is this: "no acceleration" corresponds to a straight line, but the path in @Dale's diagrams is never a straight line. It is always, as @Dale pointed out, concave inward. That means there must always be an inward acceleration, causing the path to deviate inward from a straight line. The inward acceleration simply has a magnitude which is not quite large enough to keep the path an exact circle; instead it's a spiral.

 

[Delta2]

Thanks for your replies @Dale and @PeterDonis but there is something i don't understand

The calculations that @Dale did are made in an inertial frame right?
In that inertial frame we see that the radial component $v\cdot\hat r$ of velocity is increasing. Hence there must be a radial component of acceleration (in the same direction as velocity), hence there must be a net radial force(which should be a real or interaction force because we are in an inertial frame) in the same direction as velocity. Simple as that.

 

[A.T.]

In order for the radial velocity to increase we must have radial acceleration in the same direction as radial velocity, how else does the radial velocity increases?

Consider a mass on a string in circular motion, then you cut the string and the mass flies off tangentially in a straight line. The acceleration is zero but the radial velocity increases.

 

[A.T.]

In that inertial frame we see that the radial component $v\cdot\hat r$ of velocity is increasing. Hence there must be a radial component of acceleration...

No, because the radial vector is rotating, not fixed in the inertial frame.

 

[Delta2]

Ok thanks @A.T. i think i understand now, by decomposing the velocity into radial component is like we "silently working" into a rotating non-inertial frame.

 

[DrStupid]

Their acceleration is inwards at all times.

That could erroneously be read as inward radial acceleration only. Maybe it's just me, but that would have confused me if I wouldn't know how the device works.

 

[Delta2]

That could erroneously be read as inward radial acceleration only. Maybe it's just me, but that would have confused me if I wouldn't know how the device works.

There is also tangential acceleration (in the inertial frame) if that's what you mean. This tangential acceleration is rotating in the inertial frame, (while it is not rotating in the rotating frame) and i believe is responsible for the increasing radial velocity. That's at least how i understand it. In the rotating (non inertial) frame it is the fictitious centrifugal force that is responsible for this.

 

[A.T.]

This tangential acceleration is rotating in the inertial frame, (while it is not rotating in the rotating frame) and i believe is responsible for the increasing radial velocity.

See my simple example with an inertially moving mass in post#54. There is no acceleration at all, and yet the radial velocity is increasing. So I don't think you can blame the tangential acceleration for that.

 

[Delta2]

See my simple example with an inertially moving mass in post#54. There is no acceleration at all, and yet the radial velocity is increasing. So I don't think you can blame the tangential acceleration for that.

Yes you right in post #54,however i have another example in my mind , i ll tell you what exactly i have in my mind and you tell me where i go wrong:

I am thinking of the ball inside a pipe problem:

Suppose we have a pipe of some diameter and we insert inside the pipe a ball with the same diameter. Assume that there are no frictions and that the ball is sitting stationary at the center of the pipe. Then we rotate the pipe with some angular acceleration with respect to one end of the pipe. Then the ball gets ejected from the other end with some radial velocity right? The ball starts from rest and ends up with some velocity, and hence kinetic energy. All of these in the inertial reference frame. An the question is : What force is responsible for the increase in kinetic energy of the ball? My answer is that it is the normal force from the walls of pipe that acts always in the tangential direction. Where am i wrong?

 

[DrStupid]

My answer is that it is the normal force from the walls of pipe that acts always in the tangential direction.

That's what I mean above. It doesn't work without the tangential component of the acceleration. The inward radial acceleration can't do the job. It just prevents the ball (or the weights in the centrifugal clutch) from moving outside even faster.

 

[Dale]

the radial component v⋅r^ of velocity is increasing. Hence there must be a radial component of acceleration (in the same direction as velocity),

This is simply wrong. I don’t know how it can be more clear. You should work several of these problems yourself. Sometimes there is simply no substitute for working it out.

