The position vector of the centre of mass r→cm of an symmetric uniform bar of negligible area of cross-section as shown in figure is :

r ⇀ cm = 13 8 L i ^ + 5 8 L j ^

r ⇀ cm = 13 8 L i ^ + 5 8 L j ^

r ⇀ cm = 3 8 L i ^ + 11 8 L j ^

r ⇀ cm = 5 8 L i ^ + 13 8 L j ^

r ⇀ cm = 11 8 L i ^ + 3 8 L j ^

Book Online Demo

Check Your IQ

Try Test

Courses

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

see full answer

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

$\overset{⇀}{r} cm = \frac{13}{8} L \hat{i} + \frac{5}{8} L \hat{j}$

$\overset{⇀}{r} cm = \frac{3}{8} L \hat{i} + \frac{11}{8} L \hat{j}$

$\overset{⇀}{r} cm = \frac{5}{8} L \hat{i} + \frac{13}{8} L \hat{j}$

$\overset{⇀}{r} cm = \frac{11}{8} L \hat{i} + \frac{3}{8} L \hat{j}$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

From the question,

$X_{cm} = \frac{2 mL + 2 mL + \frac{5 mL}{2}}{4 m} = \frac{13}{8} L$

$Y_{cm} = \frac{2 m \times L + m \times (\frac{L}{2}) + m \times 0}{4 m} = \frac{5 L}{8}$

Hence the correct answer is $\vec{r} cm = \frac{13}{8} L \hat{i} + \frac{5}{8} L \hat{j}$ .

Watch 3-min video & get full concept clarity