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Potential energy of a particle

  1. Oct 27, 2008 #1
    1. The problem statement, all variables and given/known data
    When a particle is a distance r from the origin, its potential energy is given by the equation U(r)=kr, where k is constant and r=(x^2+y^2+z^2)^1/2
    A. find a mathematical expression in terms of x,y, and z for the y-component of the force on the particle
    B. If U=3.00J when the particle is 2.00m from the origin, find the numerical value of the y-component of the force on this particle when it is at the point (-1.00m,2.00m,3.00m)


    2. Relevant equations



    3. The attempt at a solution
    K= ((x^2+y^2+z^2)^1/2)/U(r)
     
  2. jcsd
  3. Oct 27, 2008 #2

    cepheid

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    You haven't put anything under relevant equations. What is the relationship between force and potential energy?
     
  4. Oct 27, 2008 #3
    Force multiplied by displacement vector is equal to the energy
     
  5. Oct 27, 2008 #4

    cepheid

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    Not really. Force multiplied by the displacement vector is equal to the *work* done, which is equal to the *change in kinetic energy*. We are talking about the change in potential energy, which will be the *negative* of the change in kinetic energy (ie the negative of the work done), *provided that* the force is conservative (meaning that mechanical energy is conserved). So, IF the force were constant, we would have:

    U = -W = -Fd

    In our case, I think that the specific mathematical form of U means that the force is a constant with r. So things are simplified.

    For a general potential energy function, the force is NOT constant, hence we must resort to calculus. Have you done vector calculus. Have you seen these equations?

    [tex] \textbf{F} = -\nabla U [/tex]
    [tex] U = -\int \textbf{F} \cdot d\textbf{l} [/tex]

    This is the true relationship between force and potential energy. In this case, due to the potential being spherically symmetric, gradient reduces to dU/dr. The force is the negative of the rate of change of potential energy with position. Due to U being linear in r, that derivative reduces to a constant.
     
    Last edited: Oct 27, 2008
  6. Oct 27, 2008 #5
    I'm not following.
    F=-du/dr?
    F=-k?
     
  7. Oct 27, 2008 #6

    cepheid

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    Yes, although it's a gradient (that's what the triangle symbol is), not just a derivative, so F = -dU/dr multiplied by a radial unit vector (which I forgot to mention).

    F = -k in the radial direction, i.e. F points radially inward. In other words, if you have a constant inward attractive force, the potential energy will increase linearly as you move outward, which is what was given in the problem.

    HOWEVER you were asked for the component of the force acting in the y direction, NOT in the radial direction. ;-)
     
  8. Oct 27, 2008 #7
    I'm currently taking Calc II so I have not studied vector calculus so I don't understand gradients or vector calculus
     
  9. Oct 27, 2008 #8

    HallsofIvy

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    It's as if you had just accidently walked into the wrong class! You keep responding with either "pre-calculus" physics ("energy= force times distance" which is only true for a constant force or "F= -du/dr" which is only true in one dimension. You have been told that [itex]F= -\nabla U= -grad U[/itex]. Do you not know what the vector gradient of a scalar function is?
    If f is any function of the three variables x, y, and z, then
    [tex]\nabla f= \frac{\partial f}{\partial x}\vec{i}+ \frac{\partial f}{\partial y}\vec{j}+ \frac{\partial f}{\partial z}\vec{k}[/tex]

    You got that response in while I was typing mine! Ah, well. Old Rule: you always study the mathematics you need for this semester's physics course next semester.
     
  10. Oct 27, 2008 #9

    cepheid

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    It doesn't matter. The gradient simplifies to a derivative multiplied by a radial unit vector:

    [tex] \textbf{F} = -\frac{\partial U(r)}{\partial r} \hat{\textbf{r}} [/tex]

    Now, you need to work in cartesian coordinates, because it's asking you for the *y* component of F. What do you do? It's a change of variables. r = (some function of x,y, AND z). Therefore d/dr = ??? in terms of x, y, and z. If you're doing multivariable calculus, you should be able to handle this. Hint, change of variables in differentiation = chain rule.
     
    Last edited: Oct 27, 2008
  11. Oct 27, 2008 #10

    cepheid

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    That is SO TRUE!!! Why do they do this to us? The only instance I can remember in which we did things the right way around was: vector calc, THEN E & M. Which was very helpful, because E & M helped me to get a better foothold on the math to which I had already been introduced, so that I was no longer asking myself, "wtf is a curl?"

    Otherwise, it was always our physics profs assuming we knew stuff, and then reacting with surprise. "Oh, you, mean mean you don't know what a tensor is?" No, we DON'T! :mad:
     
  12. Oct 27, 2008 #11
    This is a problem for my Physics I class. The only requirement is Calc I.
     
  13. Oct 27, 2008 #12

    cepheid

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    This is exasperating. The only calc II you need to know is what a partial derviative IS! I mean, if you consider each component of the force separately, you get three separate derivatives in three separate directions! Which is the same as what you'd get using the definition of the gradient given by Halls! So start out in cartesian coords right off the bat!!!

    [tex] F_x = -\frac{\partial U(x,y,z)}{\partial x} [/tex]

    [tex] F_y = -\frac{\partial U(x,y,z)}{\partial y} [/tex]

    [tex] F_z = -\frac{\partial U(x,y,z)}{\partial z} [/tex]

    Are you capable of doing three derivatives? Can you see that this is equivalent to taking the gradient (you're just figuring out the three components of that gradient separately)?
     
  14. Oct 27, 2008 #13
    d/dr=(2x+2y+2z)/2(x^2+y^2+z^2)^1/2?
     
  15. Oct 27, 2008 #14

    cepheid

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    Hey, don't get smart with me. I realized that what I was saying in post 9 about first differentiating wrt r and then converting is unecessary, because you already have U as a function of x,y, and z. Read post 12. All you have to do is take the partial deriv of U(x,y,z) wrt y. I don't know how to make it any clearer than that.
     
  16. Oct 27, 2008 #15

    cepheid

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    Okay fine. I thought partial dervatives were covered in calc II. Then I realized I was thinking calc III, and that calc II is integral calculus, and you won't do calc III until second year. Which raises the question, why is your prof asking you this? It's not a 1 D problem unless you work in spherical coordinates, and you are specifically asked something that requires you to work in Cartesian coords. Anyway, my apologies. Here is what I am saying:

    First convert to cartesian

    [tex] U(x,y,z) = kr = k(x^2 + y^2 + z^2)^{\frac{1}{2}} [/tex]

    Now, in vague terms, the force is sort of, the "rate of change" of the potential energy function. E.g. gravity. The stronger the force, the more work you have to do to move upwards against it, and the more potential energy you gain as a result. So, from this, we can skip a bunch of math and take the leap to saying that the force in the Y direction depends only upon how U changes *in the y direction*. "How U changes in the y direction" is basically what the partial derivative of U with respect to y tells you. In general, I need four dimensions to plot U(x,y,z). However, in that 4D space, if I choose to move only in the y direction (*keeping x and z constant*), the rate of change of U that I measure *along that line* is the partial derivative of U with respect to y. Mathematically:

    [tex]F_y = \frac{\partial U(x,y,z)}{\partial y} = \frac{\partial}{\partial y}k(x^2 + y^2 + z^2)^{\frac{1}{2}} [/tex]

    = the derivative of U(x,y,z) with respect to y, *assuming that x and z are constants*

    This is not the total rate of change of U, but only the rate of change you'd measure if you confined yourself to movement in the y direction. Hence, it's a partial derivative.
     
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