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Get force from kinetic energy

  1. Jul 5, 2009 #1
    1. The problem statement, all variables and given/known data

    Ok, I know this might rather be a pure vector problem rather than a true physics problem, but I was wondering what is the easiest way to get the force vector from the equation for the kinetic energy, if you have the equation for the kinetic energy, the displacement vector, velocity vector and accleration vector. The question specifically states that you should get the force from the kinetic energy, thus force = mass x accleration won't work.

    2. Relevant equations

    The only relevant equation I know is [tex]\dot{T}=\dot{\bar{r}} \cdot \bar{F}[/tex]

    3. The attempt at a solution

    It would seem to get the force you have to "divide" the derivative of the kinetic energy by the derivative of the displacement (velocity), but I'm not too sure how to divide one vector by another vector and if it is the correct inverse opperation for the scalar product of vectors. Is there perhaps also another way of solving this problem?
    Last edited: Jul 5, 2009
  2. jcsd
  3. Jul 5, 2009 #2


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    Without knowing the specifics of your problem, there is the work energy relationship, such that the work, i.e. the integral of the Force, as a function of displacement, over the distance will result in a change in kinetic energy.
  4. Jul 6, 2009 #3
    Ok, that might work, because the equation for the force is given, you just have to prove it, so it would basically be going from the given force back to the calculated kinetic energy, which can be done by the equation [tex]T = \int \bar{F} \cdot d\bar{r}[/tex] if you accept that the initial kinetic energy is zero, but how would you evaluate [tex]\bar{F} \cdot d\bar{r}[/tex] in terms of cylindrical polar coordinates without converting to cartesian coordinates first? Is [tex]d\bar{r} = d\rho\hat{\rho} + dz\hat{k}[/tex] since this is derived from the position in terms of cylindrical polar coordinates? Can the scalar product of vectors even be applied in terms of cylindrical polar coordinates and if so, what is the dot product of the different components and unit vectors? I know for cartesian coordinates it would be:

    [tex]\bar{F} \cdot d\bar{r} = \left(F_{x}\hat{i} + F_{y}\hat{j} + F_{z}\hat{k}\right) \cdot \left(dx\hat{i} + dy\hat{j} + dz\hat{k}\right)[/tex]
    [tex]\bar{F} \cdot d\bar{r} = F_{x}dx + F_{y}dy + F_{y}dy[/tex]

    But what would this look like in cylindrical polar coordinates? I can't seem to find anything to help me with this. Please also tell me if any of my assumptions are incorrect.
  5. Jul 6, 2009 #4


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    I am disadvantaged in knowing exactly what you are trying to do. But coordinate transform is just a mechanical step and not so much a conceptual one. If your Force equation is in polar coordinates, I suppose it doesn't really matter how one does a path integral to determine the work that gets converted to changing kinetic energy. It's just a matter of determining what the force is in the direction of motion right?
  6. Jul 6, 2009 #5
    Ok, sorry, I thought it was just something small that I didn't understand, that's why I didn't give the full question.

    The full question is as follows:

    "Write down the kinetic energy of a particle in cylindrical polar coordinates in a frame rotating with angular velocity [tex]\omega[/tex] about the z-axis. Show that the terms proportional to [tex]\omega[/tex] and [tex]\omega^{2}[/tex] reproduce the Coriolis force and centrifugal force respectively." (It's problem 10.5 from the textbook Classical Mechanics, 5th edition, by Tom W.B. Kibble and Frank H. Berkshire)

    I started the problem off with using Cartesian coordinates, since I'm more sure about the vector operations in Cartesian coordinates and then only converted the results to cylindrical polar coordinates right at the end.

