Centrifugal forces don't exist in reality?

[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.