Questions

Three identical rods each of length L and mass m are joined to form a U-shape as shown in figure .The distance of center of mass of the combined system from the corner 'O' is

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\frac{L}{2}

\frac{L}{3}

\frac{\sqrt{8} L}{6}

\frac{\sqrt{13} L}{6}

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detailed solution

Correct option is D

$m_{1} = m, m_{2} = m, m_{3} = m$

$x_{cm} = \frac{m_{1} x_{1} + m_{2} x_{2} + m_{3} x_{3}}{m_{1} + m_{2} + m_{3}}$

$= \frac{0 \times m \times \frac{L}{2} \times m \times L}{m + m + m}$

$= \frac{m (\frac{3 L}{2})}{3 m} = \frac{3 L}{6} = \frac{L}{2}$

$y_{cm} = \frac{m_{1} y_{1} + m_{2} y_{2} + {my}_{3}}{m_{1} + m_{2} + m_{3}}$

$= (\frac{m \times \frac{L}{2} + m \times 0 + m \times \frac{L}{2}}{3 m})$

$= \frac{\frac{mL}{2} + \frac{mL}{2}}{3 m} = \frac{\frac{2 mL}{2}}{3 m} = \frac{L}{3}$

$r_{cm} = x_{cm} i + y_{cm} j$

$= \frac{L}{2} i + \frac{L}{3} j$

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

$= \sqrt{\frac{L^{2}}{4} + \frac{L^{2}}{9}} = \sqrt{\frac{9 L^{2} + 4 L^{2}}{36}}$

$= \sqrt{\frac{13 L^{2}}{36}} = \frac{\sqrt{13} L}{6}$

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