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
$\vec r\,cm = \frac{{13}}{8}L\hat x + \frac{5}{8}L\hat y$
$\vec r\,cm = \frac{{5}}{8}L\hat x + \frac{13}{8}L\hat y$
$\vec r\,cm = \frac{{3}}{8}L\hat x + \frac{11}{8}L\hat y$
$\vec r\,cm = \frac{{11}}{8}L\hat x + \frac{3}{8}L\hat y$
For determining centre of mass of a body why a body is considered as composited of multiple of small mass elements ?
A uniform thin rod $AB$ of length $L$ has linear mass density $\mu \left( x \right) = a + \frac{{bx}}{L}$ , where $x$ is measured from $A$. If the $CM$ of the rod lies at a distance of $\left( {\frac{7}{12}} \right)L$ from $A$, then $a$ and $b$ are related as
What are the position of centre of mass of symmetrical and homogeneous bodies?
Two masses $M$ and $m$ are attached to a vertical axis by weightless threads of combined length $l$. They are set in rotational motion in a horizontal plane about this axis with constant angular velocity $\omega $. If the tensions in the threads are the same during motion, the distance of $M$ from the axis is
The $(x -y)$ co-ordinates $(in\ cm)$ of the centre of mass of letter $E$ relative to the origin $O$ , whose dimensions are shown in the figure is :
(Take width of the letter $2\ cm$ every where) :