Acceleration of clubhead

[nmgolfer]

Mandrin

In virtually all mechanics texts (and I have a whole bunch of them) alpha which you show squared in equation 1 is angular acceleration. Now you've changed it and you have "an answer". You don't show how you arrived at that answer... messy details I suppose but who cares you have an answer and pretty graphics ta-boot.

Yet mandrin you stubbornly cling to the misguided notion that there exists a real centrifugal force acting on the bead. There is not... the only force the bead feels is the rod pushing on it. Your mistake is what forced me to speak up (again).

Mandrin, don't you agree that words (and symbols) have meaning and that meaning is important? When people can make up new meaning for words at their pleasure the language and communication suffers.

There are only two valid uses for that "f" word in classical mechanics and you yourself agreed that wikipedia had those two definitions right.

1) a reaction to centripetal acceleration (which the hub of your rod experience)
2) as a fictitious force that is required when using non-inertial coordinate systems in order to arrive at correct results.

Neither one of those two situation applies here and that in essence is why you are wrong. Your interpretation of results is wrong (again... still).

The radial motion of the bead can be easily explained without needing to use the "f" word. Click here and see for yourself.

 

[nmgolfer]

nmgolfer,

#202 "Also in your equation 1 you should have omega squared not alpha squared"

#211 " ...the formula is wrong... Your formulas are wrong too."

Why do I have that funny feeling that I have to redo your education?

Uh.. because you're confused? BTW The feeling is mutual...

Nice of you to show your work. You did the right thing by fixing it. BTW you have a bad habit of leaving out constants of integration which you've done above. This tendency will continue to bite you in the backside as I pointed out before.

nmgolfer, hopefully you recognize your obvious error now as (1) is expressed in a way which you are more likely to recognize. Anyhow it is rather childish to have to explain all this to someone claiming to be a scientist.

That's exactly the kind of disrespectful snipe that makes me want to smack you up side the head with my seven iron

I am really glad that you got hcw as an enthusiatic fan, he is a real asset for this forum contributing often with a multitude of original ideas.

I suspect he knows more than you but wisely choose not to wrestle with pigs

I enjoy your scientific contributions but just try to be a bit more accurate the next time around and don't blame without at least a minimum of substance in your arguments.

Do you? I don't believe that for a nanosecond. You don't like dissension an it shows through in your tone. This is your playground isn't it mandrin... nobody else allowed... we get it.

 

[JeffM]

NMGolfer

I looked at your website link, and of course, I couldn't understand your mathematical exposition!

I intuitively suspect that you are correct, but my intuitive sense that you are correct would probably become more substantive if you could also provide a child-like explanation that I could understand.

Jeff.

 

[nmgolfer]

NMGolfer

I looked at your website link, and of course, I couldn't understand your mathematical exposition!

I intuitively suspect that you are correct, but my intuitive sense that you are correct would probably become more substantive if you could also provide a child-like explanation that I could understand.

Jeff.

Well Jeff...

I already did that to the best of my (current) ability in post 203. mandrin's graphic is (probably) correct (my main beef is with his use of the term "centrifugal force"). This is a situation were the force on the side of the bead, which the rod is solely responsible for and grows in magnitude as a function of radius (which in turn is a function of angular velocity squared), is "pushing" the bead on a spiral path.

That's it... this dead horse has done been beaten. Folks can believe whomever/whatever they choose.

 

[JeffM]

nmgolfer

If you call post 203 a child-like explanation, then I need a triple-child-like (child-younger child-yet younger child) explanation.

Jeff.

 

[daniell]

JeffMann,

In post 203, look at all the forces at the various instances. Theses forces, F, is the force that the rod applies on the beads. Forces have direction and the bead wants to move in the direction of the force. Because the bead is restrained by the rod, it thus moves out along the rod. Notice that F is changing direction all the time.

In this explanation, Nmgolfer is 100% correct. And he is 100% correct that centrifugal force is a result of the bead's inertia. Thus CF is also known as an inertia force (?? not sure).

