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Total Energy of Pendulum

  1. Jul 2, 2009 #1
    The total energy of a pendulum is calculated either by it's maximum height (Gravitational Potential), or by the lowest point with maximum velocity (Kinetic Energy).

    For the significant formulas of a pendulum see http://en.wikipedia.org/wiki/Pendulum_(mathematics)" [Broken]wiki page

    If you use the highest point to figure out the total energy you'll probably use E = M*g*h - right?
    then h = A*sin([tex]\theta[/tex]0) where A is the Amplitude, and [tex]\theta[/tex]0 is the angle of swing. & A = L*sin([tex]\theta[/tex]0) with small angles, and L is length of the rod.

    So E = M*g*L*[tex]\theta[/tex]0^2 (sin([tex]\theta[/tex]0) = [tex]\theta[/tex]0)

    But if we use the maximum kinetic energy to figure out the total energy, then it would be E = [tex]\frac{M*v^2}{2}[/tex].
    Now if v = A*[tex]\sqrt{\frac{g}{L}}[/tex] where L is length of the rod.
    Then E = [tex]\frac{M*A^2*g}{2*L}[/tex]
    A = L*sin([tex]\theta[/tex]0) with small angles

    E = (M*g*L*[tex]\theta[/tex]0^2)/2 (sin([tex]\theta[/tex]0) = [tex]\theta[/tex]0)

    Whoever my have noticed, the last calculation is half the energy of the first.
    I haven't managed to find any wrong assumption or miss-calculation.
    Did anybody else catch one?

    Thanks in advance - Gothican
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jul 2, 2009 #2
    The correct total energy is E = mgL[1 - cos(θ0)]

    Don't use small angle approximations yet.

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  4. Jul 2, 2009 #3

    By the way, do you know that the pendulum energy is not conserved in a horizontally moving reference frame? Take, for example, a reference frame with V=Vmax. At the lowest pendulum position there is no potential energy nor kinetic one. At any other pendulum position the energy is positive.
    Last edited: Jul 2, 2009
  5. Jul 2, 2009 #4
    Bob for short-
    This a very interesting and thought-provoking statement. If the reference frame is moving along the z axis, and the pendulum is in the y,z plane, then the length of the pendulum at its lower limit of travel is still L, and the maximum elevation gain is still h (both transverse to the motion), In the limit of high γ the opening angle θ0 shrinks as follows

    θ0prime = atan[ tan(θ0)/γ]

    At high γ, the bob is bouncing up and down vertically with neglibile motion along z, due to the relativistic contraction. The apparent vertical length of the pendulum is then L = maximum, and L-h = minimum.

    So maybe the all the potential energy is stored in a spring of maximum length L and minimum length L - h.

    Bob S
    Last edited: Jul 3, 2009
  6. Jul 3, 2009 #5
    No, non conservation of the total pendulum energy is a purely CM effect.

    It proves that not all inertial reference frames are equivalent.:yuck:
  7. Jul 3, 2009 #6
    I think it only appears not to be equivalent because the ceiling is given infinite mass. In any real situation, the ceiling will oscillate a little?

    The idea that inertial reference frames (in a Galilean/Newtonian universe) are not equivalent is just too unpalatable for me. I suspect that if we accounted for the external forces needed to keep the ceiling in place, we'd find they were equivalent?
  8. Jul 3, 2009 #7


    Staff: Mentor

    You don't need to calculate the effect on the ceiling etc. in order to get conservation of energy. All you need to do is to realize that the pivot does work in any reference frame where it is moving. So if the pendulum is swinging from the right to the left and we are in a reference frame where the pivot is moving to the right, then the pivot f.d is negative and so the pivot is taking energy out of the pendulum. If you keep track of how much work has has been done on or by the pivot then you will get energy conservation.
  9. Jul 3, 2009 #8
    No, it is wrong. The total pendulum energy is not conserved in any moving system. It is so and it is right.

  10. Jul 3, 2009 #9
    Ah yes. Much simpler to consider the work being done by the pivot than to consider the forces needed to keep the pivot/ceiling moving at constant velocity. Nice!
  11. Jul 3, 2009 #10
    Are you actually proposing that a simple mechanical system violates energy conservation? Or are you merely suggesting that the "pendulum energy" is just one place in the system where the energy lives, and it moves around among the parts of the system? (Or something like that.)
  12. Jul 3, 2009 #11
    Everything is much simpler. We calculate the pendulum energy as usual: kinetic plus potential energies. It conserves in the resting reference frame but it gets variable in a moving reference frame. To see it it's sufficient to perform a Galileo transformation to any moving system. The energy becomes time-dependent since the potential energy becomes time dependent. In other words, in a moving RF, the force is not conservative any more.
  13. Jul 3, 2009 #12


    Staff: Mentor

    And it varies by exactly the amount of work done on or by the pivot. The conservation of energy does not say that the energy of a system is constant, it says that the change in the energy of the system is equal to the amount of work done on the system. Only an isolated system has constant energy (since no work can be done on an isolated system by definition), and a pendulum is not an isolated system. You are fundamentally misunderstanding the principle of the conservation of energy, it applies in all inertial reference frames and even in many non-inertial frames.
    Last edited: Jul 3, 2009
  14. Jul 3, 2009 #13
    No, pivot is frictionless and the work is done by the gravity.

