• DLS

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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  • DLS

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.

Physics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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  • DLS
|dw:1361001336066:dw| (I suppose......)
Can you draw?
  • DLS
:P

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Other answers:

wow you read my mind :D
  • DLS
:D
but. think.. the NET force = centripetal force !!
  • DLS
yeah I know thats the ending part
  • DLS
\[\LARGE F_{ca}=\frac{GM^2}{4R^2}\]
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
ok that makes sense.. !! proceed with your calculation!
  • DLS
\[\LARGE F_{BC}=\frac{GM^2}{2R^2}\] if im right........
  • DLS
\[\LARGE F_{CD}=\frac{GM^2}{2R^2}\]
yea correct.. now take vector sum of them!
  • DLS
now we gotta find resultant if we are correct till here
i guess so!.. proceed
  • DLS
2 at a time?
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
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
sheehs what is that?? :O :O :O
|dw:1361001955102:dw|
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
don't.. simply make it so complicated!!!
  • DLS
directions \m/
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
its ok.. symmertry always works out :D.. so as long as it is symmetrical we are fine :D
  • DLS
??
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
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
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
: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?
nope.. two vecotrs.. mag F.. making 90 deg to each other.. ?? what is the value?
  • DLS
F/2?
  • DLS
wait
  • DLS
\[\LARGE \sqrt{A^2+A^2+2AAcos90}\]
yes.. what does that yield??
  • DLS
\[\LARGE \sqrt{2}A\]
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 .-.
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
til then @thivitaa can help! :D
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
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
@thivitaa haha. :P @DLS where are you stuck?
  • DLS
Finding the resultant pellet
hold on. brb in 2 mins.
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.
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
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
|dw:1361004736715:dw|
  • DLS
confused
where are you confused?
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 \[F = GM/2R^{2}\]
^ 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?
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.
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
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?
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
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
yay one more medal :D
  • DLS
come to next one :P

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