Question

Three identical uniform rods of the same mass $M$ and length $L$ are arranged in $xy$ plane as shown in the figure. A fourth uniform rod of mass $3M$ has been placed as shown in the $xy$ plane. What should be the value of the length of the fourth rod such that the centre of mass of all the four rods lie at the origin?

Answer 1

Let the length of forth rod is L_{1}

coordinate of the center of mass of combined rod is at the origin.

so, x_{cm} and y_{cm}is at the origin.

$x_{cm}=\frac{M_1{x}_1+M_2{x}_2+M_3{x}_3+M_4{x}_4}{M_1+M_2+M_3+M_4}=0$

$M\left(0\right)+M\frac{L}{2}+M\frac{L}{2\sqrt{2}}+3M\frac{-L_1}{2\sqrt{2}}=\frac{3L_1}{2\sqrt{2}}=\frac{L(\sqrt{2}+1)}{2\sqrt{2}}$

= $L_1=\frac{L(\sqrt{2}+1)}{3}$

this length is required for the center of mass is at the origin.

similarly, you can also do $y_{cm}.$