A rod of length 3 m and its mass per unit length is directly proportional to distance $x$ from one o

Question

A rod of length 3 m and its mass per unit length is directly proportional to distance $x$ from one of its end then its centre of gravity from that end will be at\

Professor PC Thomas Classes · Answer

(b) : Let us consider an elementary length $d x$ at a distance $x$ from one end.\ Its mass $=k \cdot x \cdot d x$\[0pt] [ $k=$ proportionality constant]\ Then centre of gravity of the\ $\operatorname{rod} x_{c}$ is given by\ $$ x_{c}=\frac{\int_{0}^{3} k x d x \cdot x}{\int_{0}^{3} k x d x}=\frac{\int_{0}^{3} x^{2} d x}{\int_{0}^{3} x d x}=\frac{\left.\frac{x^{3}}{3}\right|_{0} ^{3}}{\left.\frac{x^{2}}{2}\right|_{0} ^{3}} $$ or $\quad x_{c}=\frac{27 / 3}{9 / 2}=2$.\ $\therefore \quad$ Centre of gravity of the rod will be at distance of 2 m from one end.\