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DLS

  • 3 years ago

Four particles of of equal masses M move along a circle of radius R under the action of their mutual gravitational attraction.Find the speed of each particle.

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  1. DLS
    • 3 years ago
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    |dw:1361001336066:dw| (I suppose......)

  2. Mashy
    • 3 years ago
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    Can you draw?

  3. DLS
    • 3 years ago
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    :P

  4. Mashy
    • 3 years ago
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    wow you read my mind :D

  5. DLS
    • 3 years ago
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    :D

  6. Mashy
    • 3 years ago
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    but. think.. the NET force = centripetal force !!

  7. DLS
    • 3 years ago
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    yeah I know thats the ending part

  8. DLS
    • 3 years ago
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    \[\LARGE F_{ca}=\frac{GM^2}{4R^2}\]

  9. Mashy
    • 3 years ago
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    how did you get that bold equation???? and so big?? :O :O

  10. DLS
    • 3 years ago
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    |dw:1361001487972:dw| (if im right...)

  11. DLS
    • 3 years ago
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    use the keyword large :P

  12. Mashy
    • 3 years ago
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    ok that makes sense.. !! proceed with your calculation!

  13. DLS
    • 3 years ago
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    \[\LARGE F_{BC}=\frac{GM^2}{2R^2}\] if im right........

  14. DLS
    • 3 years ago
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    \[\LARGE F_{CD}=\frac{GM^2}{2R^2}\]

  15. Mashy
    • 3 years ago
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    yea correct.. now take vector sum of them!

  16. DLS
    • 3 years ago
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    now we gotta find resultant if we are correct till here

  17. Mashy
    • 3 years ago
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    i guess so!.. proceed

  18. DLS
    • 3 years ago
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    2 at a time?

  19. Mashy
    • 3 years ago
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    yea.. call two of them as F, and the other one as F/2.. and work out.. easier work :D.. then finally substitute for F :D

  20. Mashy
    • 3 years ago
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    and your directions are wrong.. you are considering force ON c.. so opposite direction!

  21. DLS
    • 3 years ago
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    |dw:1361001778925:dw| will it work out like this?

  22. DLS
    • 3 years ago
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    ah yeah sorry

  23. Mashy
    • 3 years ago
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    sheehs what is that?? :O :O :O

  24. Mashy
    • 3 years ago
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    |dw:1361001955102:dw|

  25. Mashy
    • 3 years ago
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    think of a way to intelligently solve this!!

  26. DLS
    • 3 years ago
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    |dw:1361001846196:dw| resultant of those two |dw:1361001866511:dw| resultant of those 2 and finally of those 2 :P #fclogic

  27. Mashy
    • 3 years ago
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    don't.. simply make it so complicated!!!

  28. DLS
    • 3 years ago
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    directions \m/

  29. Mashy
    • 3 years ago
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    and for the record.. once you find the resultant of two vectors say R .. you get the resultant of the third vector and R.. as your final answer

  30. DLS
    • 3 years ago
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    not sure with the vector resultant part tbh i got confused there only and i suppose we are wrong with those calculated values of forces too :P

  31. Mashy
    • 3 years ago
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    its ok.. symmertry always works out :D.. so as long as it is symmetrical we are fine :D

  32. DLS
    • 3 years ago
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    ??

  33. Mashy
    • 3 years ago
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    now do the vector.. which two vectors if you take first.. it ll be easier to resolve??

  34. DLS
    • 3 years ago
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    WHY aren't we getting anything like this? http://prntscr.com/sxhpz

  35. DLS
    • 3 years ago
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    we have diff values

  36. Mashy
    • 3 years ago
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    its the same don't worry!! ... just proceed ... i gurantee yu.. its the same what we are doing!!

  37. DLS
    • 3 years ago
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    hmm kay so resultant thing..lets take AC and AB

  38. Mashy
    • 3 years ago
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    WHY!!???... that ll complicate it.. whenever you see two vectors with equal mag. and making 90 degrees.. you resolve them first :O :O

  39. DLS
    • 3 years ago
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    okay lets study vectors first :P

  40. Mashy
    • 3 years ago
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    :X.. go do that .. first then!

  41. DLS
    • 3 years ago
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    i meant "lets study" hmm nvm okay just tell me again :P

  42. DLS
    • 3 years ago
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    equalmag :o

  43. DLS
    • 3 years ago
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    so resultant is F/2?

  44. Mashy
    • 3 years ago
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    nope.. two vecotrs.. mag F.. making 90 deg to each other.. ?? what is the value?

  45. DLS
    • 3 years ago
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    F/2?

  46. DLS
    • 3 years ago
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    wait

  47. DLS
    • 3 years ago
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    \[\LARGE \sqrt{A^2+A^2+2AAcos90}\]

  48. Mashy
    • 3 years ago
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    yes.. what does that yield??

  49. DLS
    • 3 years ago
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    \[\LARGE \sqrt{2}A\]

  50. Mashy
    • 3 years ago
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    yes.. now what will be the direction

  51. DLS
    • 3 years ago
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    :D :D Im Einstein 8|

  52. DLS
    • 3 years ago
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    |dw:1361002507803:dw|

  53. DLS
    • 3 years ago
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    let me know when ur done with skype .-.

