Question |
| Asked by: |
Luis Gonzalez |
| Subject: |
Riddle about Spin Smarts |
| Question: |
This puzzle took me a couple of hours, and I was only able to resolve it after I wrote down the problem. Other brighter contributors to this forum will probably resolve this challenge much quicker, or may already know the answer beforehand from their experience with spinning objects. Which are you?
A baton twirling Clockwise (CW) simultaneously releases 2 mass weights, which were attached at both ends.
Centrifugal force sends both masses off in the directions in which the baton-ends were pointing when the masses became detached.
Repeat the baton experiment but this time while the baton is twirling in the opposite direction, counterclockwise (CCW).
An observer, whom cannot see the direction of baton’s spin knows that one spin was CW and the other was CCW.
How can the observer determine which masses were released from the CW event, and which ones were released from the CCW event?
Best Regards,
Luis G |
| Date: |
19 November 2011
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Answers (Ordered by Date)
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| Answer: |
Glenn Hawkins - 20/11/2011 03:49:41
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| | You might wish to rethink this.
"Centrifugal force sends both masses off in the directions in which the baton-ends were pointing when the masses became detached."
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| Answer: |
l - 20/11/2011 15:01:29
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| | Thanks Glenn,
First thing I questioned was whether the spin direction would cause a slight deviation.
However this was an error that cost me the good part of an hour.
Centripetal/Centrifugal rules are not deviated by the direction of spin.
Regards,
Luis G
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| Answer: |
Glenn Hawkins - 20/11/2011 16:54:12
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| | Hey Luis,
That one is a bit of a mind twister. I once had an interesting week investigating several of those conditions at a fair grounds. Excuse me as I explain several things that you already know, but this is the only way I know how to do it.
I though I would supply an example and was surprised I could find nothing clear enough in the internet.
http://en.wikipedia.org/wiki/Centripetal_force
Look to he blue circle illustration at the right just below the top. That arrow points the way the weighted end would travel if the string broke. It is not what one might expect.
Objects wish to continue in a straight path. The force (string) that pulls them toward the center of rotation is a fictitious force call centripetal. Centripetal is the reason the weighted end must travel at a curving right angle to the pivot point-- instead of a straight outward. Centripetal is the force that curves the momentum from continuing forward-- not outward. As the momentum is forced to curve it resist in the direction opposite of centripetal (equal and opposite) and this resistance is called centrifuge. The instant centripetal (inward) force ceases, centrifuge (outward) force ceases for there is no longer an equal and opposite condition and the object is allowed to continue in a straight line. Again this line is at a right angle to the pivot, not in a straight line outward from it.
You could definitely tell whether or not the rotation had been clock wise, or counter clockwise.
If this is useful, you are very welcome,
Glenn
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| Answer: |
Luis Gonzalez - 20/11/2011 17:30:17
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| | Hi Glen,
The instantaneous angular velocity (and momentum) of the satellite (or object on string) is always depicted by the tangent (blue arrow). (This direction is forced by the centripetal force.)
However the Centrifugal force is exactly opposite to the Centripetal force.
When the string breaks, the object’s motion follows the direction of the applied force (as the centrifugal force converts into motion). Think of a sling shot; where would you release?
(By the way, in my question/example, the equal and opposite reaction (E&OR) is the motion of the other mass, which was released in the opposite direction.)
We can disagree, as we often do (each knowing we are correct…).
Perhaps others can chime in with their perception of the truth.
Regards,
Luis G
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| Answer: |
Glenn Hawkins - 20/11/2011 19:16:18
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| | Very Good Luis,
Disagreements with a smile and an “I don't care much” work for me. We've matured a bit with age.
My brother had a 'David and the Philistine Giant' sling. He could almost knock down a wooden fence in the back yard with it. He was a big man. I took your advice anyway and whorled a compass I have tied to a long string. It travels right angle to release just like my brother's swing and aim. That is, if you released the weight and string directly at Joe standing directly in front of you, it would hit Mack standing beside you. Poor Mack. Give it a try if you like. It's easy to determine. It releases exactly in the direction indicted by the arrow in the diagram.
Otherwise, how's life treating you these days?
Regards,
Glenn
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| Answer: |
Luis Gonzalez - 20/11/2011 20:10:40
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| | Thanks for your stories Glenn.
Life is good.
I am still building and testing devices to test the various aspects and components of my design.
Whishing you well,
Regards,
Luis G
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| Answer: |
Glenn Hawkins - 21/11/2011 01:36:10
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| | Very good Luis,
Glad to hear it and someday you must tell us of your design. Good to hear from you.
Keep on keeping on.
