The position vector of the centre of mass rcm of an asymmetric uniform bar of negligible area of cross-section as shown in figure is

Question

TuteeHUB · Accepted Answer

${\rm{r}} = \frac{{11}}{8}L\hat x + \frac{3}{8}L\hat y$

TuteeHUB · Answer

${\rm{r}} = \frac{{13}}{8}{\rm{L}}\hat x + \frac{5}{8}{\rm{L}}\hat y$

TuteeHUB · Answer

${\rm{r}} = \frac{3}{8}L\hat x + \frac{{11}}{8}L\hat y$

TuteeHUB · Answer

${\rm{r}} = \frac{5}{8}L\hat x + \frac{{13}}{8}L\hat y$

1.	The position vector of the centre of mass rcm of an asymmetric uniform bar of negligible area of cross-section as shown in figure is
A.	\({\rm{r}} = \frac{{13}}{8}{\rm{L}}\hat x + \frac{5}{8}{\rm{L}}\hat y\)
B.	\({\rm{r}} = \frac{{11}}{8}L\hat x + \frac{3}{8}L\hat y\)
C.	\({\rm{r}} = \frac{3}{8}L\hat x + \frac{{11}}{8}L\hat y\)
D.	\({\rm{r}} = \frac{5}{8}L\hat x + \frac{{13}}{8}L\hat y\)
Answer» B. \({\rm{r}} = \frac{{11}}{8}L\hat x + \frac{3}{8}L\hat y\)

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