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DLS

Derive an expression for the G.field due to a uniform rod of Length L and mass M at a point on its perpendicular bisector at a distance d from the centre

  • one year ago
  • one year ago

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  1. DLS
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    |dw:1361012885773:dw|

    • one year ago
  2. DLS
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    \[\LARGE \frac{2GM}{\sqrt{4L^2+d^2}}\] is what I got one term seems missing

    • one year ago
  3. yrelhan4
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    again, seems like you didnt take the components right. E will be given by dE= 2dE1 sintheta where dE1=G(dm)/(d^2 + x^2) where dm=m/l dx sintheta=d/(d^2 + x^2) integrate with respect to x from 0 to l/2 |dw:1361012228614:dw|

    • one year ago
  4. Mashy
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    lol.. in such cases its better to take 2 diametrically opposite mass elements!

    • one year ago
  5. DLS
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    ??

    • one year ago
  6. Mashy
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    wait wait..lemme think!

    • one year ago
  7. yrelhan4
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    @DLS you didnt get what i said?

    • one year ago
  8. DLS
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    no ..u wrote scary symbols :P

    • one year ago
  9. yrelhan4
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    hmm. haha :P

    • one year ago
  10. Mashy
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    |dw:1361026132081:dw| something liek this??.. cause if you consider two elements at a time, then the net field due to them becomes perpendicular to the rod.. and hence you can now add up all the elements.. else you couldn't add up directly.. cause, the fields would be in different directions!

    • one year ago
  11. Mashy
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    |dw:1361026342375:dw|

    • one year ago
  12. DLS
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    ek se kyun nahi kar sakte O.O

    • one year ago
  13. DLS
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    |dw:1361026384089:dw|

    • one year ago
  14. Mashy
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    you need to learn more vector analysis.. its a continous mass distribution!..

    • one year ago
  15. DLS
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    :/ what

    • one year ago
  16. Mashy
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    yrelhan please explain.. dls whatever he said .. is exactly correct!

    • one year ago
  17. DLS
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    bolo na :/

    • one year ago
  18. yrelhan4
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    what should i explain?

    • one year ago
  19. yrelhan4
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    ek se kyu nhi kr skte? in these kind of questions we always take 2 elements so that1 component cancel out and we are left with one. easier to integrate. how to do with one element? i dont know.

    • one year ago
  20. Vincent-Lyon.Fr
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    You can integrate more easily using an angular variable.

    • one year ago
  21. abb0t
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    Can't you use similar triangles?

    • one year ago
  22. Vincent-Lyon.Fr
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    I have: \(\LARGE \frac{2GM}{d\sqrt{L^2+4d^2}}\)

    • one year ago
  23. DLS
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    i didnt get d :/

    • one year ago
  24. DLS
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    i mean whats wrong with my method

    • one year ago
  25. DLS
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    The mass is a distance of \[\LARGE \sqrt{L^2+\frac{d^2}{4}}\] from the rod

    • one year ago
  26. Vincent-Lyon.Fr
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    ???? the mass is a distance \(\Large d\) from the rod. What is your method? I have not seen it in your posts.

    • one year ago
  27. Vincent-Lyon.Fr
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    It's like a problem of electric field. You have to integrate Newton's law of universal gravitation in the same way as you would use Coulomb's law for charges.

    • one year ago
  28. DLS
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    so cant we do like this integrate from \[\LARGE \sqrt{L^2+\frac{d^2}{4}}\] to L/2 and multiply it by 2 :O

    • one year ago
  29. DLS
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    that would be complicated but eh

    • one year ago
  30. DLS
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    why are we considering elemental mass?

    • one year ago
  31. Vincent-Lyon.Fr
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    This is why I told you to use angle as integration variable.

    • one year ago
  32. DLS
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    okay..thanks :O but thats correct right?just complex..thanks!

    • one year ago
  33. Vincent-Lyon.Fr
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    Wait, I'll send the derivation.

    • one year ago
  34. Vincent-Lyon.Fr
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    Here it is.

    • one year ago
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