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DLS

  • one year ago

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

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  1. DLS
    • one year ago
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    |dw:1361012885773:dw|

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

  3. yrelhan4
    • one year ago
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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|

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

  5. DLS
    • one year ago
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    ??

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

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

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

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

  10. Mashy
    • one year ago
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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!

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

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

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

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

  15. DLS
    • one year ago
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    :/ what

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

  17. DLS
    • one year ago
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    bolo na :/

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

  19. yrelhan4
    • one year ago
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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.

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

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

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

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

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

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

  26. Vincent-Lyon.Fr
    • one year ago
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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.

  27. Vincent-Lyon.Fr
    • one year ago
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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.

  28. DLS
    • one year ago
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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

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

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

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

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

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

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

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