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DLS

  • 2 years 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
    • 2 years ago
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    |dw:1361012885773:dw|

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

  3. yrelhan4
    • 2 years 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
    • 2 years ago
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    lol.. in such cases its better to take 2 diametrically opposite mass elements!

  5. DLS
    • 2 years ago
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    ??

  6. Mashy
    • 2 years ago
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    wait wait..lemme think!

  7. yrelhan4
    • 2 years ago
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    @DLS you didnt get what i said?

  8. DLS
    • 2 years ago
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    no ..u wrote scary symbols :P

  9. yrelhan4
    • 2 years ago
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    hmm. haha :P

  10. Mashy
    • 2 years 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
    • 2 years ago
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    |dw:1361026342375:dw|

  12. DLS
    • 2 years ago
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    ek se kyun nahi kar sakte O.O

  13. DLS
    • 2 years ago
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    |dw:1361026384089:dw|

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

  15. DLS
    • 2 years ago
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    :/ what

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

  17. DLS
    • 2 years ago
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    bolo na :/

  18. yrelhan4
    • 2 years ago
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    what should i explain?

  19. yrelhan4
    • 2 years 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
    • 2 years ago
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    You can integrate more easily using an angular variable.

  21. abb0t
    • 2 years ago
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    Can't you use similar triangles?

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

  23. DLS
    • 2 years ago
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    i didnt get d :/

  24. DLS
    • 2 years ago
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    i mean whats wrong with my method

  25. DLS
    • 2 years 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
    • 2 years 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
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    that would be complicated but eh

  30. DLS
    • 2 years ago
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    why are we considering elemental mass?

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

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

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

  34. Vincent-Lyon.Fr
    • 2 years ago
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    Here it is.

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