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.

- DLS

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- DLS

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

- anonymous

Can you draw?

- DLS

:P

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## More answers

- anonymous

wow you read my mind :D

- DLS

:D

- anonymous

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

- DLS

yeah I know thats the ending part

- DLS

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

- anonymous

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

- DLS

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

- DLS

use the keyword large :P

- anonymous

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

- DLS

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

- DLS

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

- anonymous

yea correct.. now take vector sum of them!

- DLS

now we gotta find resultant if we are correct till here

- anonymous

i guess so!.. proceed

- DLS

2 at a time?

- anonymous

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

- anonymous

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

- DLS

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

- DLS

ah yeah sorry

- anonymous

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

- anonymous

|dw:1361001955102:dw|

- anonymous

think of a way to intelligently solve this!!

- DLS

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

- anonymous

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

- DLS

directions
\m/

- anonymous

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

- 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

- anonymous

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

- DLS

??

- anonymous

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

- DLS

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

- DLS

we have diff values

- anonymous

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

- DLS

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

- anonymous

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

- DLS

okay lets study vectors first :P

- anonymous

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

- DLS

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

- DLS

equalmag :o

- DLS

so resultant is F/2?

- anonymous

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

- DLS

F/2?

- DLS

wait

- DLS

\[\LARGE \sqrt{A^2+A^2+2AAcos90}\]

- anonymous

yes.. what does that yield??

- DLS

\[\LARGE \sqrt{2}A\]

- anonymous

yes.. now what will be the direction

- DLS

:D :D
Im Einstein 8|

- DLS

|dw:1361002507803:dw|

- DLS

let me know when ur done with skype .-.

- anonymous

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

- DLS

okay :(

- DLS

90

- anonymous

til then @thivitaa can help! :D

- yrelhan4

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

- DLS

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

- DLS

yeah

- thivitaa

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

- DLS

how did u get the answer O.O

- yrelhan4

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

- DLS

Finding the resultant pellet

- yrelhan4

hold on. brb in 2 mins.

- 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.

- 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.

- DLS

yeah i know arrow thin messed up

- DLS

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

- 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

- DLS

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

- DLS

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

- yrelhan4

|dw:1361004736715:dw|

- DLS

confused

- yrelhan4

where are you confused?

- anonymous

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 \[F = GM/2R^{2}\]

- yrelhan4

^ that'll do!

- DLS

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

- 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.

- yrelhan4

and it doesnt matter because resultant is towards the centre. so that is providing the centripetal force.
ignore this if it confuses you.

- DLS

http://www.troll.me/images/atomic-rage/chepo-fuuuuuuuuuu.jpg

- yrelhan4

good luck and have fun. :P

- 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?

- anonymous

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

- DLS

:|

- DLS

tell me ijk

- DLS

meanwhile posting the next Q:P

- anonymous

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 \[a = asin( \theta) j + acos (\theta) i\]

- DLS

okay

- anonymous

yay one more medal :D

- DLS

come to next one :P

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