 

[Delta2]

This is simply wrong. I don’t know how it can be more clear. You should work several of these problems yourself. Sometimes there is simply no substitute for working it out.

Yes it is wrong and @A.T. made me realized it with his posts #54 and #55.

 

[Dale]

An the question is : What force is responsible for the increase in kinetic energy of the ball? My answer is that it is the normal force from the walls of pipe that acts always in the tangential direction. Where am i wrong?

Nothing wrong with that.

 

[etotheipi]

Suppose we have a pipe of some diameter and we insert inside the pipe a ball with the same diameter. Assume that there are no frictions and that the ball is sitting stationary at the centre of the pipe. Then we rotate the pipe with some angular acceleration with respect to one end of the pipe. Then the ball gets ejected from the other end with some radial velocity right? The ball starts from rest and ends up with some velocity, and hence kinetic energy. All of these in the inertial reference frame. And the question is : What force is responsible for the increase in kinetic energy of the ball? My answer is that it is the normal force from the walls of pipe that acts always in the tangential direction. Where am i wrong?

If the force is purely tangential, then one has $\vec{F} = F_{\theta} \hat{\theta}$. Into Newton II: $$F_{\theta} = mr\ddot{\theta} + 2m\dot{r}\dot{\theta}$$But strangely, I can also write $$(\vec{\tau})_z = (\vec{r} \times \vec{F})_z = rF_{\theta} = mr^2 \ddot{\theta}$$which seems to suggest $F_{\theta} = mr\ddot{\theta}$ (if we assume the particle starts at $r>0$). This is odd, because $\dot{r}\dot{\theta}=0$ follows; and since $\dot{\theta} > 0$, we must have $\dot{r} = 0$, which implies the ball's radial coordinate will remain fixed...

I've made a mistake somewhere (because for circular motion we evidently need a radial component of force) but I can't see where

 

[jbriggs444]

My answer is that it is the normal force from the walls of pipe that acts always in the tangential direction. Where am i wrong?

The normal force always acts in a tangential direction, yes. But the tangential direction keeps changing. As it does, tangential velocity becomes radial velocity.

That's the Coriolis effect.

 

[vanhees71]

Consider a mass on a string in circular motion, then you cut the string and the mass flies off tangentially in a straight line. The acceleration is zero but the radial velocity increases.

The velocity after the cut is constant due to Newton's 1st (and 2nd) Law. It doesn't increase (what ever this means for a vector ;-)).

 

[jbriggs444]

The velocity after the cut is constant due to Newton's 1st (and 2nd) Law. It doesn't increase (what ever this means for a vector ;-)).

The radial component of the velocity increases. Which, of course, does not contradict your correct statement.

 

[A.T.]

The velocity after the cut is constant due to Newton's 1st (and 2nd) Law. It doesn't increase (what ever this means for a vector ;-)).

I wrote "radial velocity", not just "velocity ". This is the context:

...the radial component $v\cdot\hat r$ of velocity is increasing...

 

[vanhees71]

The particle goes on with the velocity at the moment you cut the string. At this moment the velocity was perpendicular to the radius vector since the tangent of a circle is always perpendicular to the radius...

 

[etotheipi]

The particle goes on with the velocity at the moment you cut the string. At this moment the velocity was perpendicular to the radius vector since the tangent of a circle is always perpendicular to the radius...

I think we have $v_r = v\sin{\theta}$ if the string is cut when $\theta = 0$. Then $v_r$ approaches (increasing) to $v$ as $\theta \rightarrow \frac{\pi}{2}$.

 

[jbriggs444]

The particle goes on with the velocity at the moment you cut the string. At this moment the velocity was perpendicular to the radius vector since the tangent of a circle is always perpendicular to the radius...

That tells us that the radial component of the velocity is zero at the moment of release. By itself, it tells us nothing about whether the radial component is increasing or decreasing after release.

