Question

The linear density of rod of length $L$ and placed along x-axis with the lighter and at origin, is given by $\lambda =B{x}^2$ where $B$ is a constant. The the c-ordinates of centre of mass are

• A. $(\frac{3L}{4},0)$
• B. $(0,\frac{3L}{4})$
• C. $(\frac{4L}{3},0)$
• D. $\left(\frac{0,4L}{3}\right)$

Answer 1

The correct option is A $(\frac{3L}{4},0)$

Mass of the element $dx$ situated at $X$ is $dm=\lambda dx=B{x}^{2}dx$
$X_{com}=\frac{{\int }_0^{L}xdm}{{\int }_0^{L}dm}$
$=\frac{\frac{B[x^4{]}_0^L}{4}}{\frac{B[x^3{]}_0^L}{3}}=\frac{B{L}^4}{4}\times \frac{3}{B{L}^3}$
$\therefore X_{com}=\frac{3L}{4}$

So the co-ordinate of centre of mass $=(\frac{3L}{4},0)$