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Balance, pendulum and energy

  1. Mar 29, 2007 #1
    Hi. I saw this video on google:


    It's mainly a pendulum attached to a balance arm.

    It's stated that the energy needed to keep the pendulum going is far less than the energy we can extract from the balance movement.

    I also post some pictures to try to show how it 's supposed to work.

    This is how i suppose it works:

    The force of gravity on the pendulum de-composes into two components as it oscillates: one is tangential acceleration, that keeps the oscillation going and the other one is a perpendicular force to the pendulum "track".

    When the pendulum is in vertical position, tangential force is 0, and perpendicular force is at maximum value. When the pendulum is at the 2 top points of its path, the tangential force is at maximum value, and the perpendicular one is at the minimum.

    The perpendicular force doesn't affects the pendulum's oscillations. Only the tangential component maintains the pendulum oscillating. Usually, the perpendicular force is compensated with an opposite force on the "strongpoint" of the pendulum... But what happens if we put the "strongpoint" of the pendulum pending from an arm of a balance? It's exactly the same as if we put a "variable weight" on this side of the balance, so the other side of the balance (if it has a fixed weight) will move up and down, just as if we put more or less weight on the side of the pendulum.

    So the question is: Why it's needed more energy to keep the pendulum oscillating than the work we can extract from the movement of the balance by the "variable weight" that is on the pendulum's side of the balance?

    The variation of the perpendicular component of the pendulum's weight is just a consequence of the pendulum oscillation, and usually, it's not used to do anything. But this perpendicular component can take values from 100% to 10% of the total weight of the pendulum while the pendulum is oscillating. So why the input work (for maintain the oscillation) should be equal or higher than the work we can extract from the balance movement by using the "variable weight" of the perpendicular force?

    Do you know an expresion that relates the loses on the pendulum's oscillation and the work on the balance?

    Can you explain why the work the perpendicular component can do have to be less than the loses we have to restore on the pendulum's oscillation?

    Thank you.

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    Last edited: Mar 29, 2007
  2. jcsd
  3. Mar 29, 2007 #2


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    I think it is a hoax. The motions of the swinging weight are not driving the system, there is something hidden doing the work.
  4. Mar 29, 2007 #3
    I don't think it's a hoax, just someone has misinterpreted what's happening and assumed that the hammer end must be using more energy than what is supplied at the pendulum.
  5. Mar 29, 2007 #4


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    Ceisumfrog is right - it isn't a hoax, it is just a misunderstanding of high school physics.
  6. Mar 30, 2007 #5


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    A little more explanation:

    I admit I only got about 5 minutes through a 35 minute video, but unless it got a lot better, the issue is relatively clear. The guy didn't do any calculations or take any measurements, so what he has is his intuition that there is more energy expended in raising and "dropping" a weight than is lost in the pendulum. But *my* intuition (which doesn't contradict the first law of thermodynamics, btw...) tells me that there is very little energy expended in raising and lowering a weight by moving a pendulum.
  7. Mar 30, 2007 #6
    Thank you for taking time to analyze the system.

    Lets put some numbers, and maybe you can help me:

    Imagine the pendulum has a mass of 10Kg, with a length of 50cm, and the balance is 1m of length and has a fixed mass of 10 Kg at the other end of the balance.

    Imagine we manage to get an angle of oscillation of 60º...

    -How can I calculate the energy input (in an easy way) to maintain the oscillation angle of this pendulum on 60º?

    -What is the perpendicular force value on the "strongpoint" at the ends of the path and in vertical position?

    -How can i calculate how much work can be done using the movement of the balance caused by the pendulum "variable weight" on the other end? (Or maybe, what is the height difference on initial and final position, to get the potential energy difference...)

    If you can give me no simple equations or numbers, please try to explain the proccess, and why the work to keep the oscillations going should be higher than the work we can get from the balance movement.
  8. Mar 30, 2007 #7


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    I don't understand your first question: a pendulum requires no input energy to keep moving after it is started. That's the whole point of a pendulum!

    For the second, the forces at the sides are found via trigonometry (draw a free-body diagram with the weight and angle) as the speed is zero, so there is no centripedal acceleration. At the bottom, it is the weight plus centrifugal force.

