Question

A long thin bar of length $L$ is made of material whose density varies along the length of the bar. Let $x$ be the distance from one end of the bar. If mass density of bar is given by $P\left(kg/m\right)=a{x}^2,0\le x\le L;$ where $x$ is in meter, find the centre of mass of bar.

• A. $\frac{2L}{3}$
• B. $\frac{3L}{4}$
• C. $\frac{L}{2}$
• D. L

Answer 1

The correct option is B $\frac{3L}{4}$

Total mass of bar
=$\int$$P.dx$
=$\int$$a{x}^{2}dx$

$=a\frac{[x^3{]}_0^L}{3}=\frac{a{L}^3}{3}$
The centre of mass of bar $=\frac{{\int }_0^{L}x.Pdx}{m}$

$X_{com}=\frac{{\int }_0^{L}xa{x}^{2}dx}{\frac{a{L}^3}{3}}$

$X_{com}=\frac{{\int }_0^{L}a{x}^{3}dx}{\frac{a{L}^3}{3}}=\frac{a{\left[\frac{x^4}{4}\right]}_0^L}{\frac{a{L}^3}{3}}$
$=\frac{\frac{a{L}^4}{4}}{\frac{a{L}^3}{3}}$
$=\frac{3}{4}L$