Question

A rod of length L has non-uniformly distributed mass along its length. For its mass per unit length varying with distance x from one end as $\frac{m_0}{L^2}(L+x)$,find the position of centre of mass of this system.

• A. $L/4$
• B. $L/2$
• C. $3L/9$
• D. $5L/9$

Answer 1

The correct option is D $5L/9$

$X_{CM}=\frac{{\int }_0^{L}xdm}{{\int }_0^{L}dm}=\frac{{\int }_0^{L}x\frac{m_0}{L^2}(L+x)dx}{{\int }_0^L\frac{m_0}{L^2}(L+x)dx}$
$=\frac{{\int }_0^{L}x(L+x)dx}{{\int }_0^L(L+x)dx}=\frac{{\int }_0^L(xL+x^2)dx}{{\int }_0^L(L+x)dx}$
$=\frac{{\left(\frac{x^2}{2}L\right)}_0^L+{\left(\frac{x^3}{3}\right)}_0^L}{L(x{)}_0^L{\left(\frac{x^2}{2}\right)}_0^L}=\frac{\frac{L^3}{2}+\frac{L^3}{3}}{L^2+\frac{L^2}{2}}$
$=\frac{\frac{3L^3+2L^3}{6}}{\frac{3L^2}{2}}=\frac{5L^3}{6}\times \frac{2}{3L^2}=\frac{5}{9}L$