    If you take the displacement to be:

    [tex]\bar{r} = x\hat{i} + y\hat{j} + z\hat{k}[/tex]

    and thus:

    [tex]\dot{\bar{r}} = \dot{x}\hat{i} + \dot{y}\hat{j} + \dot{z}\hat{k}[/tex]

    but since the frame is rotating it implies that:

    [tex]\frac{d\bar{r}}{dt} = \dot{\bar{r}} + \bar{w} \wedge \bar{r}[/tex]

    If this is taken into account, I found the kinetic energy to be:

    [tex]T = \frac{1}{2}m\dot{x}^{2} - m\dot{x}\omega y + \frac{1}{2}m\omega^{2}y^{2} + \frac{1}{2}m\dot{y}^{2} + m\dot{y}\omega x + \frac{1}{2}m\omega^{2}x^{2} + \frac{1}{2}m\dot{z}^{2}[/tex]

    Now the Coriolis force is given by:

    [tex]-2m\bar{\omega} \wedge \dot{\bar{r}} = 2m\omega\dot{y}\hat{i} - 2m\omega\dot{x}\hat{j}[/tex]

    Now if you apply the equation [tex]T = \int \bar{F} \cdot d\bar{r}[/tex] which is valid if you take the initial kinetic energy as being zero and [tex]d\bar{r} = dx\hat{i} + dy\hat{j} + dz\hat{k}[/tex] you get:

    [tex]T = \int \left(2m\omega\dot{y}\hat{i} - 2m\omega\dot{x}\hat{j}\right) \cdot \left(dx\hat{i} + dy\hat{j} + dz\hat{k}\right)[/tex]
    [tex]T = \int 2m\omega\dot{y}dx - 2m\omega\dot{x}dy[/tex]
    [tex]T = 2m\omega\dot{y}x - 2m\omega\dot{x}y[/tex]

    If you compare the above result with the initial calculated kinetic energy, you can see that terms containing [tex]\omega[/tex] is half of the latter calculated result. Why is this so? Because when I calculated the centrifugal component in the same way, it gave the same result and not double of what was expected. If you convert this results to cylindrical polar coordinates you also get double of what is expected, so I thought if you start off with cylindrical polar coordinates you might get a different result, but that is where my previous posts came in...

    Please tell my if there is anything wrong with my reasoning, thanks.
  7. Jul 6, 2009 #6


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    Isn't the Coriolis Force perpendicular to the motion of an object? So as a force then it is not adding to kinetic energy of a particle, though it does affect the direction. So work contributing to the kinetic energy then is not going to be the way to go, as I originally suggested thinking you were wanting to derive force more generally from a particle that is undergoing a change in kinetic energy. So assuming initial kinetic energy etc doesn't seem useful, because I don't see that the kinetic energy of a particle is changing.

    I would endeavor to go in the direction suggested by the problem then and consider the kinetic energy in polar form.

    Sorry if my suggestion served to mislead you. I wasn't aware of your context. (Demonstrating the value of starting with what the whole problem is.)
  8. Jul 6, 2009 #7
    Ok, the kinetic energy in cylindrical polar form is then:

    [tex]T = \frac{1}{2}m\dot{\rho}^{2} + \frac{1}{2}m\rho^{2}\dot{\varphi}^{2} + m\omega \rho^{2}\dot{\varphi} + \frac{1}{2}m\omega^{2}\rho^{2} + \frac{1}{2}m\dot{z}^{2}[/tex]

    But how do you now prove that the component [tex]m\omega \rho^{2}\dot{\varphi}[/tex] reproduces the Coriolis force if you can't use the work method?
  9. Jul 6, 2009 #8
    i might be overlooking something but the problem is get force from kinetic energy?

    KE=1/2mv^2 , m= F/a, replace m with F/a in the formula and solve for F?

    however thats all i got i have no idea what the rest is talking about sry :(
    Last edited: Jul 6, 2009
  10. Jul 6, 2009 #9


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  11. Jul 6, 2009 #10
    Thanks for trying to help xavior, but I think my problem gets a bit more trickier than that... :smile:
  12. Jul 6, 2009 #11
    lol i imagine when i saw the actual thing, its more advanced than where I am but good luck
  13. Jul 6, 2009 #12
    Thanks for the link lowlypion, but I can't seem to see how that would help me to get from the kinetic energy to the force or vice versa, since the kinetic energy is not a vector and that is what seems to mess up the whole thing for me (just tell me if it's something stupid I'm not seeing)
  14. Jul 8, 2009 #13
    So lowlypion, do you perhaps have a suggestion for me?
  15. Jul 8, 2009 #14
    Don't worry, I found a solution somewhere else :smile:
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