  54. Mashy
    • 3 years ago
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    yea what is the angle/??? exact angle??? sorry..... ill have to have lunch.. ll deal with this late r:P.. till then you try figuring out :D

  55. DLS
    • 3 years ago
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    okay :(

  56. DLS
    • 3 years ago
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    90

  57. Mashy
    • 3 years ago
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    til then @thivitaa can help! :D

  58. yrelhan4
    • 3 years ago
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    v = (1/2) sqrt [ GM (1+ 2 sqrt2) / R] is this the answer?

  59. DLS
    • 3 years ago
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    \[\LARGE V=\sqrt{\frac{GM}{R}(\frac{2 \sqrt{2}+1}{4}})\]

  60. DLS
    • 3 years ago
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    yeah

  61. thivitaa
    • 3 years ago
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    wrong person to ask...:( juz read the whole thing....and am confused on what ur saying...RELLLL<3 can help..sowie mashy...<3

  62. DLS
    • 3 years ago
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    how did u get the answer O.O

  63. yrelhan4
    • 3 years ago
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    @thivitaa haha. :P @DLS where are you stuck?

  64. DLS
    • 3 years ago
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    Finding the resultant pellet

  65. yrelhan4
    • 3 years ago
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    hold on. brb in 2 mins.

  66. yrelhan4
    • 3 years ago
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    ok so look. according to the figure at the top. Fca=GM^2/(4r^2) Fcb=Gm^2/(2r^2) Fcd=Gm^2/(2r^2) but Fcd and Fcb are not not providing centripetal force. Fcdcos45 and Fcdcos45 are. so what you gonna do is 1/2mv^2=Fca + Fcdcos45 + Fcdcos45 solve. you have v.

  67. yrelhan4
    • 3 years ago
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    and btw. the arrows of the forces on C should be opposite. C is getting attracted towards other masses, not other masses are getting attracted towards C.

  68. DLS
    • 3 years ago
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    yeah i know arrow thin messed up

  69. DLS
    • 3 years ago
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    but Fcd and Fcb are not not providing centripetal force. Fcdcos45 and Fcbcos45 are. ellaborate? :o

  70. yrelhan4
    • 3 years ago
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    masses can only move in a circle due to centripetal force. as fcd and fcb are not directed towards centre, so they cannot provide it. i assume here you know the concept of centripetal force

  71. DLS
    • 3 years ago
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    yeah I do but applying this thing for the first time :)

  72. DLS
    • 3 years ago
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    can you show the 2 forces in the diagram? just for some more clarification

  73. yrelhan4
    • 3 years ago
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    |dw:1361004736715:dw|

  74. DLS
    • 3 years ago
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    confused

  75. yrelhan4
    • 3 years ago
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    where are you confused?

  76. Mashy
    • 3 years ago
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    just take the RESULTANT MAN The resultant of Two F's is \[Fr = \sqrt{2}F\] this will be in the same direction as that of the third force.. hence the net force is \[Fnet = F/2 + \sqrt{2}F\] finally substitue for F \[F = GM/2R^{2}\]

  77. yrelhan4
    • 3 years ago
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    ^ that'll do!

  78. DLS
    • 3 years ago
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    why did that guy confuse me then :'( why doesnt it matter?if they are not giving the centripetal force?cos45 resolutions are?

  79. yrelhan4
    • 3 years ago
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    well you got confused. i didnt confuse you. :P solve the two, you'll get the same things. fcb=fcd=f so fcbcos45 + fcdcos45= sqrt2f resultant gices sqrt2f too. so they are the same things. forget what i did, follow what mashy said.

  80. yrelhan4
    • 3 years ago
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    and it doesnt matter because resultant is towards the centre. so that is providing the centripetal force. ignore this if it confuses you.

  81. DLS
    • 3 years ago
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    http://www.troll.me/images/atomic-rage/chepo-fuuuuuuuuuu.jpg

  82. yrelhan4
    • 3 years ago
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    good luck and have fun. :P

  83. DLS
    • 3 years ago
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    @Mashy This isn't the actual way of solving the question,is it? I suppose the cos45 component and resultant both are c-inciding so..its the sme thing..otherwise?

  84. Mashy
    • 3 years ago
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    its the same.. one is by using i j k method.. one is by using mashy method lol :D

  85. DLS
    • 3 years ago
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    :|

  86. DLS
    • 3 years ago
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    tell me ijk

  87. DLS
    • 3 years ago
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    meanwhile posting the next Q:P

  88. Mashy
    • 3 years ago
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    its the same thing.. you i and j and k.. are unit vectors along x y and z axis.. |dw:1361006361576:dw| so we rite \[a = asin( \theta) j + acos (\theta) i\]

  89. DLS
    • 3 years ago
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    okay

  90. Mashy
    • 3 years ago
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    yay one more medal :D

  91. DLS
    • 3 years ago
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    come to next one :P

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