Now look at the rotation frame. The force, F, is always pependicular to the rod. But the bead moves out along the rod. Since the bead is moving, there appears to be a force, F1, pushing the bead along the rod. And guess what, this force, F1, is what is defined by science as centrifugal force. Mandrin is correct this time.

And because centrifugal force is cause by the beads's inertia and not a physical force, centrifugal force is often termed as a ficticious force. But it really depends, it depends on the reference frame.

Now do you see, without defining the reference frames, this discussion is rather silly.

cheers,

daniel

 

[JeffM]

Daniell - now that's an excellent child-like explanation that I can near-completely understand. I am getting there!

Thanks,

Jeff.

 

[mandrin]

Children want to know

daniell,

You feel the discussion to be silly and childish and Jeffmann prefers child-like type explanations, hence you are a perfect match.

I am sure that Jeffmann wants to know why in post #201 the bead seems to travel on a perfect straight line trajectory.

Any easy child-like type of explanation for Jeff ?

 

[puttmad]

Mmmmmm,

I wish (hopefully) that one of you could explain how us mere mortals can transfer this "information" into a workable exercise in hitting the golf ball more effectively...This is a golf forum after all..
Can we just have a simple explanation of how we incorporate the "information" to maximise our efficiency in playing golf and striking the ball better, without having to resort to EXEL sheets, or computer graphs, or scientific formulae...!!!!!!!

 

[nmgolfer]

Puttmad

Nobody's got a gun to your head forcing you to read this thread. If it bothers you that much then DON'T READ IT! Furthermore... Take responsibility. Nobody owes you anything. If you can't figure out how to use information that is freely provided that is your problem.

To the others'. The answer in 201 is wrong. Because the bead is still constrained to move with the rod even though no forces are acting on it, it doesn't move in a straight line. The exact path depends on the angular velocity, lengths etc. Email me if you want to see a solved example.

For the record. This has nothing to do with rotating coordinate system. Its a language issue. It is wrong to refer to a component of acceleration in an inertial coordinate system as "centrifugal force" when that usage is inconsistent with the word's clearly defined meaning(s). Misuse of the language has left thousands (possibly millions) of golfers with the misguided notion that there is a force at the end of the club which causes it to accelerate. There IS NO FORCE OUT THERE! Crazy Jack Kuykendall is not so crazy afterall. The only force (not including gravity) the club experiences is what the hands impart. This is why hand path and hand velocity profiles are SO IMPORTANT. In-fact they're everything! The answers are starring you in the face puttmad. (You can lead a horse to water but you can't make it drink)

mandrin.. we will have to agree to disagree.

 

[mandrin]

quod erat demonstrandum

nmgolfer,

You are doing me the honor to bestow upon me a large and interesting variety of insults. Moreover you are now also threatening ‘to smack me up side the head with your seven iron’, (why a 7I ?), hopefully just to test if inertial forces are fictitious or not.

I rather prefer mathematics, the universal language of true scientists, and have used it again to show the obvious errors in your thinking. This time you really have painted yourself solidly into a corner.

I did not expect such gross errors. I hope that colleagues are not aware of it as the security guard might just shovel you through the front door the next time he sees your face.

It was fun to have a closer look at your post #201 and show everyone how weak and meager your scientific arguments really are. It is nice to take on as mission to dispel myths but it presupposes some knowledge otherwise you are making quickly a fool of yourself.

 

[Biffer]

mandrin,

As I see it, in relation to the rod, the bead slides along the rod in a straight line and when it departs from the rod it continues to move in a straight line, in the direction the rod is pointing at the time, not tangent to the arc. Do I see it correctly?

 

[nmgolfer]

mandrin ... day late an dollar short

mandrin

you are a day late and a dollar short. I realized the error in that post yesterday and deleted it this am (see my post from earlier today). I can confirm that you have the right answer for a contant angular velocity problem. I'm still troubled by your constant angular acceleration solution however.