    For your third, the whole point here is that you can't calculate the work done because there isn't any in an ideal case! All of the energy lost at the weight on the other end is due to the impact and friction in the apparatus. It is heat and sound. There is no height difference because the weight starts and ends at the same place. When the weight is lifted, it takes kinetic energy from the pendulum and converts it to potential energy. When the weight is lowered, it gives that potential energy back and the kinetic energy of the pendulum is restored. It would be easy to see if you played with the device (and maybe he showed this...), that very small adjustments in the height the weight would be lifted (thus, very small change in the potential energy when it is lifted) would result in large changes in loss and thus large changes in how long the pendulum would swing. That's all just inefficiency.

    A better apparatus to demonstrate this would have a spring connecting the weight to the platform instead of having it rest on the platform. Such a device would show clearly the conservation of energy and would oscillate for a very, very long time.
    Last edited: Mar 30, 2007
  9. Mar 30, 2007 #8
    I was about to put that theoretically a pendulum needed no energy to input when I saw your answer.

    So the loses we have to restore are just friction ones (low energy indeed!).

    But i don't agree with you on the part of the balance. I mean:

    The perpendicular force of the pendulum acts like a "variable weight" on this side, that is like a "variable mass"...

    So when the weight goes increasing, gravity is doing work pulling this side of the balance down, and this work is compensed on the other side of the balance, rising the fixed weight.

    When the "variable weight" of the pendulum goes decreasing, then the gravity does the work on the fixed weight part (that goes down), and then this work is compensed on the side of the pendulum, rising the whole arm, and gaining potential energy...

    If you do the complete account of energy for the whole balance, it is clear there are no gain on energy, but if you look only on the fixed weight part, there is always something doing work!

    When the fixed weight go down, gravityt is doing work, and when the fixed weight goes up is just because something on the other side of the balance is doing work (the "variable mass" and gravity on the other side... again!)

    If you can do work with a balance in wich you put and removes weights on one side, then this device should work as stated.
  10. Mar 30, 2007 #9


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    The accounting of the fixed part looks like this:


    When the weight is lifted, it takes energy, when the weight is lowered, it gives it back.
    Your statements are incomplete/inaccurate and should read:

    "When the fixed weight goes down, gravity is doing the work"...on the weight, against the combined forces on the pendulum.
    "When the fixed weight goes up [the combined forces on the pendulum]"... are doing work on the weight, against gravity.

    As you can see, the two statements are mirror images of each other, hence "X" and "-X".

    Somehow you see gravity doing work on both sides and don't see that that makes it symmetrical. But if you just have a pendulum, you can easily see that gravity does all the work there and the work is symmetrical, can't you? This is only very slightly more complicated.
    I'm not exactly sure what you mean there, but if you remove weights while the system is operating, then you are doing the work to keep it operating.
    Last edited: Mar 30, 2007
  11. Mar 30, 2007 #10
    When i say "if you put weight, and then removes it in one side of a balance, you can extract work on the other side of the balance." then you say "you are doing the work"... and that is exactly what the pendulum oscillation is doing for free! (because the change on the perpendicular component is for free, no energy consuming)

    I mean: Imagine a balance with a weight on one side.

    Now, you put some weight on the other side, the balance rises, you put more weight, the balance keeps moving...

    Now you remove some weight, the balance goes down, you remove more weight, the balance keeps moving on same direction...

    Is there no work being done on the inital weight of the other side?

    I think the answer is yes, there is.

    The pendulum is just equivalent to put or remove weight on one side of the balance... but for free!
    Last edited: Mar 30, 2007
  12. Mar 30, 2007 #11


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    No, it isn't. The key word here is extract. You cannot extract energy from this system or it will stop moving.

    I'm not going to pursue this line of thinking because if you can't understand the original scenario, changing it is just going to confuse you more. This is a big red-flag of crackpot logic (as is the "convince me I'm wrong" mindset: no -instead, keep your mind open and make an effort to learn). Heck, the original scenario is two simple concepts put together in a way that only increase the complexity a little bit and you can't even grasp that!

