DLS
  • 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
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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DLS
  • DLS
|dw:1361001336066:dw| (I suppose......)
anonymous
  • anonymous
Can you draw?
DLS
  • DLS
:P

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More answers

anonymous
  • anonymous
wow you read my mind :D
DLS
  • DLS
:D
anonymous
  • anonymous
but. think.. the NET force = centripetal force !!
DLS
  • DLS
yeah I know thats the ending part
DLS
  • DLS
\[\LARGE F_{ca}=\frac{GM^2}{4R^2}\]
anonymous
  • anonymous
how did you get that bold equation???? and so big?? :O :O
DLS
  • DLS
|dw:1361001487972:dw| (if im right...)
DLS
  • DLS
use the keyword large :P
anonymous
  • anonymous
ok that makes sense.. !! proceed with your calculation!
DLS
  • DLS
\[\LARGE F_{BC}=\frac{GM^2}{2R^2}\] if im right........
DLS
  • DLS
\[\LARGE F_{CD}=\frac{GM^2}{2R^2}\]
anonymous
  • anonymous
yea correct.. now take vector sum of them!
DLS
  • DLS
now we gotta find resultant if we are correct till here
anonymous
  • anonymous
i guess so!.. proceed
DLS
  • DLS
2 at a time?
anonymous
  • 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
  • anonymous
and your directions are wrong.. you are considering force ON c.. so opposite direction!
DLS
  • DLS
|dw:1361001778925:dw| will it work out like this?
DLS
  • DLS
ah yeah sorry
anonymous
  • anonymous
sheehs what is that?? :O :O :O
anonymous
  • anonymous
|dw:1361001955102:dw|
anonymous
  • anonymous
think of a way to intelligently solve this!!
DLS
  • DLS
|dw:1361001846196:dw| resultant of those two |dw:1361001866511:dw| resultant of those 2 and finally of those 2 :P #fclogic
anonymous
  • anonymous
don't.. simply make it so complicated!!!
DLS
  • DLS
directions \m/
anonymous
  • 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
  • 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
  • anonymous
its ok.. symmertry always works out :D.. so as long as it is symmetrical we are fine :D
DLS
  • DLS
??
anonymous
  • anonymous
now do the vector.. which two vectors if you take first.. it ll be easier to resolve??
DLS
  • DLS
WHY aren't we getting anything like this? http://prntscr.com/sxhpz
DLS
  • DLS
we have diff values
anonymous
  • anonymous
its the same don't worry!! ... just proceed ... i gurantee yu.. its the same what we are doing!!
DLS
  • DLS
hmm kay so resultant thing..lets take AC and AB
anonymous
  • 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
  • DLS
okay lets study vectors first :P
anonymous
  • anonymous
:X.. go do that .. first then!
DLS
  • DLS
i meant "lets study" hmm nvm okay just tell me again :P
DLS
  • DLS
equalmag :o
DLS
  • DLS
so resultant is F/2?
anonymous
  • anonymous
nope.. two vecotrs.. mag F.. making 90 deg to each other.. ?? what is the value?
DLS
  • DLS
F/2?
DLS
  • DLS
wait
DLS
  • DLS
\[\LARGE \sqrt{A^2+A^2+2AAcos90}\]
anonymous
  • anonymous
yes.. what does that yield??
DLS
  • DLS
\[\LARGE \sqrt{2}A\]
anonymous
  • anonymous
yes.. now what will be the direction
DLS
  • DLS
:D :D Im Einstein 8|
DLS
  • DLS
|dw:1361002507803:dw|
DLS
  • DLS
let me know when ur done with skype .-.
anonymous
  • 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
  • DLS
okay :(
DLS
  • DLS
90
anonymous
  • anonymous
til then @thivitaa can help! :D
yrelhan4
  • yrelhan4
v = (1/2) sqrt [ GM (1+ 2 sqrt2) / R] is this the answer?
DLS
  • DLS
\[\LARGE V=\sqrt{\frac{GM}{R}(\frac{2 \sqrt{2}+1}{4}})\]
DLS
  • DLS
yeah
thivitaa
  • 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
  • DLS
how did u get the answer O.O
yrelhan4
  • yrelhan4
@thivitaa haha. :P @DLS where are you stuck?
DLS
  • DLS
Finding the resultant pellet
yrelhan4
  • yrelhan4
hold on. brb in 2 mins.
yrelhan4
  • 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
  • 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
  • DLS
yeah i know arrow thin messed up
DLS
  • DLS
but Fcd and Fcb are not not providing centripetal force. Fcdcos45 and Fcbcos45 are. ellaborate? :o
yrelhan4
  • 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
  • DLS
yeah I do but applying this thing for the first time :)
DLS
  • DLS
can you show the 2 forces in the diagram? just for some more clarification
yrelhan4
  • yrelhan4
|dw:1361004736715:dw|
DLS
  • DLS
confused
yrelhan4
  • yrelhan4
where are you confused?
anonymous
  • 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
  • yrelhan4
^ that'll do!
DLS
  • DLS
why did that guy confuse me then :'( why doesnt it matter?if they are not giving the centripetal force?cos45 resolutions are?
yrelhan4
  • 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
  • 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
  • DLS
http://www.troll.me/images/atomic-rage/chepo-fuuuuuuuuuu.jpg
yrelhan4
  • yrelhan4
good luck and have fun. :P
DLS
  • 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
  • anonymous
its the same.. one is by using i j k method.. one is by using mashy method lol :D
DLS
  • DLS
:|
DLS
  • DLS
tell me ijk
DLS
  • DLS
meanwhile posting the next Q:P
anonymous
  • 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
  • DLS
okay
anonymous
  • anonymous
yay one more medal :D
DLS
  • DLS
come to next one :P

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