Centripetal Vs. Centrifugal

[Lu Zuckerman]

Who says that old dogs never learn? From what has been stated in many of the last 100 posts there is a real force called centripetal force and an imaginary force called centrifugal force.. .. .Here is what I think is happening relative to the centrifugal clutch and I may be completely wrong but that is not news to many of you.. .. .Let’s imagine that the driving portion of the clutch consists of 4 shoes attached to a central hub by pivot pins and possibly restrained by springs. Now let’s remove the springs because that is what they did on the Bell 47 models. When the engine starts and comes up to speed the hub rotates with the engine causing to shoes to rotate with the hub. Imagine now that there is no drum or driven member. Because of the mechanical forces involved the center of mass of each shoe will align itself with the attaching pin (shoes are now disposed outward from the attaching pin) forming a crucifix.. .. .The force causing the rotation is the centripetal force and the reacting force if it could be measured is the centrifugal force, which is caused by the shoes. There are several other factors that influence the centripetal and centrifugal forces and they are the weight of the shoes and the speed of rotation.. .. .Now lets place this clutch in a Bell 47. When the engine starts the shoes will try to align themselves with the driving pivot pin but they can’t so they will move outward until the clutch drum restrains them. Since the engine is not up to speed the rotor will immediately start to turn but in the process the clutch is slipping somewhat because of rotor inertia. Looking back at the previous illustration if the drum were not there the shoes would not form a crucifix but would hang back somewhat because their inertia is too strong allow the shoes to fly outward to form the cruciform shape. As the engine speed increases the forces that want the shoes to fly outward are intensified and the friction lock is so intense that the driven member is at the same speed as the driving member is and there is no slippage.. .. .To sum it all up, Centripetal force is the force that causes objects to rotate about a central axis. Because the rotating elements have weight (mass) the rotation caused by the centripetal forces will create a force that can be measured and this measurement is a factor of speed of rotation and the weight of the object being rotated. The faster the speed the higher level of measured forces or, the higher the weight at a given rotational speed the higher the measured forces.. .. .If you are the driving member the force you feel pulling against you is centrifugal force and if you are the shoe the restraining force you feel is centripetal force. If you could measure these forces they would be the same.. .. .Now here is a problem for you scientific types. Instrument a rotor blade so that at each given blade station from the root to the tip there was a strain gage to measure the forces and then brought the blade up to speed the forces would go from max to minimum from the root outward to the tip. Is this correct or not. Why?

[Dave Jackson]

Lu,. .. .You're mixing apples and oranges. <img border="0" title="" alt="[Smile]" src="smile.gif" /> . .. .The clutch is an example of centrifugal force, since the drum is attempting to PUSH the break pads inward.. .. .The rotor is an example of centripetal force, since the inner portion of the blade is attempting to PULL the outer portion inward.. .. .The following is the answer to your 'Why'. <img border="0" title="" alt="[Wink]" src="wink.gif" /> . ._____________________ . .. .The shoe must have mass for the clutch to work.. .The drum does not need to have mass.. .. .The equation for centripetal force is:. .F = M * v^ / r. .ie. Force = Mass time velocity squared divided by radius.. .. .Therefore, because the drum does not need to have any Mass it cannot impart a Force. <img border="0" title="" alt="[Eek!]" src="eek.gif" />. . . . <small>[ 27 March 2002, 20:14: Message edited by: Dave Jackson ]</small>

[Lu Zuckerman]

To: Dave Jackson. .. .It is you that is mixing apples and oranges.. .. .Here is the question I posed.. .. .Now here is a problem for you scientific types. Instrument a rotor blade so that at each given blade station from the root to the tip there was a strain gage to measure the forces and then bring the blade up to speed the forces would go from max to minimum from the root outward to the tip. Is this correct or not. Why?. .. .Here is your response, which is confusing as it deals with blade stations and centrifugal clutches. Is the formula you posted below your answer to why?. .. .The following is the answer to your 'Why'. . ._____________________ . .. .The shoe must have mass for the clutch to work.. .The drum does not need to have mass.. .. .The equation for centripetal force is:. .F = M * v^ / r. .i.e. Force = Mass time velocity squared divided by radius.. .. .Therefore, because the drum does not need to have any Mass it cannot impart a Force.

