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20 June 2019 00:38

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Asked by: Glenn Hawkins
Subject: My Chubby Checker experiment (Do the twist)
What happens when two gyros are set to the opposite tips of a shaft and while spinning in unison, one is let to fall into gravity; while the other is caused to rotate in line with the shaft; try clockwise and counterclockwise rotation? Note; The in line rotation on the shaft side is not allowed to move upward, or downward in the vertical plane.
Date: 30 May 2014
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Answers (Ordered by Date)

Answer: Blaze - 01/06/2014 03:43:25
 If they are both spinning in the same direction and allowed to precess in the direction they want to, then the device will precess much like an overhung gyro but it would precess more slowly than an overhung gyro of the same arm length with only one flywheel at the end of the arm.

I don't know what you mean by "while the other is caused to rotate in line with the shaft; try clockwise and counterclockwise rotation".


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Answer: Glenn Hawkins - 01/06/2014 04:07:53
 Its difficult to paint the picture. . . . while the other is caused to rotate in line with the shaft;

Suppose there is a ring on the tip of one end of the shaft with the ring opening to forward and rearward, which would be at a 90 degrees angle to the shaft itself. The gyro is places inside the ring and locked to it so that it and the ring are rotated at a 90o angle to the spin of the plain of the wheel. It would be gimbles like, except the gyro is locked against precessing inside its ring. If this doesn't makes sense I will try again.
Regards Glenn,

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Answer: Blaze - 01/06/2014 17:43:24
 Sorry Glenn, I am still not quite getting it. Does the shaft start out horizontal? Is the ring vertical or horizontal? Is the gyro in the ring centered in the ring or on the edge of the ring?


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Answer: Glenn Hawkins - 02/06/2014 16:26:15
 Yes, Blaze, it is a little difficult explain and picture.

One vertically rotating flywheel sets on the end of a shaft and will be allowed to fall the same as a gyro, but the free end of the shaft is setting on a table, not a pedestal.

Another vertically rotating flywheel with ring guard is attached to the opposite 'free end' of the shaft, sets on the table spinning at supper, supper speed.

The slower speed flywheel end is dropped and the tendency to precess occurs.

The super speed flywheel resting on the table resist being twisted by the precession tendency of the dropped flywheel.

Lots of countering forces begin and I will leave this puzzle for any, primarily you, to enjoy solving. Its probably easy for you, but there are several answers all which depend on the variation of momentum assigned to each flywheels.

Cheers, Glenn

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Answer: Glenn Hawkins - 11/06/2014 00:34:01
 This is the most important question I have ever seen on here and and full of odd mechanics.

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Answer: Blaze - 12/06/2014 02:32:07
 Hi Glenn. This scenario sounds very much like your most recent question "Challenging Questions". It would be easiest to think of this as two "systems" where one of the gyros for each "system" is just considered as dead weight and not as a spinning gyro. The system with the gyro that is allowed to fall would precess at a certain rate based on the system parameters. The system with the high speed gyro would want to precess at a slower rate. The falling gyro would be the weight that is making the high speed gyro want to precess. The actual precession rate would be a compromise of the two precession rates.


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Answer: Glenn Hawkins - 12/06/2014 03:43:26
 Hi Blaze,
Yes, the questions are similar, yet different. This question would continue in stages and eventually lead to the high speed gyro attempting to precess as in a gimbles rings, yet not being allow to precess at all. It is too complicated, because it is unnecessary, as I hope to advance this question into the last question 'Challenges' I posted in this one's place. Let us forget this one and pounce on the last one, but yes to your observation that separate rates cause uneven, imbalanced results on the. Good!


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