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Question

A 2-dimensional bent rod is made out of a material with uniform density.
Its center of mass is located at which position coordinate? Refer to the diagram of the bent rod and ignore the rod's thickness in your calculations.
496054_efc867be6e714b1985c486b43e42c8a3.png


A
(0,0)
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B
(w4,l4)
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C
(w3,l3)
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D
(w2,l2)
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E
(w22(l+w),l22(l+w))
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Solution

The correct option is B $$\left(\dfrac{{w}^{2}}{2\left(l+w\right)}, \dfrac{{l}^{2}}{2\left(l+w\right)}\right)$$
Mass of horizontal rod      $$m_1 = \rho  w$$               where $$\rho$$ is the density of the rod
Mass of vertical rod      $$m_2 = \rho  l$$
Centre of mass of an individual rod lies at its geometric centre.
From figure,   $$x_1 = \dfrac{w}{2}$$                $$y_2 = \dfrac{l}{2}$$             $$x_2 = y_1 = 0$$

Horizontal direction :        $$x_{cm} = \dfrac{m_1 x_1 + m_2 x_2}{m_1 +m_2}$$
$$\therefore$$     $$x_{cm} = \dfrac{(\rho w) \times \frac{w}{2}+ 0}{\rho  w + \rho l } = \dfrac{w^2}{2(l+w)}$$

Vertical direction :        $$y_{cm} = \dfrac{m_1 y_1 + m_2 y_2}{m_1 +m_2}$$
$$\therefore$$     $$y_{cm} = \dfrac{0+ (\rho l) \times \frac{l}{2}}{\rho  w + \rho l } = \dfrac{l^2}{2(l+w)}$$
Thus centre of mass coordinates :    $$\bigg(\dfrac{w^2}{2 (l+w)}, \dfrac{l^2}{2(l+w)} \bigg)$$

547818_496054_ans_a3bf9410a4f546869a6f02fcfe8c7b9e.png

Physics

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