Question

The density of a linear rod of length L varies as
$\rho$ = A + Bx where x is the distance from the left end.
Locate the centre of mass.

• A. $\frac{2AL+3B{L}^2}{2(3A+BL)}$
• B. $\frac{3AL+4B{L}^2}{2(3A+2BL)}$
• C. $\frac{3AL+2B{L}^2}{3(2A+BL)}$
• D. $\frac{3AL+B{L}^2}{(2A+BL)}$

Answer 1

The correct option is C

$\frac{3AL+2B{L}^2}{3(2A+BL)}$

Let the cross-sectional area be $\alpha$. the mass of an element of length dx located at a distance x away from the left ends is (A + Bx) dx. The x-coordinate of the centre of mass is given by
$X_{cm}=\frac{\int xdm}{\int dm}=\frac{{\int }_O^{L}x(A+Bx)adx}{{\int }_O^L(A+Bx)adx}$
$=\frac{A\frac{L^2}{2}+B\frac{L^3}{3}}{AL+B\frac{L^2}{2}}=\frac{3AL+2B{L}^2}{3(2A+BL)}$