 

[Dale]

The particle goes on with the velocity at the moment you cut the string. At this moment the velocity was perpendicular to the radius vector since the tangent of a circle is always perpendicular to the radius...

At this moment yes. After this moment @A.T. is correct.

Due to the geometry the velocity starts out as purely tangential and becomes progressively more and more radial as it asymptotically approaches a purely radial velocity with no tangential component. All without acceleration (it is the meaning of "radial" and "tangential" that changes, not the velocity).

 

[vanhees71]

Then I don't understand what you mean by "radial". Take a uniform rotation,
$$\vec{r}(t)=a \hat{r}(t)=a (\cos(\omega t),\sin(\omega t)).$$
The velocity is
$$\vec{v}(t)=a \dot{\hat{r}}(t)=a \omega (-\sin(\omega t),\cos(\omega t)).$$
Cutting at $t=t_0$ you have for $t>t_0$
$$\vec{v}(t)=\vec{v}(t_0)=a \omega (-\sin(\omega t_0),\cos(\omega t_0))=\text{const}.$$
So what is growing here? It's constant, isn't it?

 

[Dale]

Then I don't understand what you mean by "radial". Take a uniform rotation,
$$\vec{r}(t)=a \hat{r}(t)=a (\cos(\omega t),\sin(\omega t)).$$
The velocity is
$$\vec{v}(t)=a \dot{\hat{r}}(t)=a \omega (-\sin(\omega t),\cos(\omega t)).$$
Cutting at $t=t_0$ you have for $t>t_0$
$$\vec{v}(t)=\vec{v}(t_0)=a \omega (-\sin(\omega t_0),\cos(\omega t_0))=\text{const}.$$
So what is growing here? It's constant, isn't it?

$\vec v$ is constant, but the radial component $\vec v \cdot \hat r$ is not.

 

[etotheipi]

Then I don't understand what you mean by "radial".

I think they are referring to the component of the velocity in the direction of $\hat{r}$, $v_r = \vec{v}(t_0) \cdot \hat{r}$. It is attempting to show that $v_r$ can increase without acceleration (or applied force) in the radial direction.

 

[vanhees71]

You mean you use $\hat{r}(t)$ to also describe $\vec{v}(t_0)$? That's a bit artificial a choice as a reference frame. Ok, then it's
$$\hat{r}(t) \cdot \vec{v}(t_0)=a \omega [-\cos(\omega t) \sin(\omega t_0)+\sin(\omega t) \cos(\omega t_0)]=a \omega \sin[\omega(t-t_0)],$$
which is indeed increasing first (but later on oscillating). But what's the point of this?

 

[Dale]

But what's the point of this?

It is indeed completely pointless.

 

[etotheipi]

You mean you use $\hat{r}(t)$ to also describe $\vec{v}(t_0)$? That's a bit artificial a choice as a reference frame. Ok, then it's
$$\hat{r}(t) \cdot \vec{v}(t_0)=a \omega [-\cos(\omega t) \sin(\omega t_0)+\sin(\omega t) \cos(\omega t_0)]=a \omega \sin[\omega(t-t_0)],$$ which is indeed increasing first (but later on oscillating). But what's the point of this?

I don't think it will ever oscillate, since $\hat{r}(t) = \cos{(\theta(t))}\hat{x} + \sin{(\theta(t))}\hat{y}$ where $\theta(t) \in [0, \frac{\pi}{2})$, so it would be asymptotic (i.e. after it loses contact $\theta(t) \neq \omega_0 t$).

Also whilst I am still here, I wondered whether anyone had found the flaw in post #65? I've stared at it for about 10 minutes but still can't make any sense of it .

 

[jbriggs444]

You mean you use $\hat{r}(t)$ to also describe $\vec{v}(t_0)$? That's a bit artificial a choice as a reference frame. Ok, then it's
$$\hat{r}(t) \cdot \vec{v}(t_0)=a \omega [-\cos(\omega t) \sin(\omega t_0)+\sin(\omega t) \cos(\omega t_0)]=a \omega \sin[\omega(t-t_0)],$$
which is indeed increasing first (but later on oscillating). But what's the point of this?