Best regards,
Glenn
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| Answer: |
Luis Gonzalez - 04/12/2011 03:08:55
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| | First of all I want to say, thanks for pointing me in a different direction.
None of my slingshot tests go to the side, and they don’t go straight forward either.
All my tests consistently follow a diagonal path (between straight forward and sideways (more or less 45 degrees).
I had to figure it out from my own perspective (limited as it may be).
So, after some calculating and a spreadsheets here is what I got:
We know Centrifugal Acceleration is proportional to the square of the angular velocity (Ac = WWR).
We also know that Kinetic energy is also proportional to the square of velocity (Ek = VV/2) which is equal to (WR)(WR)/2.
I confess my difficulty explaining conversions of Kinetic energy to Acceleration, so I won’t even try. But I know that if I increase the spin rate both kinetic energy and centrifugal force will increase more or less at a similar rate (exponentially).
Therefore science tells me that anything shot by a centrifuge will travel in a direction that is motivated by both the kinetic energy (sideways), and by the centrifugal force (straight ahead).
I leave it to someone smarter or better trained in physics to provide a more exact path.
Best Regards,
Luis G
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| Answer: |
Glenn Hawkins - 05/12/2011 04:29:17
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| | Dear Luis,
The questions you pose are good. If there would be no centrifugal force as Sandy’s findings state, then there would be perhaps no equal and opposite force at the two ends of the shaft of a gyroscope overhung. Equal and opposite would be that one shaft end of the gyroscope is trying to precess counter clock wise, while the other is trying to precess in the opposite direction. That is to say, the pedestal would also be attempting to rotate around the center of the gyro’s mass, but cannot because of table friction. If this is not true then I think propulsion would be possible. That is the one deciding factor to me.
As a direct answer to what you are working on try this site. About halfway down the lesson you will see the question (166) later you will see the answer (170).
http://www.mcbridehq.com/ClassHP/LAHS/APPhys/Handouts/Gravity/2%20-%20Centripetal%20Force.pdf
It is a worthy question you are working with. Glad to see it.
Best Regards, Glenn
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| Answer: |
Luis Gonzalez - 10/12/2011 02:09:36
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| | You are absolutely correct on this one Glenn, and I found the reason for my consistent test discrepancies, using a long slingshot in a short target range.
Thank you for your first explanation, which was clear right from the very start.
I also think the potential absence of centrifugal force can re-open interesting discussions.
Best Regards,
Luis G
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| Answer: |
Glenn Hawkins - 10/12/2011 04:02:04
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| | Thank you Luis. Likewise all this was very tricky for me too when I first encountered it and I spent much time playing with it in my mind. You said, “I also think the potential absence of centrifugal force can re-open interesting discussions.” It is also interesting that I came up with the same, but independent thinking. I started to write twice about this potential absence and I found that to contemplate how it might work is extremely complicated. Maybe one of us will start a new thread on the subject. I’m in no hurry. Go for it if you like. Start one when you are ready.
Best Regards, Glenn
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Luis Gonzalez - 24/12/2011 18:18:02
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| | On 3 occasions I started writing about the effects of mixing centrifugal with gyro spin, and wound-up in the same place every time. In fact this item has been previously discussed a few times in this forum. Unfortunately the explanations are virtually impossible to follow through complete conclusion, even if each one of us knows exactly what we are talking about. It’s no-wonder when there isn’t even common terminology.
All I will say is that the centrifugal force exists only as long as the centripetal force is acting (we know the directions of both these forces).
I have previously explained my opinions regarding what occurs among these CC forces (and other forces) in a “hub configuration” when spin is involved, and think am through pursuing those dynamics. Thanks to Glenn I also found conclusion my last related quandary on a more basic but complicated question (i.e. direction of items release from spin or rotation). Thanks Glenn.
I am still interested in exploring other spin configurations that have not been explored previously in this forum.
Merry Christmas to all,
Luis G
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| Answer: |
Glenn Hawkins - 25/12/2011 22:28:40
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| | Merry Christmas, Luis. May good luck shine down on you and yours.
Glenn
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| Answer: |
padmg - 29/12/2011 22:46:19
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| | The centripetal force at the very centre has no bearing what so ever on fluidicity of its neighboring enviroment...but as like a black hole all proton yeild will be absorbed with more greed of consuption as before twas black hole in its explosive repultion all heavier proton mass ejected faster on the edge of the proclastic ripple,leaving an electron enviroment ie gasseous
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| Answer: |
padmag - 06/01/2012 04:52:36
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| | louiz gonzalez
The ans is it would be slightly to the left or right of top dead centre, the next indiff would prob show 363.275 degrees x
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