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Uniform, semicircular rod of radius R and mass m in the center

  1. Jun 1, 2014 #1
    1. The problem statement, all variables and given/known data
    Pic: http://i.imgur.com/Ny4YAKf.jpg

    Rod has radius R, and mass M. Small mass has mass m.

    I have to find gravitational force exerted on mass m by rod and potential energy of the whole thing.


    2. Relevant equations



    3. The attempt at a solution

    So my idea is that soluton is 2 integrals from 0 to pi/2 of gravitational force of small masses dM, Ill use linear density p.

    [tex] p = \frac{M}{\pi R} [/tex]
    [tex] dM = ds p [/tex]
    [tex] Rd\alpha = ds [/tex]
    [tex] F= \frac{GdM}{R^2}\frac{\vec{r}}{r} [/tex]
    [tex] \vec{r} = R(cos\alpha,sin\alpha) [/tex]
    [tex] 2\int\limits_{0}^{\frac{\pi}{2}} \frac{GdM}{R^2} (cos\alpha,sin\alpha) d\alpha [/tex]

    Now substitute first 3 equations so I get

    [tex] 2\int\limits_{0}^{\frac{\pi}{2}} \frac{GMd\alpha }{\pi R^2} (cos\alpha,sin\alpha) d\alpha [/tex]
    [tex] 2\frac{GM}{\pi R^{2}} \int\limits_{0}^{\frac{\pi}{2}} d\alpha (cos\alpha,sin\alpha) d\alpha [/tex]


    And I don't know what to do with it. I am also not sure about how to calculate the total potential energy
     
  2. jcsd
  3. Jun 1, 2014 #2

    mfb

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    2016 Award

    Staff: Mentor

    Those integrals look easy to evaluate for both components. There is a sign error, however, one component has a negative sign in one of the integrals. Why did you split that at all? You can just integrate from 0 to pi.
    You can get one of the components with a symmetry argument, by the way.

    You can use a similar integral, just for the potential instead of the force.
     
  4. Jun 1, 2014 #3

    Well it's cool that they look easy to evaluate, but I don't know how to do that and I was expecting help ;d Is my work up to the last one correct?

    Sign error? I am not sure where

    Idk why I split them, It doesn't matter too much I guess.
     
  5. Jun 1, 2014 #4
    1st: Use equations to solve problems. Tip: if the symbol "=" is not present, it ain't an equation.
    2nd: Your integrals have two differential "dα" factors. All you need is one of them. The other one is redundant and makes your equation look wrong.
    3rd: Don't you know how to integrate a cosine and/or sine function(s)? That's like the second integral people learn, after learning to integrate polynomials.
     
  6. Jun 1, 2014 #5
    1. Yeah, you can add = after each integral in my op then.
    2. Yeah, but I somehow got 2 da factors - how to get rid of it, because I don't think I can just cross it?
    3. I do how to integrate cosine and sine function, but I don't know how to integrate a vector with 2 da.
     
  7. Jun 1, 2014 #6
    1. My point is that you made hard for me to figure out what you're calculating
    2. You made a mistake in one of your steps introducing a differential that wasn't there. I think it happened when you substituted the expression for r.
    3. There ain't two dα. There is only one. The other one is a mistake.

    added by EDIT: differentials are nilpotent. that means that the product of two differentials vanishes. If that ever happens, you know you've made a mistake somewhere. Don't confuse that with multiple integration in which case you're supposed to have several differential factors. That's not the case here. You're calculating a single integral.
     
    Last edited: Jun 1, 2014
  8. Jun 1, 2014 #7
    So for 2, ill explain what I do since I don't know where did second wrong da come from...

    I get formula for linear density - all mass divided by all length

    [tex] p = \frac{M}{\pi R} [/tex]

    I get formula for infinite small mass with small length - it is its length times linear density

    [tex] dM = ds p [/tex]

    I get formula for small length, as a function of α

    [tex] ds = rdα [/tex]

    Now I start with equation number 2 and put there the other two so

    [tex] dM = ds p = rdα*p = rdα * \frac{M}{\pi R} = \frac{Mdα}{\pi} [/tex]

    So now I have dM and want to calculate integral of infinite small gravitational forces from 0 to π/2. We also know that vector r with direction can be written as [tex]\vec{r} = R(cosα, sinα) [/tex]

    So that integral - [tex]\int\limits_{0}^{\frac{\pi}{2}} \frac{GdM}{r^2} \frac{\vec{r}}{r} ds = \int\limits_{0}^{\frac{\pi}{2}} \frac{G\frac{Mdα}{\pi}}{r^2} \frac{\vec{r}}{r} r dα = \frac{GM}{\pi} \int\limits_{0}^{\frac{\pi}{2}} \frac{dα \vec{r}}{r^2} dα[/tex]

    and yeah, I don't know where did I make a mistake
     
  9. Jun 1, 2014 #8
    Your starting differential is ##d\vec F = \frac{GmdM}{r^2} \frac{\vec{r}}{r}##, not ##d\vec F = \frac{GmdM}{r^2} \frac{\vec{r}}{r} ds##.
     
  10. Jun 1, 2014 #9
    Aaaah yes, thanks. Btw is it GmdM or GdM?

    So I have:

    [tex] \frac{GmM}{\pi*r^{2}} \int (cosa,sina) da [/tex]

    What can I do with this integral? Do I treat it as integral cosa da + integral sina da or something?
     
  11. Jun 1, 2014 #10
    No, you don't add the integrals. They are separate components. The result of the integration is a vector with two components. Each component is a separate integral.
     
  12. Jun 1, 2014 #11
    Ok, so I take it that you just put the integral into the vector so you have (integral of cos, integral of sin)? If so then the result is vector (1,1) which.. does it make sense? It means that equal force vector is [tex] (\frac{2GmM}{\pi*r^2}, \frac{2GmM}{\pi*r^2}) [/tex]

    Also please answer question about wheter should it be GmM/.. or GM/...
     
  13. Jun 1, 2014 #12
    Isn't it that all forces on x axis are 0?
     
  14. Jun 1, 2014 #13
    No, (1,1) isn't right. What are the limits of the integral?
     
  15. Jun 1, 2014 #14
    If you're calculating the force than its GmM/r2, isn't it?
     
  16. Jun 1, 2014 #15
    Ah okay I dun goofed, with limits of 0 to pi it's actually (0,2) and so we get [tex](0,\frac{2GmM}{\pi*r^{2}})[/tex] Which appears to be correcT?
     
  17. Jun 1, 2014 #16
    That seems right.
     
  18. Jun 1, 2014 #17
    What about potential energy? Is it the same integrals but with just formula -GmM/r instead of what I used here?
     
  19. Jun 3, 2014 #18

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Right.
     
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