Question

The density of a linear rod of length $L=1\text{m}$ varies as $\rho =A+Bx$, where $A={10}^4{\text{kg/m}}^3\&B={10}^3{\text{kg/m}}^2$ are constant, and $x$ is the distance from the left end of the rod. Locate the center of mass of the rod.

• A. $\frac{32}{63}\text{m}$
• B. $\frac{23}{63}\text{m}$
• C. $\frac{5}{9}\text{m}$
• D. $\frac{40}{63}\text{m}$

Answer 1

The correct option is A $\frac{32}{63}\text{m}$


Let us consider a small elemental mass $dm$ of length $dx$ and area of cross-section $a$ at a distance $x$ away from the left end of the rod as shown in figure.

Mass of the element is,

$m=\rho dV=(A+Bx)$ $adx$.

The $x$-coordinate of the centre of mass of the rod is,

$X_{CM}=\frac{\int xdm}{\int dm}=\frac{\int x(A+Bx)adx}{\int (A+Bx)adx}$

On integrating the above relation, with limit $0$ to $L$ we get,

$X_{CM}=\frac{{\int }_0^{L}x(A+Bx)adx}{{\int }_0^L(A+Bx)adx}=\frac{A\frac{L^2}{2}+B\frac{L^3}{3}}{AL+B\frac{L^2}{2}}$

$\therefore X_{CM}=\frac{3AL+2B{L}^2}{3(2A+BL)}$

Substituting the values,

$X_{CM}=\frac{3\times {10}^4\times 1+2\times {10}^3\times 1^2}{3(2\times {10}^4+{10}^3\times 1)}=\frac{32}{63}\text{m}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(A\right)$ is the correct answer.