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Proving that central force is conservative and getting the energy

  1. Jun 1, 2014 #1
    1. The problem statement, all variables and given/known data
    Hey, sorry in advance if something I write is unclear but I am not native English speaker.

    I have a central force that is defined as [tex]F=-f(r) \frac{\vec{r}}{r}[/tex] where [tex]f(r)[/tex] is some function of distance and [tex]\vec{r}=(x,y,z)[/tex]. I have to calculate potential energy when [tex]f(r)=\frac{a}{r^{2}}[/tex] and I have to use [tex]\vec{r}d\vec{r}=\frac{dr^{2}}{2}[/tex]


    2. Relevant equations



    3. The attempt at a solution

    I wrote Force F as [tex]\left( -\frac{f(r)x}{r}, -\frac{f(r)y}{r},-\frac{f(r)z}{r} \right)[/tex] can I or can I not do it?
    Calculated the curl, it was 0 so it is conservative. Now I substituted f(r) and got [tex]\left( -\frac{ax}{r^{3}}, -\frac{ay}{r^{3}},-\frac{az}{r^{3}} \right)[/tex]

    To calculate potential energy I have 3 equations:

    [tex]-\frac{ax}{r^{3}} = \frac{dE}{dx}[/tex]
    [tex]-\frac{ay}{r^{3}} = \frac{dE}{dy}[/tex]
    [tex]-\frac{az}{r^{3}} = \frac{dE}{dz}[/tex]

    and the result is [tex]-\frac{a}{2r^{3}}\left(x^{2}+y^{2}+z^{2}\right) + C[/tex]. I didn't use the [tex]\vec{r}d\vec{r}=\frac{dr^{2}}{2}[/tex] so I guess I did something wrong


    When you reply please tell me where and why did I make a mistake and how to correct it.
     
    Last edited: Jun 1, 2014
  2. jcsd
  3. Jun 1, 2014 #2

    CAF123

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    Gold Member

    I think you missed negatives in your calculations when solving for E. (Since dE/dx = -Fx etc..)

    Note also ##x^2 + y^2 + z^2 = r^2##.

    Did you mean to write ##\vec r \cdot d\vec r = dr^2/2##?

    Then $$U = - \int \vec F \cdot d\vec r = - \int \frac{f(r)}{r} \vec r \cdot d\vec r \dots$$
    And $$\int \frac{dr^2}{r^3} \propto \frac{1}{r}.$$

    So yes, your method is fine, but maybe they wanted you to use the approach above.
     
  4. Jun 1, 2014 #3
    Yes I forgot about the minuses, thanks for noticing that. Yes I meant what you wrote with the r*dr.

    Thanks, I thought my method is incorrect :)
     
  5. Jun 1, 2014 #4
    The only mistake I can find is an extra factor of (1/2) in your final answer that shouldn't be there. It is possible to solve the problem the way you did (after fixing that mistake), but it is easier to solve it using the identity provided in the problem. Try it...
     
  6. Jun 1, 2014 #5
    Why is 1/2 a mistake?

    When I have the [tex]-\frac{ax}{r^{3}} = \frac{dE}{dx}[/tex] it turns into

    [tex]-\frac{ax}{r^{3}}dx = dE [/tex]

    [tex]-\int\frac{ax}{r^{3}}dx = \int dE[/tex]

    [tex] E=-\frac{ax^{2}}{2r^{3}} + C[/tex]
     
  7. Jun 1, 2014 #6
    The integral is incorrect because r is a function of x. It cannot be treated as a constant. r2 = x2 = y2 = z2.
     
  8. Jun 1, 2014 #7
    but [tex] \vec{r} [/tex] is a function of x, [tex] r [/tex] is is simply a scalar ? I would have tried the method mentioned here but I have no idea how to even start to be honest ;p I've missed some lectures about how to solve integrals with vectors and such
     
  9. Jun 1, 2014 #8
    Missing lectures is a bad idea. r is indeed a scalar, but it is not a constant scalar. It is a function of x, y, and z. Try to change the variable of integration from x to r.
     
  10. Jun 1, 2014 #9
    Okay so I can see how r^2=x^2+y^2+z^2, but I don't know how to change the variable of integration, I can't just substitute sqrt(r^2-y^2-z^2) for x...
     
  11. Jun 1, 2014 #10
    Yes you can. What do you get for dx?

    Tip: It's easier if you use implicit differentiation.
     
  12. Jun 1, 2014 #11
    Okay, so
    [tex] x=\sqrt{r^{2}-y^{2}-z^{2}} [/tex]

    then

    [tex] dx=\frac{r}{\sqrt{r^{2}-y^{2}-z^{2}}} dr[/tex]

    to integral now

    [tex] ax∫\frac{1}{r^{2}\sqrt{r^{2}-y^{2}-z^{2}}} dr [/tex]

    ???
     
  13. Jun 1, 2014 #12
    You made the same mistake again. You pulled the x out of the integral. You can't do that because x is a function of r. Just substitute the expression for x provided at the top of your post.
     
  14. Jun 1, 2014 #13
    Ahaaa! Thanks :)

    [tex] a∫\frac{x}{r^{2}\sqrt{r^{2}-y^{2}-z^{2}}} dr =[/tex]

    [tex] a∫\frac{\sqrt{r^{2}-y^{2}-z^{2}}}{r^{2}\sqrt{r^{2}-y^{2}-z^{2}}} dr =[/tex]

    [tex] -\frac{a}{r} + C [/tex]

    So that would mean that [tex] E = -\frac{3a}{r}+ C [/tex]

    Is it correct? Or should it be just [tex]-\frac{a}{r}+ C [/tex]
     
    Last edited: Jun 1, 2014
  15. Jun 1, 2014 #14
    it's just -a/r + C.
     
  16. Jun 1, 2014 #15

    ehild

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    You can choose the potential zero at infinity, integrate from ##\vec r_## to ∞.


    [tex]V(\vec {r})=\int_{\vec r}^{∞}{\frac{a}{r'^3}\vec r' \cdot \vec{dr'}}[/tex]

    You can write r'dr= 0.5 d(r2) and introduce the variable u=r2.

    The integral becomes[tex] \int{0.5 \frac{a}{u^{3/2} }du}[/tex] What do you get?

    Substitute back u=r2 and take the limits.


    ehild
     
    Last edited: Jun 1, 2014
  17. Jun 1, 2014 #16
    So it's [tex]-(0-\frac{a}{r}) = \frac{a}{r} [/tex] ? Why do we integrate from r1 to r2, and why do we take r2 as infinity?
     
  18. Jun 1, 2014 #17

    ehild

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    The potential difference is the integral of the negative force between two points, r1 and r2. If you choose r2 at infinity, where the potential is zero, you get the potential at r1.
    I edited my post a bit.

    ehild
     
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