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(D) Let the mass per unit length of the rod be $\lambda$. The total length is $5L$,so the total mass is $M = 5L\lambda$. The two segments have lengths

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$4 \hat{i} + 9 \hat{j}$

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(D) Let the mass per unit length of the rod be $\lambda$. The total length is $5L$,so the total mass is $M = 5L\lambda$. The two segments have lengths $2L$ and $3L$,with masses $m_1 = 2L\lambda$ and $m_2 = 3L\lambda$.For the horizontal segment of length $2L$ along the $x$-axis,the center of mass is at $(x_1, y_1) = (L, 0)$.For the vertical segment of length $3L$ along the $y$-axis,the center of mass is at $(x_2, y_2) = (0, 1.5L)$.Given $L = 10 \ cm$,we have $m_1 = 20\lambda$ and $m_2 = 30\lambda$. The coordinates are $(x_1, y_1) = (10, 0)$ and $(x_2, y_2) = (0, 15)$.The $x$-coordinate of the center of mass is $x_{com} = \frac{m_1x_1 + m_2x_2}{m_1 + m_2} = \frac{20\lambda(10) + 30\lambda(0)}{50\lambda} = \frac{200}{50} = 4 \ cm$.The $y$-coordinate of the center of mass is $y_{com} = \frac{m_1y_1 + m_2y_2}{m_1 + m_2} = \frac{20\lambda(0) + 30\lambda(15)}{50\lambda} = \frac{450}{50} = 9 \ cm$.Thus,the position vector is $\vec{r}_{com} = 4 \hat{i} + 9 \hat{j}$.

$A$ rod of length $5L$ is bent at a right angle,keeping one side length as $2L$. Find the position of the center of mass of the system. (Consider $L = 10 \ cm$)

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