    Then it is not conservation.

    [/QUOTE]...it says that the change in the energy of the system is equal to the amount of work done on the system.[/QUOTE]

    In the resting reference system it is gravity that makes work. In a moving RF it is the same.

    I violently disagree that my pendulum is not isolated.
  15. Jul 3, 2009 #14

    Doc Al

    User Avatar

    Staff: Mentor

    Ask yourself: Does the tension in the cord do work on the pendulum bob?
  16. Jul 3, 2009 #15
    In a resting reference frame, no, the tension does not do any work. The total energy is constant there. And in any moving RF the total energy is oscillating.
  17. Jul 3, 2009 #16

    Doc Al

    User Avatar

    Staff: Mentor

    Right. But in the moving frame it does.
    Due to the work done by the tension (or the pivot force, as you like). That's what DaleSpam was saying.
  18. Jul 3, 2009 #17
    If the cord does not change its length, no work is done by it.

    Another thing it serves as an intermediary between the Earth and the bob. In a moving RF the Earth force becomes non conservative, and that is the only right explanation.

    E=T+U(r) - conservative force

    E'=T'+U(r+Vt) - non conservative force
    Last edited: Jul 3, 2009
  19. Jul 3, 2009 #18


    Staff: Mentor

    I am also talking about a frictionless pivot, but because the pivot is moving in the moving frame there is work done by/on it. Remember the definition of work: W=f.d There is clearly a force at the pivot. In the rest frame d=0, so there is no work regardless of the force at the pivot, but in any other frame both f and d are non-zero so W is non-zero.

    That is wrong on multiple levels, and a little creepy.

    No, if the cord does not change its length then no elastic potential energy is stored in it, that does not imply that it does not do any work.
    Last edited: Jul 3, 2009
  20. Jul 3, 2009 #19
    If you replace "the pivot" by "the Earth", it will be better.

    The pivot as well as the cord are intermediary things in this problem.

    Consider a squash ball reflecting from a wall. In the resting frame the ball energy is conserved because the wall potential does not depend on time. In a moving RF (for example, where the ball is initially at rest) it is the wall that moves and hits the ball, so the ball acquires some energy (the ball energy is not conserves there). It is described as a t-dependence of an external potential.
    Last edited: Jul 3, 2009
  21. Jul 3, 2009 #20


    Staff: Mentor

    Sure, you can always expand your system until you get an isolated system, but it is not necessary and it doesn't change anything I said about the work done on the pendulum itself.
  22. Jul 4, 2009 #21
    That's either a totally brilliant or terribly misleading way to model the collision. I'm not smart enough to tell which.

    It seems to me that the ball energy is conserved (assuming the collision is elastic) because none of its kinetic energy is transferred to the wall. If the wall had less than infinite mass, then the ball's energy would drop a bit. The total energy of the wall and ball would be conserved however.

    It's hard to understand what can be meant by the potential of the wall. What if the ball never collides with it. Then what would its potential be then? Describing the wall as having a time-dependent potential seems like an arcane way to describe that the wall is exerting a force on the ball through a distance (in the case of a moving frame), which means that work is being done on the ball. The mathematics of W=Fd seem so much simpler than trying to write down a formula for such a t-dependent potential.

    In any case, the distance through which the wall is in contact with the ball (in the moving frame) is a property of the elastic nature of the ball. Again, it's hard to see how it's helpful to describe that as the potential of the wall. I have to say though, that this is an enjoyable discussion, and I feel that I am learning from it.
  23. Aug 24, 2009 #22


    Staff: Mentor

    Btw, I was reviewing Leonard Susskind's lectures on Classical Mechanics and was reminded of this thread. Since we had discussed the conservation or non-conservation of energy I was curious how Noether's theorem would apply. So I set up the Lagrangian for the moving pendulum
    [tex]L = \frac{m v^2}{2}+m r \cos (\theta ) \theta ' v+\frac{1}{2} m r^2
    \theta '^2+g m r \cos (\theta )[/tex]

    And since it is not explicitly dependent on time there must be a conserved energy per Noether's theorem. So I found the energy by:

    [tex]E = H = \frac{\partial L}{\partial \theta '} \theta ' - L = -\frac{m v^2}{2}+\frac{1}{2} m r^2 \theta '^2-g m r \cos (\theta

    Which reduces to the normal expression for a pendulum's energy for v=0, and has a nice intuitive form. So energy is conserved for a moving pendulum.
    Last edited: Aug 24, 2009
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