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Where do the input energy go? |
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| Jan6-13, 01:48 PM | #18 |
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Where do the input energy go? |
| Jan6-13, 02:56 PM | #19 |
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I will use a more simple example.
I have a modified powerball that is fixed to a concrete floor. Assume that I can turn the flywheel in it perpendicular to the flywheels rotation. Initially the flywheel does not spin. It starts to spin only when I turn it perpendiculary. If I stop turning it, its spin also stop. A gear mechanism makes sure of this. So, the flywheel will under all circumstances spin in the vertical plane, and at the same time can turn in the horizontal plane - without exceptions. I assume I will feel a counterforce as soon as I try to turn the flywheel perpendicular to its rotation, since the flywheel at the same time starts to spin perpendicular to the turn. Is that correct? Say the flywheel is spinning constant at 10 000 rpm. And it turns around perpendiculary at a constant 60 rpm. If I successfully maintan this constant cycle over time, would that also mean that I have to apply energy to sustain the rpm in the vertical and horizontal plane except for making up for friction? Vidar |
| Jan6-13, 04:41 PM | #20 |
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| Jan6-13, 05:51 PM | #21 |
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Therfor I guess I must do work while turning the gyro around perpendicular to its rotation. (Suppose it is the same as fighting agains the angular momentum - applying energy to do so) 2. Which torques lead to what changes in angular momentum, and which do work, and why? Not really sure where you're heading with these questions. Vidar |
| Jan6-13, 07:57 PM | #22 |
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Use this page as a reference http://en.wikipedia.org/wiki/Torque Especially the section on The relation to angular momentum and the section on The relation between torque power and Energy Other sources are http://en.wikiversity.org/wiki/Torqu...r_acceleration And http://bolvan.ph.utexas.edu/~vadim/C...11s/linang.pdf http://www2.cose.isu.edu/~hackmart/torque.pdf |
| Jan6-13, 09:41 PM | #23 |
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Try to analyze it on your own first, but here are some hints.
Spoiler
Assuming a very high gear ratio we can assume that the angular momentum, L, is almost entirely along the axis of the gyroscope. The gearing makes it so that L precesses slowly about a circle. Since L is changing that means that there is a torque according to [itex]\tau=\frac{dL}{dt}[/itex].
If L is precessing in a horizontal plane then τ is also in the horizontal plane, but 90° "ahead" of L. So τ tends to make L rotate vertically, but L is mechanically constrained to remain in the horizontal plane. Therefore the rotation about the axis of τ is 0°. Therefore according to [itex]W = τ θ[/itex] the work done by the torque that makes the gyroscope precess is 0. |
| Jan7-13, 06:47 AM | #24 |
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I have now learned that precession will slowly turn the gyroscope in the same direction as the gyroscope is spinning. That precession will occour if a force, like gravity, is trying to overturn the gyroscope. Will it be possible to manually force the "precession" the opposite way if the gyro is still spinning the same way as before? If so, what happens with the speed of the gyro? Vidar |
| Jan7-13, 06:52 AM | #25 |
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| Jan7-13, 08:38 AM | #26 |
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Thanks. I'm much wiser now :-)
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| Jan7-13, 09:47 AM | #27 |
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Excellent! You are welcome.
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| Jan8-13, 06:15 AM | #28 |
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Sorry - not finished yet
![]() I would think that the flywheel mass that constantly changes direction, and thus experiences a form of "pseudo" acceleration or G-forces, would require a certain amount of energy in order to do this. Such as the "Coriolis" effect (Might be incorrectly expressed) of a centrifugal pump. Although the liquid can be returned to the pump input (a loop where the liquid repeats the same cycle), the acceleration of the fluid tangentially to the rotation as it flows away from the center of the pump will still try to prevent rotation. Where does the energy in such a system go? Will the liquid heat up? Vidar |
| Jan8-13, 07:16 AM | #29 |
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![]() |
| Jan8-13, 09:35 AM | #30 |
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OK, thanks. I just visualized the experiment where I place a red dot on the flywheels circumference. I rotate the flywheel 1 rps, and the flywheel have a circumference of 1 meter. The red dot travels now at 1m/s. If I turn the flywheel perpendicular to its rotation 1 complete turn per second, the red dot will cover a longer distance per revolution of the flywheel, and increase velocity by how much? Do you have an equation for this?
Vidar |
| Jan8-13, 10:49 AM | #31 |
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I don't have that equation, although it shouldn't be too hard to derive.
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| Jan10-13, 04:11 PM | #32 |
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I have drawn an illustration in Google ScetchUp - hope you understand it.
It is a fast spinning flywheel (red) on a shaft (Blue). Two green guides prevents the shaft and the flywheel to move sideways. Gravity is pulling on the heavy flywheel. Question: As the spinning flywheel starts falling from the top (Not illustrated). Will the velovity of it be greatest at point A, B or C? What velocity can we expect at point A, B and C? Will the shaft with its spinning flywheel behave as longer pendulum (Kind of a pendulum in slow motion)? Vidar |
| Jan11-13, 08:18 AM | #33 |
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Anyone? Vidar |
| Jan11-13, 10:51 AM | #34 |
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That is a really annoying system to analyze. The weight will cause a torque about a horizontal axis which will try to get the gyroscope to precess horizontally. It will then run into the green guide which will exert a force to prevent its horizontal precession. This force will also generate a torque, this time about the vertical axis. The torque about the vertical axis will cause the gyroscope to precess either up or down until it reaches the top or the bottom. At that point gravity will no longer be exerting a torque and it will just stay there. It will not act as a pendulum.
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