    I don't know how to convince you of this and I'm not inclined to waste much more of my time trying to teach someone who doesn't want to learn. You're going to wast a lot of time and effort chasing somthing that doesn't exist if you insist on keeping your blinders on here. People have wasted their lives in search of perpetual motion. Don't waste yours. Make an effort and learn some real physics.
    Last edited: Mar 30, 2007
  13. Mar 31, 2007 #12
    I'm sorry, but still don't know what is the answer to the question:
    Changes on a variable weight on a balance can do work on the other side of the balance?
  14. Mar 31, 2007 #13
    ask yourself, what is happening on the pendullum side.
    KE is being turned into PE and the pe reaches a maximum value when the KE equals zero at the top of the oscillation.

    this PE would then be equal to mg delta h which would also equal the total energy of the system.

    if you extract energy from this system the maximum height that the pendullum would reach changes. Thus on the next cycle there is less energy to extract from the system, you can continue this until the pendullum stops moving (as there would be zero energy in the system).
  15. Mar 31, 2007 #14
    the uy in the video also periodically pushes on the balance in order to maintain the oscillation (thus adding energy)

    and he seems to think he has a new way of amplifying mechanical force, when in reality he's dealing with a lever of sorts and an oscillation.
  16. Apr 1, 2007 #15
    Thank Luke, good analisys.

    But why the movement of the balance should quit energy to the pendulum?

    Imagine that the pendulum is not connected to the balance. It transforms KE to PE, and it gets a potential energy with a height of "h".

    OK. Now, connect this pendulum to the balance. We should get the same "h" difference from oscillation, plus more height, due to the work the balance (from gravity) do on the fixed weight part. If you "extract" this work, then the fixed weight will do not rise the balance on the pendulum side, but the initial "h" of the pendulum maintains in the same value. So we have the same "h" always, and more or less height due to the balance movement. This aditional height is a consequence of the oscillation, but does not affect the oscillation.

    That is why it's stated that the balance doesn't affect the pendulum.

    Don't you think so? Is there some error?
  17. Apr 1, 2007 #16
    honestly, please stop analyzing physics using words. Learn how to describe a physical system qualitatively. Either draw a free body digram, work out the forces, or get a Lagrangian.

    descriptive physics gets you nowhere.
  18. Apr 1, 2007 #17
    tim lou is absolutely correct, this problem would easily be resolved for you if you worked out the forces.

    however in response to your previous question, the balance is set up such that when it raises the pendullum (when the system is at rest) in a way such that the pendullum rises perpendicularly, this stores energy in the form of gravitational potential. When the balnce lowers the energy is returned in te form of kinetic enery. This kind of motion will result in no net increase of energy in the oscillator.

    furthermore even in the case when the pendullum wasn't at rest andsomeone did work on the beam, it only raises the pendullum a tiny amount, adding very little energy to the system, whereas by moving the pendullum you are free to add a lot more eergy to the system.

    the entire system works like a lever really
  19. Apr 1, 2007 #18
    also a mathematial analysis yields

    -mgh +mgr cos theta

    where h is the height difference between that the lever arm oscillates through, as the total energy of the system when the pendullum is at the botom of its swing

    that mgh term is the energy gotten from the pendullum going dropping down through the lever height, and the mgrcos theta is the energy gotten from swinging from its intital height, (note this assumes that both parts of the system are set up such that the lever reaches its maximum height when the pendullum is at the bottom of its swing

    now lets say that on every oscillation the lever sucks out an amount of energy equal to E0, then the next oscillation would have a total energy equal to

    E-E0=mgh-E0+mgrcos theta=mgh1+mgrcos theta1

    and I'm sure there's another equation lurking somwhere that will allow you to solve that equation, however you can see that h and theta will both decrease, and this will continue until they are 0.
    which for some reason unbeknownst to me always yields E0 equals zero, I haven't had a full intermediate mechanics course so maybe someone could point out the error here for me.
  20. Apr 2, 2007 #19
    I've seen the video and:
    1) the system needs energy. Only one finger can do the job, but needs energy supply.
    2) The pendulum is heavy and attached to the short arm. So an small movement on the short arm wil be greater on the other side. This is not magic. But is more common to use in the other way: wide and weak movement produces a short and powerful one.
    3) Gravity is working all the time on both parts. You should not use gravity where you like.

    You could obtain a 25Kg hammer activated by a finger hanging it from a spring.
    The device performs an interesting hammering in an easy and convenient way, but again, it's not magical, it is not a continuous movement machine.
  21. Apr 2, 2007 #20
    It's not stated as a perpetual motion machine,

    It is suposed to give more output than input.

    I think the correct answer is NO... but the video seems to show the machine working that way...

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