The position vector of the centre of mass vec | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

Three parts of rod can be considered as point masses.

${\overrightarrow r _{cm}} = \frac{{2m\,{{\overrightarrow r }_1} + m{{\overrightarrow r }_2}

Vedclass · Accepted Answer

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

Vedclass · Answer

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$

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

Similar Questions

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?

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) :