# Where do the input energy go?

by Low-Q
Tags: energy, input
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 PF Gold P: 244 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
Mentor
P: 16,942
 Quote by Low-Q 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?
Based on the above discussion and your knowledge of physics, what would you guess and why? Think about torque and angular momentum. Think about torque and work. Which torques lead to what changes in angular momentum, and which do work, and why?
PF Gold
P: 244
 Quote by DaleSpam Based on the above discussion and your knowledge of physics, what would you guess and why? Think about torque and angular momentum. Think about torque and work. Which torques lead to what changes in angular momentum, and which do work, and why?
1. I would guess it takes energy to sustain the cycle, because the gyro don't want to turn in the same direction as the applied torque (As shown in the youtube video earlier where the two gyros are spinning in the same direction) - but the gyro in the powerball experiment has no other options other than turning in the same direction as the applied force.
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
Mentor
P: 16,942
 Quote by Low-Q 1. I would guess it takes energy to sustain the cycle,
Since that leads to a failure of conservation of energy we know that this guess is wrong.

 Quote by Low-Q because the gyro don't want to turn in the same direction as the applied torque (As shown in the youtube video earlier where the two gyros are spinning in the same direction) - but the gyro in the powerball experiment has no other options other than turning in the same direction as the applied force. Therfor I guess I must do work whileturning the gyro around perpendicular to its rotation. (Suppose it is the same as fighting agains the angular momentum - applying energy to do so)
Why? Why do you think that work must be done in this case? What is the formula for work in angular motion? Does it apply here?

 Quote by Low-Q 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.
I am just trying to get you to actually analyze the device using sound physical reasoning.

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
 Mentor P: 16,942 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 $\tau=\frac{dL}{dt}$. 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 $W = τ θ$ the work done by the torque that makes the gyroscope precess is 0.
PF Gold
P: 244
 Quote by DaleSpam Try to analyze it on your own first, but here are some hints. Assuming a very high gear ratio we can assume that the angular momentum, L, is almost entirely along the axis of the gyroscope.
Does "almost entirely along the axis" mean that the gear ratio isn't infinitely high? If the gear ratio was infinetily high, the angular momentum would be entirely along the axis?

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
Mentor
P: 16,942
 Quote by Low-Q Does "almost entirely along the axis" mean that the gear ratio isn't infinitely high? If the gear ratio was infinetily high, the angular momentum would be entirely along the axis?
Yes.

 Quote by Low-Q 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?
You can make the precession go either way simply by changing the direction of the external torque. Nothing happens to the speed of the gyro unless there are losses.
 PF Gold P: 244 Thanks. I'm much wiser now :-)
 Mentor P: 16,942 Excellent! You are welcome.
 PF Gold P: 244 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
Mentor
P: 16,942
 Quote by Low-Q Sorry - not finished yet
No worries. Questions and answers are not rationed here

 Quote by Low-Q 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.
Neglecting losses precession does not require energy in general. This is because the torque is perpendicular to the axis of rotation. Just like a force perpendicular to motion does no work, for the same reason a torque perpendicular to the rotation also does no work. We have analyzed this specifically for your geared gyroscope, but it applies to precession in general.

 Quote by Low-Q 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?
Yes, in this case the system is not frictionless. For fluids, this kind of friction is called viscosity.
 PF Gold P: 244 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
 Mentor P: 16,942 I don't have that equation, although it shouldn't be too hard to derive.
 PF Gold P: 244 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 Attached Thumbnails
PF Gold
P: 244
 Quote by Low-Q 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
Ofcourse, velocity cannot be known with unknown RPM of the flywheel, its weight, and length of the shaft. Forget that question.

Anyone?

Vidar
 Mentor P: 16,942 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.
PF Gold
P: 244
 Quote by DaleSpam 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.
Very annoying actually. At risk of getting a lot of lag error, I venture to ask some questions, and write down some thoughts:

What is the reason why the weight will move upwards? One would think at first glance that the vertical torque is solely caused by the weight and gravity.

