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Three particles of equal mass m are situated at verticles of a equilateral triangle of side l. What should be velocity of each particle so that they move on a circular path without changing l

\[(a)\;\sqrt {\frac{GM}{2l}} \quad (b)\;\sqrt {\frac{GM}{l}} \quad (c)\;\sqrt {\frac{2GM}{l}} \quad (d)\;\sqrt {\frac{GM}{3l}} \]

1 Answer

The particle to move in a circular path without chaging distance 'l' will happen if the resultant gravitational force on each particle provides the centripetal force necessary for the motion
$\large\frac{mv^2}{r}$$=\sqrt {F^2+F^2+2 F^2 \cos 60}$
$\large\frac{mv^2}{r}$$=\sqrt {3 }\; F$
Radius of the path is
$r= \large\frac{\sqrt 3}{2}$$l \times \large\frac{2}{3}$
$\quad= \large\frac{l}{3}$
$\therefore v= \sqrt {\large\frac {GM}{l}}$
Hence b is the correct answer.


answered Aug 30, 2013 by meena.p
edited Feb 17, 2014 by meena.p

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