Three identical metal balls each of radius ' $r$ ' are placed such that an equilateral triangle is formed…
Three identical metal balls each of radius ' $r$ ' are placed such that an equilateral triangle is formed when centres of three ball are joined. The centre of mass of the system is located at
centre of one of the balls.
point of intersection of medians.
line joining centres of any two balls.
on the circumference of any one of the balls.
Solution
Balls are identical and of same radius, so volume and density of balls are the same.
$\therefore \quad$ Masses of the ball will be same
$\begin{aligned}
& \mathrm{X}_{\mathrm{cm}}=\frac{\mathrm{mx}_1+\mathrm{mx}_2+\mathrm{mx}_3}{\mathrm{~m}+\mathrm{m}+\mathrm{m}}=\frac{\mathrm{x}_1+\mathrm{x}_2+\mathrm{x}_3}{3} \\
& \mathrm{Y}_{\mathrm{cm}}=\frac{\mathrm{my}_1+\mathrm{my}_2+\mathrm{my}_3}{\mathrm{~m}+\mathrm{m}+\mathrm{m}}=\frac{\mathrm{y}_1+\mathrm{y}_2+\mathrm{y}_3}{3}
\end{aligned}$
$\mathrm{X}_{\mathrm{cm}}, \mathrm{Y}_{\mathrm{cm}}$ is the centroid of the equilateral triangle.