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
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|dw:1361001336066:dw|
(I suppose......)
Mashy
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Can you draw?
DLS
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:P
Mashy
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wow you read my mind :D
DLS
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:D
Mashy
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but. think.. the NET force = centripetal force !!
DLS
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yeah I know thats the ending part
DLS
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\[\LARGE F_{ca}=\frac{GM^2}{4R^2}\]
Mashy
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how did you get that bold equation???? and so big?? :O :O
DLS
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|dw:1361001487972:dw|
(if im right...)
DLS
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use the keyword large :P
Mashy
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ok that makes sense.. !! proceed with your calculation!
DLS
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\[\LARGE F_{BC}=\frac{GM^2}{2R^2}\]
if im right........
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\[\LARGE F_{CD}=\frac{GM^2}{2R^2}\]
Mashy
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yea correct.. now take vector sum of them!
DLS
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now we gotta find resultant if we are correct till here
Mashy
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i guess so!.. proceed
DLS
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2 at a time?
Mashy
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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
Mashy
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and your directions are wrong.. you are considering force ON c.. so opposite direction!
DLS
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|dw:1361001778925:dw|
will it work out like this?
DLS
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ah yeah sorry
Mashy
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sheehs what is that?? :O :O :O
Mashy
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|dw:1361001955102:dw|
Mashy
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think of a way to intelligently solve this!!
DLS
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|dw:1361001846196:dw|
resultant of those two
|dw:1361001866511:dw|
resultant of those 2
and finally of those 2 :P
#fclogic
Mashy
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don't.. simply make it so complicated!!!
DLS
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directions
\m/
Mashy
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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
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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
Mashy
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its ok.. symmertry always works out :D.. so as long as it is symmetrical we are fine :D
DLS
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??
Mashy
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now do the vector.. which two vectors if you take first.. it ll be easier to resolve??
DLS
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we have diff values
Mashy
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its the same don't worry!! ... just proceed ... i gurantee yu.. its the same what we are doing!!
DLS
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hmm kay so resultant thing..lets take AC and AB
Mashy
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WHY!!???... that ll complicate it.. whenever you see two vectors with equal mag. and making 90 degrees.. you resolve them first :O :O
DLS
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okay lets study vectors first :P
Mashy
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:X.. go do that .. first then!
DLS
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i meant "lets study" hmm nvm okay just tell me again :P
DLS
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equalmag :o
DLS
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so resultant is F/2?
Mashy
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nope.. two vecotrs.. mag F.. making 90 deg to each other.. ?? what is the value?
DLS
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F/2?
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wait
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\[\LARGE \sqrt{A^2+A^2+2AAcos90}\]
Mashy
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yes.. what does that yield??
DLS
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\[\LARGE \sqrt{2}A\]
Mashy
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yes.. now what will be the direction
DLS
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:D :D
Im Einstein 8|
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|dw:1361002507803:dw|
DLS
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let me know when ur done with skype .-.
Mashy
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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
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okay :(
DLS
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90
Mashy
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til then @thivitaa can help! :D
yrelhan4
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v = (1/2) sqrt [ GM (1+ 2 sqrt2) / R]
is this the answer?
DLS
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\[\LARGE V=\sqrt{\frac{GM}{R}(\frac{2 \sqrt{2}+1}{4}})\]
DLS
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yeah
thivitaa
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wrong person to ask...:( juz read the whole thing....and am confused on what ur saying...RELLLL<3 can help..sowie mashy...<3
DLS
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how did u get the answer O.O
yrelhan4
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@thivitaa haha. :P
@DLS where are you stuck?
DLS
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Finding the resultant pellet
yrelhan4
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hold on. brb in 2 mins.
yrelhan4
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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
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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
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yeah i know arrow thin messed up
DLS
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but Fcd and Fcb are not not providing centripetal force.
Fcdcos45 and Fcbcos45 are.
ellaborate? :o
yrelhan4
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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
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yeah I do but applying this thing for the first time :)
DLS
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can you show the 2 forces in the diagram?
just for some more clarification
yrelhan4
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|dw:1361004736715:dw|
DLS
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confused
yrelhan4
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where are you confused?
Mashy
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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
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^ that'll do!
DLS
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why did that guy confuse me then :'( why doesnt it matter?if they are not giving the centripetal force?cos45 resolutions are?
yrelhan4
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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
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and it doesnt matter because resultant is towards the centre. so that is providing the centripetal force.
ignore this if it confuses you.
yrelhan4
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good luck and have fun. :P
DLS
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@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?
Mashy
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its the same.. one is by using i j k method.. one is by using mashy method lol :D
DLS
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:|
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tell me ijk
DLS
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meanwhile posting the next Q:P
Mashy
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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
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okay
Mashy
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yay one more medal :D
DLS
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come to next one :P