The position vector of the centre of mass r , cm of an asymmetric… - Vidyadip

MCQ

The position vector of the centre of mass $\vec r\, cm$ of an asymmetric uniform bar of negligible area of cross-section as shown in figure is

✓
$\vec r\,cm = \frac{{13}}{8}L\hat x + \frac{5}{8}L\hat y$
B
$\vec r\,cm = \frac{{5}}{8}L\hat x + \frac{13}{8}L\hat y$
C
$\vec r\,cm = \frac{{3}}{8}L\hat x + \frac{11}{8}L\hat y$
D
$\vec r\,cm = \frac{{11}}{8}L\hat x + \frac{3}{8}L\hat y$

✓

Answer

Correct option: A.

$\vec r\,cm = \frac{{13}}{8}L\hat x + \frac{5}{8}L\hat y$

a
Three parts of rod can be considered as point masses.

${\overrightarrow r _{cm}} = \frac{{2m\,{{\overrightarrow r }_1} + m{{\overrightarrow r }_2} + m{{\overrightarrow r }_3}}}{{4\,m}}$

${\overrightarrow r _{cm}} = \frac{{2m\left( {L\hat i + L\hat j} \right) + m\left( {2L\hat i + \frac{L}{2}\hat j} \right) + m\left( {\frac{{5L}}{2}\hat i} \right)}}{{4\,m}}$

$ = \frac{{\frac{{13}}{2}L\hat i + \frac{5}{2}L\hat j}}{4}$

$ = \frac{{13}}{8}L\hat i + \frac{5}{8}L\hat j$

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