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(C) Let the three rods be $R_1, R_2, R_3$ and the fourth rod be $R_4$. $R_1$ is along the positive $X$-axis: mass $m$,center of mass $(L/2, 0)$. $R_2$

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let the three rods be $R_1, R_2, R_3$ and the fourth rod be $R_4$. $R_1$ is along the positive $X$-axis: mass $m$,center of mass $(L/2, 0)$. $R_2$ is along the positive $Y$-axis: mass $m$,center of mass $(0, L/2)$. $R_3$ is at $45^\circ$ to the $X$-axis: mass $m$,center of mass $(L/2 \cos 45^\circ, L/2 \sin 45^\circ) = (L/2\sqrt{2}, L/2\sqrt{2})$. $R_4$ has mass $3m$ and length $L_4$. It is placed in the third quadrant at $45^\circ$ to the negative $X$-axis. Its center of mass is at $(-L_4/2 \cos 45^\circ, -L_4/2 \sin 45^\circ) = (-L_4/2\sqrt{2}, -L_4/2\sqrt{2})$. For the center of mass to be at the origin $(0,0)$,the sum of the moments must be zero: $\sum m_i x_i = 0$ and $\sum m_i y_i = 0$. For $X$-coordinate: $m(L/2) + m(0) + m(L/2\sqrt{2}) + 3m(-L_4/2\sqrt{2}) = 0$. $L/2 + L/2\sqrt{2} = 3L_4/2\sqrt{2}$. Multiply by $2\sqrt{2}$: $L\sqrt{2} + L = 3L_4$. $L(\sqrt{2}+1) = 3L_4$. $L_4 = \frac{L(\sqrt{2}+1)}{3}$.

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