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Three particles of the same mass are kept at the vertices of an equilateral triangle. The mass of each particle is m and the length of an arm of the triangle isl. Due to the the mutual gravitational force of attraction, the particles revolve along the circumcircle of the triangle. Find the velocity of each particle. |
Answer» Solution :ABC is an equilateral triangle . At its vertices three particles A,B and C each of mass m , are kept [Fig.1.4].  The point of intersection of the medians of ABC is O, which is also the centre of the circumcircle of the triangle. HENCE, radius of the circumcircle, `r=OA =2/3AD =2/3AB sin 60^@= 2/3*l*sqrt(3)/2=l/sqrt(3)`. Hence the centripetal force needed by each particle to revolve along the circumcircle with velocity v is `F_1=(mv^2)/r=(mv^2)/(l/sqrt(3))=(sqrt(3)mv^2)/l` Now the gravitational force acting on the particle at point can be calculated. The force of gravitation on the particle A due to the particle at B along AB, `F=(G*m*m)/((AB)^2)=(Gm^2)/(l^2)` COMPONENT of this force along AE =`cos 60^@` and that along `AD=Fsin 60^@` Again , gravitational attraction onthe particle at A due to the particle kept at C is F along AC. Components ofthis force along `AH =F cos 60^@` and that along `AD =F sin 60^@`. Clearly, components along AH an AE cancel each other Hence , the resultant force `F_2` on the particle kept at A =sum of the components along AD. `therefore F_2=F sin 60^@ +Fsin 60^@ =2Fsin 60^@` `=2F*(sqrt(3))/(2)=sqrt(3)F=(sqrt(3)Gm^2)/(l^2)` As PAR the question, this force of attraction due to gravitation supplies the necessary centripetal force. Hence `F_1=F_2 or, (sqrt(3)mv^2)/l =(sqrt(3)Gm^2)/(l^2) or, v^2 =(Gm)/lor, v=sqrt((Gm)/l)`. |
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