You are assuming here that $\hat{r}(t)$ is uniformly rotating. The intent by @A.T. is pretty clearly that it always points toward the object.

Equivalently, one is considering the position of the object over time using polar coordinates $(r,\theta)$ and looking at how r evolves over time.

 

[PeterDonis]

I wondered whether anyone had found the flaw in post #65?

I would suggest working the problem in an inertial frame first, using Cartesian coordinates.

 

[etotheipi]

I would suggest working the problem in an inertial frame first, using Cartesian coordinates.

Okay, I'll try Cartesian first. I had been doing it in an inertial frame, just in polar coordinates with $\vec{a} = (\ddot{r} - r\dot{\theta}^2)\hat{r} + (r\ddot{\theta} + 2\dot{r}\dot{\theta})\hat{\theta}$.

 

[PeterDonis]

I had been doing it in an inertial frame, just in polar coordinates

If you do that, you have to include the rate of change of $\hat{r}$ and $\hat{\theta}$, since their directions are not constant because the object is moving. The great advantage of Cartesian coordinates is not having to do that since the coordinate basis vectors are unchanging.

 

[etotheipi]

If you do that, you have to include the rate of change of $\hat{r}$ and $\hat{\theta}$, since their directions are not constant because the object is moving. The great advantage of Cartesian coordinates is not having to do that since the coordinate basis vectors are unchanging.

I thought this part was already accounted for in the derivation of $\vec{a} = (\ddot{r} - r\dot{\theta}^2)\hat{r} + (r\ddot{\theta} + 2\dot{r}\dot{\theta})\hat{\theta}$, where we must treat the unit vectors as $\hat{r}(t)$ and $\hat{\theta}(t)$. However, it seems okay to then say $\vec{F} = m\vec{a} = F_{\theta} \hat{\theta} + F_{r} \hat{r} = ma_{\theta}\hat{\theta} + ma_{r} \hat{r} \implies F_{\theta} = ma_{\theta}$ by decomposing in the instantaneous basis.

I figured out that the resolution is that $\vec{\tau} = mr\hat{r} \times (2\dot{r}\dot{\theta} + r\ddot{\theta})\hat{\theta}$ in the general case, which indeed produces the correct result. We only get $\tau = I\alpha$ if no particles have non-zero $\dot{r}$!

 

[Dale]

You mean you use $\hat{r}(t)$ to also describe $\vec{v}(t_0)$? That's a bit artificial a choice as a reference frame. Ok, then it's
$$\hat{r}(t) \cdot \vec{v}(t_0)=a \omega [-\cos(\omega t) \sin(\omega t_0)+\sin(\omega t) \cos(\omega t_0)]=a \omega \sin[\omega(t-t_0)],$$
which is indeed increasing first (but later on oscillating).

Hmm, I don't get any oscillation. It looks like you are neglecting the change in position and the resulting change in $\hat r$. $$\vec v(t) = a \omega \left( -\sin (\omega t_0), \cos (\omega t_0) \right) = \text{const.}$$ $$\vec r(t)= a \left( \cos(\omega t_0), \sin(\omega t_0) \right)+t a \omega \left( -\sin (\omega t_0), \cos (\omega t_0) \right) \ne \text{const.}$$ $$\hat r = \frac{\vec r}{|\vec r|}$$ so $$\vec v \cdot \hat r = \frac{a t \omega^2}{\sqrt{1+t^2 \omega^2}}$$ which smoothly approaches $a\omega$ in the limit as $t$ goes to $\infty$.

Of course, it is all rather pointless as mentioned earlier.

 

[vanhees71]

Ok, that makes of course some sense. Sorry for the confusion, which was entirely on my side :-(.

 

[Riichiro Mizoguchi]

This is my first post to the forum.
This is not a reply to any of the past posts but is related to the main topic of centrifugal force.

