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DLS
Group Title
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
 one year ago
DLS Group Title
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
 one year ago

This Question is Closed

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

Mashy Group TitleBest ResponseYou've already chosen the best response.1
Can you draw?
 one year ago

Mashy Group TitleBest ResponseYou've already chosen the best response.1
wow you read my mind :D
 one year ago

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

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

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

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

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

DLS Group TitleBest ResponseYou've already chosen the best response.0
use the keyword large :P
 one year ago

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

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

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

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

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

Mashy Group TitleBest ResponseYou've already chosen the best response.1
i guess so!.. proceed
 one year ago

Mashy Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

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

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

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

Mashy Group TitleBest ResponseYou've already chosen the best response.1
dw:1361001955102:dw
 one year ago

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

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

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

DLS Group TitleBest ResponseYou've already chosen the best response.0
directions \m/
 one year ago

Mashy Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

DLS Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

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

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

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

DLS Group TitleBest ResponseYou've already chosen the best response.0
we have diff values
 one year ago

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

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

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

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

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

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

DLS Group TitleBest ResponseYou've already chosen the best response.0
so resultant is F/2?
 one year ago

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

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

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

DLS Group TitleBest ResponseYou've already chosen the best response.0
\[\LARGE \sqrt{2}A\]
 one year ago

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

DLS Group TitleBest ResponseYou've already chosen the best response.0
:D :D Im Einstein 8
 one year ago

DLS Group TitleBest ResponseYou've already chosen the best response.0
dw:1361002507803:dw
 one year ago

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

Mashy Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

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

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

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

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

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

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

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

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

yrelhan4 Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

yrelhan4 Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

DLS Group TitleBest ResponseYou've already chosen the best response.0
yeah i know arrow thin messed up
 one year ago

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

yrelhan4 Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

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

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

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

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

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

yrelhan4 Group TitleBest ResponseYou've already chosen the best response.0
^ that'll do!
 one year ago

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

yrelhan4 Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

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

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

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

DLS Group TitleBest 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?
 one year ago

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

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

Mashy Group TitleBest ResponseYou've already chosen the best response.1
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 \[a = asin( \theta) j + acos (\theta) i\]
 one year ago

Mashy Group TitleBest ResponseYou've already chosen the best response.1
yay one more medal :D
 one year ago

DLS Group TitleBest ResponseYou've already chosen the best response.0
come to next one :P
 one year ago
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