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 one year 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.
 one year 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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DLS
 one year ago
Best ResponseYou've already chosen the best response.0dw:1361001336066:dw (I suppose......)

Mashy
 one year ago
Best ResponseYou've already chosen the best response.1but. think.. the NET force = centripetal force !!

DLS
 one year ago
Best ResponseYou've already chosen the best response.0yeah I know thats the ending part

DLS
 one year ago
Best ResponseYou've already chosen the best response.0\[\LARGE F_{ca}=\frac{GM^2}{4R^2}\]

Mashy
 one year ago
Best ResponseYou've already chosen the best response.1how did you get that bold equation???? and so big?? :O :O

DLS
 one year ago
Best ResponseYou've already chosen the best response.0dw:1361001487972:dw (if im right...)

Mashy
 one year ago
Best ResponseYou've already chosen the best response.1ok that makes sense.. !! proceed with your calculation!

DLS
 one year ago
Best ResponseYou've already chosen the best response.0\[\LARGE F_{BC}=\frac{GM^2}{2R^2}\] if im right........

DLS
 one year ago
Best ResponseYou've already chosen the best response.0\[\LARGE F_{CD}=\frac{GM^2}{2R^2}\]

Mashy
 one year ago
Best ResponseYou've already chosen the best response.1yea correct.. now take vector sum of them!

DLS
 one year ago
Best ResponseYou've already chosen the best response.0now we gotta find resultant if we are correct till here

Mashy
 one year ago
Best ResponseYou've already chosen the best response.1yea.. 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

Mashy
 one year ago
Best ResponseYou've already chosen the best response.1and your directions are wrong.. you are considering force ON c.. so opposite direction!

DLS
 one year ago
Best ResponseYou've already chosen the best response.0dw:1361001778925:dw will it work out like this?

Mashy
 one year ago
Best ResponseYou've already chosen the best response.1sheehs what is that?? :O :O :O

Mashy
 one year ago
Best ResponseYou've already chosen the best response.1think of a way to intelligently solve this!!

DLS
 one year ago
Best ResponseYou've already chosen the best response.0dw:1361001846196:dw resultant of those two dw:1361001866511:dw resultant of those 2 and finally of those 2 :P #fclogic

Mashy
 one year ago
Best ResponseYou've already chosen the best response.1don't.. simply make it so complicated!!!

Mashy
 one year ago
Best ResponseYou've already chosen the best response.1and 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
 one year ago
Best ResponseYou've already chosen the best response.0not 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

Mashy
 one year ago
Best ResponseYou've already chosen the best response.1its ok.. symmertry always works out :D.. so as long as it is symmetrical we are fine :D

Mashy
 one year ago
Best ResponseYou've already chosen the best response.1now do the vector.. which two vectors if you take first.. it ll be easier to resolve??

DLS
 one year ago
Best ResponseYou've already chosen the best response.0WHY aren't we getting anything like this? http://prntscr.com/sxhpz

Mashy
 one year ago
Best ResponseYou've already chosen the best response.1its the same don't worry!! ... just proceed ... i gurantee yu.. its the same what we are doing!!

DLS
 one year ago
Best ResponseYou've already chosen the best response.0hmm kay so resultant thing..lets take AC and AB

Mashy
 one year ago
Best ResponseYou've already chosen the best response.1WHY!!???... that ll complicate it.. whenever you see two vectors with equal mag. and making 90 degrees.. you resolve them first :O :O

DLS
 one year ago
Best ResponseYou've already chosen the best response.0okay lets study vectors first :P

Mashy
 one year ago
Best ResponseYou've already chosen the best response.1:X.. go do that .. first then!

DLS
 one year ago
Best ResponseYou've already chosen the best response.0i meant "lets study" hmm nvm okay just tell me again :P

Mashy
 one year ago
Best ResponseYou've already chosen the best response.1nope.. two vecotrs.. mag F.. making 90 deg to each other.. ?? what is the value?

DLS
 one year ago
Best ResponseYou've already chosen the best response.0\[\LARGE \sqrt{A^2+A^2+2AAcos90}\]

Mashy
 one year ago
Best ResponseYou've already chosen the best response.1yes.. what does that yield??

