## DLS 2 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.

1. DLS

|dw:1361001336066:dw| (I suppose......)

2. Mashy

Can you draw?

3. DLS

:P

4. Mashy

wow you read my mind :D

5. DLS

:D

6. Mashy

but. think.. the NET force = centripetal force !!

7. DLS

yeah I know thats the ending part

8. DLS

$\LARGE F_{ca}=\frac{GM^2}{4R^2}$

9. Mashy

how did you get that bold equation???? and so big?? :O :O

10. DLS

|dw:1361001487972:dw| (if im right...)

11. DLS

use the keyword large :P

12. Mashy

ok that makes sense.. !! proceed with your calculation!

13. DLS

$\LARGE F_{BC}=\frac{GM^2}{2R^2}$ if im right........

14. DLS

$\LARGE F_{CD}=\frac{GM^2}{2R^2}$

15. Mashy

yea correct.. now take vector sum of them!

16. DLS

now we gotta find resultant if we are correct till here

17. Mashy

i guess so!.. proceed

18. DLS

2 at a time?

19. Mashy

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

and your directions are wrong.. you are considering force ON c.. so opposite direction!

21. DLS

|dw:1361001778925:dw| will it work out like this?

22. DLS

ah yeah sorry

23. Mashy

sheehs what is that?? :O :O :O

24. Mashy

|dw:1361001955102:dw|

25. Mashy

think of a way to intelligently solve this!!

26. DLS

|dw:1361001846196:dw| resultant of those two |dw:1361001866511:dw| resultant of those 2 and finally of those 2 :P #fclogic

27. Mashy

don't.. simply make it so complicated!!!

28. DLS

directions \m/

29. Mashy

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

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

its ok.. symmertry always works out :D.. so as long as it is symmetrical we are fine :D

32. DLS

??

33. Mashy

now do the vector.. which two vectors if you take first.. it ll be easier to resolve??

34. DLS

WHY aren't we getting anything like this? http://prntscr.com/sxhpz

35. DLS

we have diff values

36. Mashy

its the same don't worry!! ... just proceed ... i gurantee yu.. its the same what we are doing!!

37. DLS

hmm kay so resultant thing..lets take AC and AB

38. Mashy

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

okay lets study vectors first :P

40. Mashy

:X.. go do that .. first then!

41. DLS

i meant "lets study" hmm nvm okay just tell me again :P

42. DLS

equalmag :o

43. DLS

so resultant is F/2?

44. Mashy

nope.. two vecotrs.. mag F.. making 90 deg to each other.. ?? what is the value?

45. DLS

F/2?

46. DLS

wait

47. DLS

$\LARGE \sqrt{A^2+A^2+2AAcos90}$

48. Mashy

yes.. what does that yield??

49. DLS

$\LARGE \sqrt{2}A$

50. Mashy

yes.. now what will be the direction

51. DLS

:D :D Im Einstein 8|

52. DLS

|dw:1361002507803:dw|

53. DLS

let me know when ur done with skype .-.

54. Mashy

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

okay :(

56. DLS

90

57. Mashy

til then @thivitaa can help! :D

58. yrelhan4

v = (1/2) sqrt [ GM (1+ 2 sqrt2) / R] is this the answer?

59. DLS

$\LARGE V=\sqrt{\frac{GM}{R}(\frac{2 \sqrt{2}+1}{4}})$

60. DLS

yeah

61. thivitaa

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

how did u get the answer O.O

63. yrelhan4

@thivitaa haha. :P @DLS where are you stuck?

64. DLS

Finding the resultant pellet

65. yrelhan4

hold on. brb in 2 mins.

66. yrelhan4

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

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

yeah i know arrow thin messed up

69. DLS

but Fcd and Fcb are not not providing centripetal force. Fcdcos45 and Fcbcos45 are. ellaborate? :o

70. yrelhan4

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

yeah I do but applying this thing for the first time :)

72. DLS

can you show the 2 forces in the diagram? just for some more clarification

73. yrelhan4

|dw:1361004736715:dw|

74. DLS

confused

75. yrelhan4

where are you confused?

76. Mashy

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

^ that'll do!

78. DLS

why did that guy confuse me then :'( why doesnt it matter?if they are not giving the centripetal force?cos45 resolutions are?

79. yrelhan4

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

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
82. yrelhan4

good luck and have fun. :P

83. DLS

@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

its the same.. one is by using i j k method.. one is by using mashy method lol :D

85. DLS

:|

86. DLS

tell me ijk

87. DLS

meanwhile posting the next Q:P

88. Mashy

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

okay

90. Mashy

yay one more medal :D

91. DLS

come to next one :P