You claim (eqn. 4) the radius function is e raised to a power of (-1/2t^2 alpha) then the software doing the thinking for you spits out some hypergeometricF1 function mumbo jumbo. What the bleep is that? Are we supposed to know? Heres the problem... When you take the second derivative of the e^(-1/2t^2 alpha) you get TWO terms. What happened to the second term mandrin? You are not fudging your answers are you?

If you would show your work we could inspect to our own satisfaction whether or not what you say is true. Since I already know you are prone to making simple mistakes please understand why I have serious doubts regarding the veracity of your results.

On another note... you like this little game don't you... pose a problem that you spent weeks pondering and have already answered then sit back expecting people to bow down before you and kiss your toes. I'm not impressed. After-all anyone that does not understand centrifugal force doesn't understand mechanics. I think the truth is you would be crippled with out your math package and that's why you never show the details.

PS I was kidding about using my seven Iron on your noggin. I make a habit never to hit rocks with it.

nmgolfer,

You are doing me the honor to bestow upon me a large and interesting variety of insults. Moreover you are now also threatening ‘to smack me up side the head with your seven iron’, (why a 7I ?), hopefully just to test if inertial forces are fictitious or not.

I rather prefer mathematics, the universal language of true scientists, and have used it again to show the obvious errors in your thinking. This time you really have painted yourself solidly into a corner.

I did not expect such gross errors. I hope that colleagues are not aware of it as the security guard might just shovel you through the front door the next time he sees your face.

It was fun to have a closer look at your post #201 and show everyone how weak and meager your scientific arguments really are. It is nice to take on as mission to dispel myths but it presupposes some knowledge otherwise you are making quickly a fool of yourself.

 

[nmgolfer]

mandrin,

As I see it, in relation to the rod, the bead slides along the rod in a straight line and when it departs from the rod it continues to move in a straight line, in the direction the rod is pointing at the time, not tangent to the arc. Do I see it correctly?

I know you are asking mandrin biffer but no... the bead has two velocity components: tangential and radial. The corresponding magnitude of each determines the angle it moves off in a straight line ---once it reaches the end of the rod---. My answering you gives mandrin yet another chance to try to "decorticate" my posts. Its my act of charity since we all know he so revels in such opportunities.

 

[Biffer]

nmgolfer,

Thanks for answering my query.

 

[Bronco Billy]

F=ma

For those being intrigued by nmgolfer's strong denial of centrifugal force I propose a simple thought experiment. Look up in the dictionary the word 'accelerometer'. It will tell you that it is a device used to measure force.

If one can measure something we will be definitely forced to admit its existence, that is inherent in a logical common sense approach necessary and required in any scientific endeavour.

So let's imagine to purchase two micro wireless accelerometers and we subsequently glue them on the bead, taking care to align their measuring axes respectively parallel to the rod and perpendicular to the rod.

The measurements obtained will be as shown in Fig2, ie., zero force perpendicular to the rod and a net varying centrifugal force, corresponding to the colored arrow, along the rod.

Hi There

To Me If the Above is True it is Indisputable that There is a Parallel Force and No Perpendicular Force as Depicted by the Diagram.... As Far as I Know Most Call this Parallel Force-CENTRIFUGAL.....

Cheers

 

[mandrin]

mandrin,

As I see it, in relation to the rod, the bead slides along the rod in a straight line and when it departs from the rod it continues to move in a straight line, in the direction the rod is pointing at the time, not tangent to the arc. Do I see it correctly?

Biffer,

The bead leaves the rod exactly with the end conditions it has at that precise moment of time. It is hence moving neither tangentially nor radially. A picture is always nice to look at and it shows this clearly.

It is the same approach as for the case of a particle circling around a center and cutting the restraint; the mass will continue with the conditions at that moment, hence moves away tangentially.

 

[nmgolfer]

Deception... last bastion of scoundrels

BB,

Are you talking about mandrin's fig 2 in his attempted decortication of my 201 post? If so then you should know that, although he conveniently failed to mention it, mandrin switched to a rotating (non-inertial) coordinate system for that so called "analysis". The coriolis acceleration (2 omega X Vrel... which I might add, mandrin shows pointed in the wrong direction) only comes into play in certain a rotating coordinate systems. Futhermore in a rotating coordinate system one MUST impose the fictitious centrifugal force, as mandrin has correctly shown, in order for the results to be correct.

