Question

The centre of gravity (centre of mass) of a rod (of length $L$) whose linear mass density varies as the square of the distance from one end is at

• A. $\frac{L}{3}$
• B. $\frac{3L}{4}$
• C. $\frac{3L}{5}$
• D. $\frac{2L}{5}$

Answer 1

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

Center of the gravity of a rod of length $L$
$x_{\text{cm}}=\frac{\underset{0}{\overset{L}{\int }}xdm}{\underset{0}{\overset{L}{\int }}dm}=\frac{\underset{0}{\overset{L}{\int }}x\lambda dx}{\underset{0}{\overset{L}{\int }}\lambda dx}$
Let linear mass density $\lambda$ which is directly proportional to square of distance,
therefore
$\Rightarrow x_{\text{cm}}$$=\frac{\underset{0}{\overset{L}{\int }}x\cdot k\cdot x^{2}dx}{\underset{0}{\overset{L}{\int }}k\cdot x^{2}dx}$
$\Rightarrow x_{\text{cm}}$$=\frac{\underset{0}{\overset{L}{\int }}{x}^{3}dx}{\underset{0}{\overset{L}{\int }}{x}^{2}dx}=\frac{\frac{L^4}{4}}{\frac{L^3}{3}}=\frac{3L}{4}$