[heedm]

Flight Safety,. .. .The drum is applying a force to the pads. That force causes the pads to accelerate towards the center of rotation.. .. .If you throw a baseball, prior to it leaving your hand there is friction between the baseball and your hand that prevents it from slipping (not rolling) across your palm. Your hand is pushing on the ball because you created a force that is transmitted through your hand. The ball does not create a force. Yes, you feel the reaction on your hand, but that isn't a 'real' force. In analogy, the drum is your hand the pad is the baseball.. .____________. .. .Lu, . .. .You're getting there.. .. .______________. .. .Dave,. .. .Push pull doesn't matter. They are both inwards. They are both centripetal.. .. .Whether the drum has mass or not is irrelevant because we aren't considering this as being accelerated inwards (we can...but we're not), just the pads. In reality the drum has mass. If it didn't it wouldn't be there. It is not always valid to extend physics to a singularity.

[Dave Jackson]

Sorry Lu,. .. .My posting wasn't meant to be taken too seriously.. .. .It was an attempt at light humor and the second part should have been directed toward the more knowledgeable people in this field such as heedm et al.. .______________________. .. .The following is serious. <img border="0" title="" alt="[Smile]" src="smile.gif" /> No joke.. .. .You said ".... the forces would go from max to minimum from the root outward to the tip. Is this correct or not. Why?". .. .Yes, it is correct.. .. .The table and related notes on <a href="http://www.unicopter.com/1069.html#Axial_Force" target="_blank">Centrifugal Force</a> should give the answer. The forces shown are for each blade element, then they are summed at the bottom of the table.. .. .You will notice that individual element forces are greater in the outer elements (-&gt;tip) then they are in the inner elements (-&gt;root). BUT, the strain gauges read the SUM of all the element forces outside of their location. Therefore, the strain gages near the root will show higher values then those near the tip.. . . . <small>[ 28 March 2002, 01:34: Message edited by: Dave Jackson ]</small>

[helmet fire]

Now we have two new ones, bringing the grand total to 24 examples of the exact same thing. The new ones are:. .23. Centrifugal clutch.. .24. Baseball in your hand.. .. .Flight Safety: HAVE YOU CARRIED OUT MY EXPERIMENT? I shall atempt a final answer for you. I believe your example of the clutch is a great demonstration of centripetal (like the other 23). The flaw in your arguement is that the pads are not "flung outward". Despite 8 pages of why this is incorrect, you remain committed to this viewpoint. It is the crucial element to a centrifugal theory.. .. .The reality remains that the pads are not flung outwards. They attempt to move in a staight line but cannot because the drum is forcing them to turn. Turning equals a change in direction, which is a change in velocity, which requires the application of an acceleration, which can be considered a force. Acceleration is directional, therefore to turn, must accelerate, and accelerations act toward the change, therefore it must act toward the direction of turn. In a turn that continues into a complete circle, the centre of that circle is where the acceleration must act. IE the force is centripetal.. .. .There is no "flung outward". There is only the force required to turn the mass of the pads.. .. .The experiment I outlined is VERY SIMPLE, and takes less than 1 minute. PLEASE carry it out before thinking up another example.. . <img border="0" title="" alt="[Confused]" src="confused.gif" /> <img border="0" title="" alt="[Confused]" src="confused.gif" /> . .. .The experiment reposted:. .Everyone whom reads this thread need only carry out one simple experiment to see if the world’s physicists are correct in believing centripetal is real and centrifugal is not. Please carry this experiment out before posting again on this thread.. .. .Get a large easy to see object with some weight, such as a full soft drink can or water bottle. Go out side with a friend. Face a very obvious marker, such as a distant white building or even use your friend. Hold the object out at arms length from your body, DIRECTLY TO YOUR FRONT, so that as you turn your body, your arm stays straight out in front. Spin around. When you and your hand are exactly facing your marker, open your hand.. .. .PLEASE DO THIS BEFORE POSTING AGAIN ON THIS THREAD. <img border="0" title="" alt="[Smile]" src="smile.gif" /> <img border="0" title="" alt="[Smile]" src="smile.gif" />

