Question

The linear mass density $(\lambda)$ of a rod of length $L$ kept along $x$-axis varies as $\lambda=\alpha+\beta x$; where $\alpha$ and $\beta$ are positive constants. The centre of mass of the rod is at

• A. $\frac{(2 \beta+3 \alpha L) L}{2(2 \beta+\alpha L)}$
• B. $\frac{(3 \alpha+2 \beta L) L}{3(2 \alpha+\beta L)}$
• C. $\frac{(3 \beta+2 \alpha L) L}{3(2 \beta+\alpha L)}$
• D. $\frac{(3 \beta+2 \alpha L) L}{3 \beta+2 \alpha}$

Answer 1

Answer: (B)

$\lambda=\alpha+\beta x$
$d m=(\alpha+\beta x) d x$
$x_{c m}=\frac{\int\limits_{0}^{L} x(\alpha+\beta x) d x}{\int\limits_{0}^{L}(\alpha+\beta x) d x}$
$=\frac{\alpha \int\limits_{0}^{L} x d x+\beta \int\limits_{0}^{L} x^{2} d x}{\alpha \int\limits_{0}^{L} d x+\beta \int\limits_{0}^{L} x d x}$
$x_{c m}=\frac{\frac{\alpha L^{2}}{2}+\frac{\beta L^{3}}{3}}{\alpha L+\frac{\beta L^{2}}{2}}$