# Centrifugal forces don't exist in reality?

• Aeronautic Freek
In summary: Centripetal force... Causes compression in the blade. What centrifugal force does is it causes the blade to rotate in a circle. As the blade rotates, the centrifugal force straightens out the blade, and because the tension is in the blade, the centrifugal force also causes the blade to stay in that position.
vanhees71 said:
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}##.

Dale
vanhees71 said:
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.

vanhees71 said:
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).

jbriggs444
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?

vanhees71 said:
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.

Ibix and etotheipi
vanhees71 said:
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.

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). But what's the point of this?

vanhees71 said:
But what's the point of this?
It is indeed completely pointless.

vanhees71 said:
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 .

vanhees71 said:
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.

etotheipi said:
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
PeterDonis said:
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}##.

etotheipi said:
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.

PeterDonis said:
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}##!

Last edited by a moderator:
vanhees71 said:
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 :-(.

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.

PeroK
Riichiro Mizoguchi said:
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.

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.

weirdoguy
Riichiro Mizoguchi said:
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.

Riichiro Mizoguchi said:
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.

Riichiro Mizoguchi said:
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.

Last edited:
jbriggs444 said:
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, ...
jbriggs444 said:
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?

Last edited by a moderator:
etotheipi, Motore, PeroK and 1 other person
Riichiro Mizoguchi said:
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|

Riichiro Mizoguchi said:
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.

Last edited:
Dale said:
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.

#### Attachments

• Theorem II.docx
33 KB · Views: 154
weirdoguy
Riichiro Mizoguchi said:
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.

Replies
2
Views
8K
Replies
1
Views
1K
Replies
2
Views
1K
Replies
2
Views
955
Replies
15
Views
2K
Replies
1
Views
2K
Replies
23
Views
4K
Replies
4
Views
2K
Replies
18
Views
2K
Replies
4
Views
2K