  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. A
(0,0)  B
(w4,l4)  C
(w3,l3)  D
(w2,l2)  E
(w22(l+w),l22(l+w))  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 rodMass 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 :    $$\bigg(\dfrac{w^2}{2 (l+w)}, \dfrac{l^2}{2(l+w)} \bigg)$$ Physics

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