Mashy
 one year ago
Best ResponseYou've already chosen the best response.1yes.. now what will be the direction

DLS
 one year ago
Best ResponseYou've already chosen the best response.0let me know when ur done with skype ..

Mashy
 one year ago
Best ResponseYou've already chosen the best response.1yea 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

Mashy
 one year ago
Best ResponseYou've already chosen the best response.1til then @thivitaa can help! :D

yrelhan4
 one year ago
Best ResponseYou've already chosen the best response.0v = (1/2) sqrt [ GM (1+ 2 sqrt2) / R] is this the answer?

DLS
 one year ago
Best ResponseYou've already chosen the best response.0\[\LARGE V=\sqrt{\frac{GM}{R}(\frac{2 \sqrt{2}+1}{4}})\]

thivitaa
 one year ago
Best ResponseYou've already chosen the best response.1wrong person to ask...:( juz read the whole thing....and am confused on what ur saying...RELLLL<3 can help..sowie mashy...<3

DLS
 one year ago
Best ResponseYou've already chosen the best response.0how did u get the answer O.O

yrelhan4
 one year ago
Best ResponseYou've already chosen the best response.0@thivitaa haha. :P @DLS where are you stuck?

DLS
 one year ago
Best ResponseYou've already chosen the best response.0Finding the resultant pellet

yrelhan4
 one year ago
Best ResponseYou've already chosen the best response.0hold on. brb in 2 mins.

yrelhan4
 one year ago
Best ResponseYou've already chosen the best response.0ok 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
 one year ago
Best ResponseYou've already chosen the best response.0and 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
 one year ago
Best ResponseYou've already chosen the best response.0yeah i know arrow thin messed up

DLS
 one year ago
Best ResponseYou've already chosen the best response.0but Fcd and Fcb are not not providing centripetal force. Fcdcos45 and Fcbcos45 are. ellaborate? :o

yrelhan4
 one year ago
Best ResponseYou've already chosen the best response.0masses 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
 one year ago
Best ResponseYou've already chosen the best response.0yeah I do but applying this thing for the first time :)

DLS
 one year ago
Best ResponseYou've already chosen the best response.0can you show the 2 forces in the diagram? just for some more clarification

yrelhan4
 one year ago
Best ResponseYou've already chosen the best response.0dw:1361004736715:dw

yrelhan4
 one year ago
Best ResponseYou've already chosen the best response.0where are you confused?

Mashy
 one year ago
Best ResponseYou've already chosen the best response.1just 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 \[F = GM/2R^{2}\]

DLS
 one year ago
Best ResponseYou've already chosen the best response.0why did that guy confuse me then :'( why doesnt it matter?if they are not giving the centripetal force?cos45 resolutions are?

yrelhan4
 one year ago
Best ResponseYou've already chosen the best response.0well 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
 one year ago
Best ResponseYou've already chosen the best response.0and it doesnt matter because resultant is towards the centre. so that is providing the centripetal force. ignore this if it confuses you.

DLS
 one year ago
Best ResponseYou've already chosen the best response.0http://www.troll.me/images/atomicrage/chepofuuuuuuuuuu.jpg

yrelhan4
 one year ago
Best ResponseYou've already chosen the best response.0good luck and have fun. :P

DLS
 one year ago
Best ResponseYou've already chosen the best response.0@Mashy This isn't the actual way of solving the question,is it? I suppose the cos45 component and resultant both are cinciding so..its the sme thing..otherwise?

Mashy
 one year ago
Best ResponseYou've already chosen the best response.1its the same.. one is by using i j k method.. one is by using mashy method lol :D

DLS
 one year ago
Best ResponseYou've already chosen the best response.0meanwhile posting the next Q:P

Mashy
 one year ago
Best ResponseYou've already chosen the best response.1its the same thing.. you i and j and k.. are unit vectors along x y and z axis.. dw:1361006361576:dw so we rite \[a = asin( \theta) j + acos (\theta) i\]
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