So you see BB, all though he (may be) technically correct in some round-about obscure way, and since he as usual provided virtually no details (which are needed to falsify or validate lest one want to waste countless hours attempting to recreate the mental machinations of a well....), he achieved that result (a valid use of the words centrifugal force) only through deception and trickery.

When I say (it looks like) mandrin has the right answer, I refer specifically to his figure 1 picture.

Its too bad mandrin has already stated in this forum that there is no need for non-inertial (rotating) coordinate systems in the analysis of a golf swing otherwise he might well have found a ligitimate use for the words "centrifugal force" in some obscure analysis of release and been able to extricate himself from the hole he dug himself into during the Ringer debate.... but I still seriously doubt it.

snip....

Hi There

To Me If the Above is True it is Indisputable that There is a Parallel Force and No Perpendicular Force as Depicted by the Diagram.... As Far as I Know Most Call this Parallel Force-CENTRIFUGAL.....

Cheers

 

[Bronco Billy]

BB,

Are you talking about mandrin's fig 2 in his attempted decortication of my 201 post? If so then you should know that, although he conveniently failed to mention it, mandrin switched to a rotating (non-inertial) coordinate system for that so called "analysis". The coriolis acceleration (2 omega X r) only comes into play in certain a rotating coordinate systems. Futhermore in a rotating coordinate system one MUST impose the fictitious centrifugal force, as mandrin has correctly shown, in order for the results to be correct.

So you see BB, all though he (may be) technically correct in some round-about obscure way, and since he as usual provided virtually no details (which are needed to falsify or validate lest one want to waste countless hours attempting to recreate the mental machinations of a well....), he achieved that result (a valid use of the word centrifugal force) only through deception and trickery.

When I say (it looks like) mandrin has the right answer, I refer specifically to his figure 1 picture.

Its too bad mandrin has already stated in this forum that there is no need for non-inertial (rotating) coordinate systems in the analysis of a golf swing otherwise he might well have found a ligitimate use for the word "centrifugal force" in some obscure analysis of release and been able to extricate himself from the hole he dug himself into during the Ringer debate.... but I still seriously doubt it. Centripetal/centrifugal by definition is perpendicular to path.

Hi nm

Reread my Post(The Part I Quoted from Mandrin's Post) I was saying Nothing About His Graphs or Mathematics...... All I said was that if the Accelerometer Read as he Said in his Little Experiment then Centrifugal Force Exists..... Can You Dispute Centrifugal Force if the the Accelerometer Readings measured what they did in His Little Experiment?

Cheers

 

[nmgolfer]

Hi nm

Reread my Post(The Part I Quoted from Mandrin's Post) I was saying Nothing About His Graphs or Mathematics...... All I said was that if the Accelerometer Read as he Said in his Little Experiment then Centrifugal Force Exists..... Can You Dispute Centrifugal Force if the the Accelerometer Readings measured what they did in His Little Experiment?

Cheers

I don't dispute the bead accelerates. I showed you why mathematically it must acceleratehere. I also showed you thatin that case there is only one force acting on the bead, that being due to the rod.

In the current case (constant angular velocity) the bead accelerates, because it has mass and because mass has interia and because it is rotating but there are no forces acting on it. Accelerometers measure acceleration despite what some web page says.

Its a diction thing. If you believe like I do that words have meaning and that its not appropriate to pervert those meanings to ones personal advantange then we must work with the words (language) we're given. In an earlier post mandrin agreed like I do with the only two valid uses for the words "centrifugal force" that are listed in wikipedia. That means he cannot turn around and use the term inappropriately (and then excoriate those who do use it appropriately like J. Kuykendall et al).

You see BB the words acceleration and force are not synonymous. The bead accelerates but there's no CF.

PS.. he didn't conduct an experiment... it was purely hypothetical