[Lu Zuckerman]

To: Helmet Fire. .. .First of all, I’m 71 years old and I can’t spin as described in your post. If I could spin my French Canadian neighbors would think me an idiot. I assume that if I were to release the water bottle when facing my target the bottle would not move towards the target but at a tangent in the direction of rotation. Am I correct?. .. .Now please explain at what point an individual (David or some young Palestinian kid) must release his slingshot in order to hit a target standing in front of him. Remember also (if it makes any difference) the sling shot is not rotating in a perfect circle but in an oval shaped pattern which adds momentum to the rock being slung outward.

[helmet fire]

Lu,. .Perhaps you could ask your French Canadian neighbours wife to spin for you in a short skirt so that you dont feel like 71 anymore! . . <img border="0" title="" alt="[Big Grin]" src="biggrin.gif" /> Mind you, how would you keep your eyes on the object?. .. .You are right according to my results - the object does not fly outwards at all, it flys tangentially. Thus David would need to release the rock when the giant was in line with the tangent, ie when the rock is to the side of David (assuming a horizontal rotation - oval or not is irrelevant).

[Flight Safety]

Helmet Fire, I guess I don't need to introduce any more examples, but I'd like to comment on your experiment. You stated it this way...

Get a large easy to see object with some weight, such as a full soft drink can or water bottle. Go out side with a friend. Face a very obvious marker, such as a distant white building or even use your friend. Hold the object out at arms length from your body, DIRECTLY TO YOUR FRONT, so that as you turn your body, your arm stays straight out in front. Spin around. When you and your hand are exactly facing your marker, open your hand.

OK, instead of performing this experiment, let's just diagram it. We'll assume that the direction of rotation is counter-clockwise when looking from above. The diagram will consist of 3 points, 1 circle, and 3 lines. The explanation that follows may sound a little complicated, but actually it's very easy to free hand draw the diagram.

First, on a sheet of paper (oriented landscape style in front of you), draw a point (point A) in the horizontal center about 2/3 down from the top. This represents where your feet will be when you spin. Draw a circle around point "A" with a radius that represents a scale of the distance to the hand on your outstreched arm. Lets say the radius is 2 feet for simplicity. Next, draw another point (point B) in the horizontal center, towards the top of the page directly above point "A". This point will represent your friend at a scale distance from you of say 10 feet.

Next, draw a line (line A) that is vertical on the page, starting at point "A" and passing through point "B". The point (point C) where this line (line A) intersects the circle, is the location of the release point for the soft drink (or bottle of water), directly in front of your friend. Next, draw a line (line B) that is horizontal on the page and passing through point "A". This line then, should pass through the center of the circle, and also be perpendicular to line "A". Finally, draw a line (line C) that is horizontal on the page and begins at point "C" (on the circle) and draw it towards the left edge of the page. This line should be parallel to line "B".

Now you've finished the diagram.

You said in a response to Lu...

You are right according to my results - the object does not fly outwards at all, it flys tangentially.

You are partly right and partly wrong. If when you said "the object does not fly outward at all", you meant that the object does NOT follow line "A" directly towards your friend, then yes you are correct. No force exists in this experiment to cause the soft drink to turn 90 degrees to the right, immediately upon release, in order for it to impact your unsuspecting friend.

You (and Lu) are also correct when you say that the soft drink will fly "tangentially". However, you are wrong if when you said "the object does not fly outwards at all", you meant that the object does not display ANY motion away from the center of rotation. It does in fact display this motion, because it's new path of motion, immediately after release, will be along line "C". Remember that this line is parallel to line "B", the line that runs directly through the center of the circle (of rotation). While there is "tangentail" motion relative to the center of rotation, the motion is still "largely" away from that center. The object's resistance to "directional" acceleration (caused by its own inertia) is still trying to move the object away from the center of rotation, even if that motion is along line "C" and is partly "tangential".

