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Conservation of momentum 
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#91
May1414, 08:54 AM

P: 125

Then at the beginning (I mean, when the wheel is spinning vertically) the total angular momentum of the system (system = platform + girl + wheel) is equal to the angular momentum of the wheel, which is nonzero, something like (L, 0, 0) with respect to that XYZframe. If she were isolated in outer space (no need of the platform in this case), as she tries to change the plane of rotation (just as she does in the video) she herself would start to spin, both with a Zcomponent (negative or downwards) and with a Xcomponent (positive). Why? Because as she is trying to change the plane of rotation of the wheel, what is happening with the angular momentum of the wheel is that it is changing from something like (L,0,0) to something like (La, 0, b) with "a" and "b" being positive given numbers. So the angular momentum of the girl (in outer space we don't need platform at all) will change in exactly the opposite sense, from (0,0,0) at the beginnig to something like (a, 0, b). In the real case of the video, she is not in outer space, so the Earth (the friction of her feet with the Earth surface) does not allow her that Xcomponent of her spin ( the "a" in (a, 0, b) ). 


#92
May1514, 02:00 AM

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#93
May1514, 02:43 AM

P: 4,069




#94
May1514, 05:02 AM

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#95
May1514, 05:25 AM

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#96
May1514, 10:49 AM

Mentor
P: 17,330

For example, sliding a block to a different location on a frictionless level surface. Changing direction at a constant speed on a level surface. Circular orbits. Etc. In such cases you cannot treat the moment of inertia as a scalar and you have to use the whole tensor. I recommend that you learn the simple cases before the complicated ones. Please concentrate on the equations I posted in post 72. Do you understand those? If so, then try to apply them to some simple scenarios involving simple rigid bodies spinning about their axis of symmetry. Please stop simply posting random videos to analyze. The number of videos available on the internet far outstrips my desire to analyze them. Furthermore, you need to study simple cases and not seek complicated ones. Look to textbooks or coherent presentations, not random videos. 


#97
May1614, 11:24 PM

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P: 579

Please, just tell me when to stop. For future posters' sake I'd hate do cause the closure. Thanks 


#98
May1714, 02:31 AM

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P: 579

I have repeatedly and clearly explained what I mean : if an object is rotating in one plane it has k KE, if you make it rotate in 2 different planes it has undeniably KE > k, somebody must have given it some KE and therefore must have spent some energy. Is that vague to you? Is this wrong or arguable in any case? The link I gave in post # 92 is not a random video, I posted to show visually what seemed so hard to get across: the gyro (after the jerk/twist by Laithwaite,) spins in two different planes at the same time. Can we say that it required no energy? In this case it goes on spinning longer than in the previous (post #80:https://www.youtube.com/watch?v=IaOIWXqH9Io at 1:45/49) because friction/inertia/ or other force does not slow it down, (or as Dalespam says : you can use a scalar and not the whole tensor), but can we say that in the previous case it required less or no energy? In both cases the force has been applied perpendicularly to the plane of rotation. That is what I was talking about. Thanks , anyway, for your efforts. 


#99
May1714, 04:59 AM

P: 963

But rotation in three dimensions is complicated. The rotation vector of an object can change even when it is under no external torques. An easy experiment is a pencil that is rapidly spinning about its long axis and given a slow rotation from end to end as it is tossed into the air. From an untutored perspective, one could describe it as having "rotation in two planes" (which both change over time). The perspective that Dale would use would describe it as rotating about a single plane not aligned with any axis of symmetry. That plane can change over time. Despite the change in the rotation vector over time, angular momentum is conserved. Because in this regime, the objects moment of inertia is not just a simple scalar. It is a tensor (as Dale pointed out in post 90). I have been careful to talk about what can happen as we approach an ideal case where the applied force is gentle and and at right angles to the momentary axis of rotation. Dale has been careful to talk about what does happen in the ideal limiting case where the applied force is gentle and at right angles to the momentary axis of rotation. You've been asked before to become comfortable with ordinary linear mechanics rather than going on about gyroscopes. In linear mechanics the analogue to a jerk/twist is that of an impulsive force. That is usually discussed in the context of collisions. Suppose that you have a ball moving from west to east on a pool table as it is hit by the cue ball moving from south to north and striking the target ball exactly at right angles. The target ball will continue to move eastward at the same rate as before. But it also acquires a velocity component in the northward direction. The impulsive force was applied at right angles to the ball's motion. No work should have been done (you suggest). So where did the extra kinetic energy come from? Try to answer this riddle before we come back to jerks and gyroscopes. 


#100
May1714, 05:12 AM

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It is no riddle, the ball will move in a northeast direction and the angle will be determined by a parallelogram and the M/m ratio , and the its speed and KE will increase accordingly. What is the problem? Changing the direction and increasing KE of the vector required KE/work/energy. Now suppose your ball hits perpendicularly the edge of a spinning wheel on gimbals like in video #18, it starts to rotate or not, does it gain speed and KE in that direction or not, does the total KE of the gyro increase or not, has work been done or not? 


#101
May1714, 05:23 AM

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#102
May1714, 07:15 AM

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Surely it is not constructive. This unwieldy thread might have ended at post #11 if you had given me a direct, exaustive reply (we are not in the HW forum). And I think it is not fair that you ask for my (stupid, I know) opinion/explanation first and then you refuse to give yours. And you seem to get patience by fits and starts. Before you further hurt my feelings, I beg you to read these threads and tell me if I can deal with linear momentum and if you would have been able to give such elegant solutions: http://www.physicsforums.com/showthr...58#post4586458 http://www.physicsforums.com/showthr...53#post4562489 Thanks for your attention 


#103
May1714, 07:59 AM

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P: 17,330

But mostly I am telling you exactly what I said: "study simple cases". You have a strong tendency which I have noticed over multiple threads to ask about a halfdozen random and extremely complicated cases before you have even mastered the simplest case. That simply is not an effective way to learn. If you find a halfdozen cases, then look at them for the simplest one, and ask questions about that one case (never introducing the others) until you fully understand it. And if people tell you that there is an even simpler case you should study first, then do that. Approach learning stepbystep, instead of trying to run before you can even crawl. 


#104
May1714, 08:03 AM

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P: 17,330

After you have done that you have learned to crawl, so then try to stand up: Calculate the angular momentum during the precession using the 2nd formula (this is the hardest step and I will be glad to help). Remember that the torque is, at all times, perpendicular to the axis of rotation. Then, once you are comfortable with that, try to walk: Calculate the work done during the precession by using the 3rd formula on the results from the last part. Compare that to the results you got in the first part. 


#105
May1714, 08:43 AM

P: 963

I have not tried to characterize your attitude and will not do so now. I beg you to address the riddle. A pool ball is subject to an impulsive collision at right angles to its path. It is clear that its kinetic energy has increased. Since the impulse was at right angles, it seems clear that no work can have been done on the ball. We are assured by the work energy theorem that if no work has been done then kinetic energy cannot increase. And yet the kinetic energy has increased. How can one reconcile these things? That is the simple problem that contains (my guess at) the kernel of the misunderstanding that we are currently faced with. 


#106
May1714, 08:45 AM

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#107
May1814, 02:24 PM

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#108
May1814, 09:30 PM

Mentor
P: 17,330

Since this thread is getting a little tense, and has already gone over 100 posts, the other mentors and I have decided that it is past time to close it.
Bobie, please look at the great advice that you have received from many people and try to actually work out some of the details on the very simplest possible scenarios. Don't assume that you know the answer until you have actually worked out the math. 


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