Wether the precess wants to be clockwise or counterclockwise perpendicular to the guides wouldn't matter as there is no physical precess present due to the guides, right?
If horizontal precess really is an important factor of which vertical direction the weight will move in this system, I will accept that - but have trouble in understanding why.

In my mind, the spinning itself will cause increased "inertia" that is parallell with the force which pulls on it. However, I might not use the word inertia, but I couldn't find a better word.
As this system will not act as a pendulum but stop at the bottom or the top, the weight will slowly move down or up, and rest there so the weight is spinning perpendicular to the force of gravity.

I think: In this particular system the "inertia" would appear to be an artificial friction (As it would look like if one watched it - A pendulum in syrup), but it would not cause heat as the weight moves up or down in the guides. So instead of generating heat, the spinning weight must slow down. Energy must be conserved.

If the force that pulls on the weight is allways perpendicular to the motioin of it (As if the gravity turns around the system in the direction of the weight), the weight will finally stop spinning. That applied force could likely be my finger trying to move the spinning weight up and down inside the guides - regardless if the guides is aligned vertical or horizontal.

I tested something similar with a toy car that you push along the floor, and hear that iron flywheel spins up to high speed (Except I was using a powerful drill and almost destroied the gears inside). If I then turn that car left and right back and forth very fast perpendicular to the flywheel, that flywheel stops much earlier than if I don't. The needle bearings does not provide much friction anyways, so in case of increased friction I don't think this is the main reason why the flywheel stops faster.

So, then I am back to the initial question in this thread. Analyzed the above system, I now put a gear to the weight so the weight is spinning because I push it along the guides. Whould't that mean that I try to accelerate the RPM of the weight at the same time as the shift of the wheel force to stop it's spin? While doing this, the force from my finger is moving a given distance. This will appearently end up with an energy consume from my side that is not going anywhere - not heat, not increased kinetic energy ???

Vidar
Mentor
P: 16,942
 Quote by Low-Q What is the reason why the weight will move upwards?
Torque.

 Quote by Low-Q One would think at first glance that the vertical torque is solely caused by the weight and gravity.
No. Gravity cannot cause a vertical torque. Remember, torque is always perpendicular to the force, so torque from gravity is always horizontal.

 Quote by Low-Q Wether the precess wants to be clockwise or counterclockwise perpendicular to the guides wouldn't matter as there is no physical precess present due to the guides, right? If horizontal precess really is an important factor of which vertical direction the weight will move in this system, I will accept that - but have trouble in understanding why.
Let's say that the system is in front of me with the gyroscope facing directly towards me and spinning counter-clockwise. This is an angular momentum pointing horizontally directly towards me.

The weight will cause a torque to the right so the gyro will want to precess to the right. Instead it will run into the guide which will prevent any precession to the right.

The force that the guide exerts will in turn make a torque vertically down. Since the guides allow motion down the gyroscope will precess downwards.

Once it reaches the bottom there will be no more torque due to gravity and therefore no more force from the guide and therefore no more torque from the guide and it will stop precessing.

Why don't you try the analysis if the gyro is spinning clockwise (angular momentum directly away), and see if it precesses up or down. I don't actually know.

 Quote by Low-Q instead of generating heat, the spinning weight must slow down. Energy must be conserved.
Yes.

 Quote by Low-Q So, then I am back to the initial question in this thread. Analyzed the above system, I now put a gear to the weight so the weight is spinning because I push it along the guides. Whould't that mean that I try to accelerate the RPM of the weight at the same time as the shift of the wheel force to stop it's spin? While doing this, the force from my finger is moving a given distance. This will appearently end up with an energy consume from my side that is not going anywhere - not heat, not increased kinetic energy ???
Obviously this is incorrect: energy is conserved. I have dealt with this question multiple times already and am not interested in repeating it yet again over and over. Please re-read my previous comments, if they are not sufficient then go get a textbook because apparently my communication style is not clear enough.

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