I understand we agree that at least in the context of contact force when an object O is in a circular motion a reactive centrifugal force exerting BY O exists in every frame. The reactive centrifugal force forms a pair of action and reaction forces with the centripetal force exerted on O.

I want to know your opinion on an additional claim of mine shown below.

My claim:
Whenever a reactive centrifugal force exerted BY O exists, a force pointing away from the center of rotation exerting ON O exists in every frame in the context of contact force.

The following is an intuitive summary of the proof of the above claim.

Situation:
Imagine O is in a circular motion because it is placed in a train in a circular motion. O is contacting with the wall of the train. Assume no friction between O and the floor of the train. Newtonian physics (NP) says a centripetal force F is exerted ON O by the wall and the reactive centrifugal force -F is exerted on the wall BY O.

Intuitive proof:
Suppose you slice O into two pieces O1 and O2 so that |F1|=|F2| where -(F1+F2)=-F and -F1 is exerting on O2 by O1 and -F2 is on the wall by O2. Here, I assume the same mechanism works at the boundary between O1 and O2 as that between O and the wall. Now, you find an exciting fact that -F1 which is half of -F is exerted ON O2 which is a part of O.

You next divide both O1 and O2 into two pieces to have F11, F12, F21, and F22, respectively. If you continue this operation recursively (forever), you eventually see that a force equivalent to -F is exerted on O itself
by summing up all the sub forces (say, -Fij) exerted on every part of O.

Although this proof is just an intuitive summary, it says that when a reactive centrifugal force -F is exerted on the wall BY O, a force equivalent to -F is exerted ON O itself as well. How do you interpret this force? Is it what people call "centrifugal force"? I'm afraid you would say it is not. Then, how about another example which, iI hope, supports the above proof as follows:

Imagine a book is put on a table. You observe the fact that force mg is exerted on the table BY the book and assume you don't know the gravity force. If you slice the book horizontally in half, then you observe the fact that force mg/2 is exerted on the lower half of the book BY the upper half of the book because of the same mechanism as that the book pushes down the table by force mg. In the same manner, as the above proof, you can easily find a force equivalent to mg is exerted ON the book without knowing (using) the gravity force, and you will be able to know the force exerted ON the book is the gravity force by the earth.My claim can be generalized as follows:

In the case of contact force between two (non-agentive) objects O1 and O2, whenever a force F is exerted on O2 by O1, a force equivalent to F is exerted on O1 as well.

In high school education where point mass is employed, it would be easy to accept the above claim, since both kinds of forces coincide with each other for a point mass.

I'm afraid I might have made a big mistake somewhere, but if not, the implication of this conclusion is not small because it suggests a force exerted BY O and the force exerted ON O co-occur in every frame in the
context of contact force.

 

[Nugatory]

and the reactive centrifugal force -F is exerted on the wall BY O.

That’s not right. There is a force on the wall exerted by O, and this force is the second law pair for the force exerted in the wall... but it is not the centrifugal force, it’s acting on the wrong body. The centrifugal force acts on the object, not the wall.

 

[Riichiro Mizoguchi]

Thanks for your reply. OK, if the situation I imagine is not appropriate, then please imagine another one which shows a pair of centripetal and reactive centrifugal forces exist and try to evaluate my explanation of the intuitive proof.

 

[jbriggs444]

Whenever a reactive centrifugal force exerted BY O exists, a force pointing away from the center of rotation exerting ON O exists in every frame in the context of contact force.

You claim that if O is exerting an outward force on something then it follows that something else must be exerting an outward force on O.

This is, I think, what @Nugatory had in mind when he spoke of a "second law pair". The invocation of Newton's second law only works to produce the desired result if O is either massless or is not accelerating so that its momentum remains constant.

Thanks for your reply. OK, if the situation I imagine is not appropriate, then please imagine another one which shows a pair of centripetal and reactive centrifugal forces exist and try to evaluate my explanation of the intuitive proof.