While the object is in rotation, point "C" (the release point) will keep moving around the cicle until a release actually occurs. The object's resistance to "directional" acceleration will cause it to continually try to follow this path (represented by line "C"), and the force this creates against the "centripetal" pull (or force) is very real, and is in fact the "centrifugal" force.

So in the end there is still enough of this motion away from the hub (or center) to make devices like the centrifugal clutch work.

(edited for spelling and typos)
(edited again to reformat the text into normal paragraphs)

[helmet fire]

Wow.
We may be getting somewhere.

It seems your reason for centrifugal is that a tangential travel along line C appears to take the object away from the center at point A. In fact, you are incorrect in terms of rotational dynamics. Travel along line C may be farther away from point A, but it is not increasing radius distance at all.

Here we go with frames of reference again. Remember above we discussed the fact that frames of reference are irrelevant to the reality of the situation: i.e. just because you view it from a different point, reality is not suddenly altered.

So, if you were standing STILL at point A (not spinning), and facing line B, and simply threw the ball out in front of you, what happens? The ball moves away from point A doesnt it? In fact, if your arm action released the ball two feet from you, it would even follow line C wouldn't it? In other words, the ball follows the same path away from point A, even though no rotation was applied. SO WHERE IS THE CENTRIFUGAL FORCE?

Just because you change your reference point (spinning or not spinning), reality has not changed. The ball moves along line C for the same reason in both cases: it is released with the same motions applied to it. If it had different forces on it, the ball would necessarily follow different paths, but in our case it will not.

So, the only thing you are changing by spinning around is your frame of reference, not the forces on the ball at immediately after release - thus centrifugal cannot be supported by your arguement.

An easier way to understand this would be to imagine if you were on a moving car whilst carrying out the experiment (ie to change your frame of reference). If the car was moving parralel to line C it should be very easy for you to see that the ball does not get farther away from point A after release - thus disproving your theory.

Hopefully by now you can now see that frames of reference do not alter the forces acting on the object.

Back to your clutch now. If I get a large elastic band and wrapped it around a huge fat chick, joined the ends together and tried to pull her along, the rubber band stretches. Lets say this chick is so fat that the rubber band breaks before I can move her, can you really say that SHE applied a force to the rubber band? Obviously not. Now lets say she is running along the road and I am running along parrallel with her. Just for fun, I suddenly swerve and run away from her, and stretch the rubber band. Again the rubber band will break before she alters her course. In other words there is no difference between the two cases is there?

Now - when we were running along, did she apply ANY force to the rubber band? No. Same as the first case. The fact that she was running along a straight line is irrelevant.

Now lets say I get a rubber band that will not break. When she stands still, I pull on the rubber band and she moves toward me. Has she applied any force to the rubber band? No.

Now we are running again, but when I pull the rubber band and it doesnt break, but in fact she begins to move toward me, has she applied any force to the rubber band? Again - NO. Same forces, different frames of reference.

Now lets say I elect to stand in the same point and continue to pull the rubber band whilst she continues try and run in a straight line, and I do it four times so she has travelled in a sort of square around me. This is no different than the case of one quick pull as above is it? So has she magically started to exert a force on the rubber band just because I apply my pulling force at four seperate times? Of course not, because each time is a seperate occurence of the example as if she was standing still and I pulled her toward me - but we have just changed the frame of reference.

Now I pull her twenty times and she travels around me in a many sided twentyagonal kind of shape. Has she exerted any force on the rubber band? No. Still no. And if I pulled continuously, she would travel around me in a circle. Has this suddenly made her apply any force to the rubber band? NO NO NO NO. Therefore, centrifugal is NOT a force. Substituting the fat chick for your clutch pads, you can see that the pads apply no force of their own to the outer rim (elastic band).

But what force am I applying to the rubber band when she is standing still? I am pulling her TOWARD me. Now that you know frames of reference do not change reality, when we are turning am I not still pulling her TOWARD me (into the center)? Therefore Centripetal IS a force.

Simple hey?

Sorry for example 26. I try not to pull fat chicks now that I am older.....


edited for new format and some semblance of relevance (!!)