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.

Question

dls · Answer

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(I suppose......)

anonymous · Answer

Can you draw?

dls · Answer

:P

anonymous · Answer

wow you read my mind :D

dls · Answer

:D

anonymous · Answer

but. think.. the NET force = centripetal force !!

dls · Answer

yeah I know thats the ending part

dls · Answer

$$\LARGE F_{ca}=\frac{GM^2}{4R^2}$$

anonymous · Answer

how did you get that bold equation???? and so big?? :O :O

dls · Answer

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(if im right...)

dls · Answer

use the keyword large :P

anonymous · Answer

ok that makes sense.. !! proceed with your calculation!

dls · Answer

$$\LARGE F_{BC}=\frac{GM^2}{2R^2}$$
if im right........

dls · Answer

$$\LARGE F_{CD}=\frac{GM^2}{2R^2}$$

anonymous · Answer

yea correct.. now take vector sum of them!

dls · Answer

now we gotta find resultant if we are correct till here

anonymous · Answer

i guess so!.. proceed

dls · Answer

2 at a time?

anonymous · Answer

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 · Answer

and your directions are wrong.. you are considering force ON c.. so opposite direction!

dls · Answer

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will it work out like this?

dls · Answer

ah yeah sorry

anonymous · Answer

sheehs what is that?? :O :O :O

anonymous · Answer

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