Stable particle ring - JEE

Since the motion of the 4 particles is constrained, we can simply find their respective speeds through the radial acceleration equation,

${ a }_{ radial }\quad =\quad \frac { v^{ 2 }\quad }{ R }$

From that, we only need to find ${ a }_{ radial }$ and we're good to go.

So first, use the given image to imagine the force vectors of a single particle (just focus on one to save you the hassle), in my case, the leftmost particle will be the point of focus. I believe you'll agree with me that the three force vectors are one that's point ingdirectly to the right, one that's 45° upward and the third one's 45° downward. From there, you only need to find the +x components of each vectors (radial force).

${ F }_{ radial }\quad =\quad { F }_{ 1 }cos45\quad +\quad { F }_{ 2 }\quad +\quad F_{ 3 }cos45\\ M{ a }_{ radial }\quad =\quad G\frac { { M }^{ 2 } }{ 2{ R }^{ 2 } } cos45\quad +\quad G\frac { { M }^{ 2 } }{ 4R^{ 2 } } \quad +\quad G\frac { { M }^{ 2 } }{ 2{ R }^{ 2 } } cos45\\ { a }_{ radial }\quad =\quad G\frac { { M } }{ 4R^{ 2 } } (1\quad +\quad 2\sqrt { 2 } )$

After which, just substitute ${ a }_{ radial }$ to the first equation above and solve for v.

Particles are not equidistant to each other as stated...

easy1, mere calculation

can u give the discription for this answer pls

Net force on any one particle will be sum of gravitational forces from other three particles and this force will be equal to centripetal force so \frac { M{ u }^{ 2 } }{ { R } } =\frac { G{ M }^{ 2 } }{ { R }^{ 2 } } \left[ \frac { 1+2\sqrt { 2 } }{ 4 } \right] and you get the answer u=\frac { 1 }{ 2 } \sqrt { \frac { GM }{ R } \left( 1+2\sqrt { 2 } \right) }

The question is wrong......the particles are not equidistant from each other as stated

particles should have been given to be on vertices of a square