Since the claim you make is false, finding a sound proof will be... challenging.

You next divide both O1 and O2 into two pieces to have F11, F12, F21, and F22, respectively. If you continue this operation recursively (forever), you eventually see that a force equivalent to -F is exerted on O itself
by summing up all the sub forces (say, -Fij) exerted on every part of O.

The proof you attempt sub-divides O into a bunch of little pieces, looks at the outward forces between the various pieces, ignores the inward forces between the various pieces and concludes that the sum of the forces that are examined is as desired.

If one is permitted to ignore forces at will, reactionless drives become possible, Perpetual motion becomes possible and further posting is prohibited.

 

[Riichiro Mizoguchi]

You claim that if O is exerting an outward force on something then it follows that something else must be exerting an outward force on O.

No. I mean if O is exerting a reactive outward force, ...

The proof you attempt sub-divides O into a bunch of little pieces looks at the outward forces between the various pieces, ignores the inward forces between the various pieces, and concludes that the sum of the forces that are examined is as desired.

No. I don't ignore them because my goal is to show the existence of a particular kind of force. There is no cancelation of forces when we are not talking about motion caused by forces. Imagine you are pushed by two persons in opposite directions of the same strength. You would say these two forces cancel each other. You don't move but it is different from a situation where no one is pushing you. You are exerted on two forces.

By the way, what we see at every boundary between pieces of objects is the same at the boundary between O and the other object generating centripetal force both of which would form action/reaction pairs?

Or, are you going to say, in the case of a book on the table, my proof doesn't work because the downward force exerted on the book is canceled by the upward force exerted on the book by the table so that there is no downward force exerted on the book?

 

[Dale]

Suppose you slice O into two pieces O1 and O2 so that |F1|=|F2| where -(F1+F2)=-F and -F1 is exerting on O2 by O1 and -F2 is on the wall by O2.

I could really use a drawing here because from your description I don’t think that this combination of facts is possible. If F2 is the force on the wall then F2=F (cutting O into pieces internally doesn’t change anything externally). So then you cannot have both F1+F2=F and |F1|=|F2|

You would say these two forces cancel each other. You don't move but it is different from a situation where no one is pushing you. You are exerted on two forces.

The force does cancel. What is different is the stress, not the force. When no one is pushing you the net force is zero and the stress is zero. When two people are pushing in opposite directions the net force is still zero but the stress is non-zero.

O does have a nonzero stress. It does not have an additional force.

 

[Riichiro Mizoguchi]

I could really use a drawing here because from your description I don’t think that this combination of facts is possible. If F2 is the force on the wall then F2=F (cutting O into pieces internally doesn’t change anything externally). So then you cannot have both F1+F2=F and |F1|=|F2|

The force does cancel. What is different is the stress, not the force. When no one is pushing you the net force is zero and the stress is zero. When two people are pushing in opposite directions the net force is still zero but the stress is non-zero.

O does have a nonzero stress. It does not have an additional force.

I'm afraid there might be a misunderstanding. Not F but -F. F is the force the wall pushes O and -F is the force O pushes the wall.

The notion of the net force is defined for describing motion caused by forces. It doesn't affect the existence of those forces. In the case of two persons push me, even if the two forces are canceled, the existence of those two forces is never canceled. They are there.

Allow me to attach a detailed proof of my claim. I hope my translation from Japanese to English works well and hope this clarifies my proof.

 

[Nugatory]

hope this clarifies my proof.

It does, and you are repeating the same error. The only outwards forces in your diagram are the second law partners of the inwards centripetal force - there is no centrifugal force. The best way to see this is to consider the forces acting on the innermost object: there is an inwards-directed force pushing it onto a curved path instead of the straight-line path it would follow in the absence of any net force, and no outward force acting on it at all.

This thread has reached the point where we're propagating misinformation under the guise of a proof, so it is closed.