Question

If the linear density of the rod of length $L$ varies as $\lambda =A+Bx$, then its centre of mass is given by:

• A. $X_{cm}=\frac{L(2A+BL)}{3(3A+2BL)}$
• B. $X_{cm}=\frac{L(3A+2BL)}{3(2A+BL)}$
• C. $X_{cm}=\frac{L(3A+2BL)}{3}$
• D. $X_{cm}=\frac{L(2A+3BL)}{3}$

Answer 1

The correct option is B $X_{cm}=\frac{L(3A+2BL)}{3(2A+BL)}$

Consider an element $dx$ at a distance $x$ from one end of the rod of length $L$.

The center of mass of the rod is $X_{cm}=\frac{{\int }_0^{L}x\lambda dx}{{\int }_0^L\lambda dx}$

or $X_{cm}=\frac{{\int }_0^{L}x(A+Bx)dx}{{\int }_0^L(A+Bx)dx}$

$=\frac{[A{x}^2/2+B{x}^3/3{]}_0^L}{[Ax+B{x}^2/2{]}_0^L}=\frac{A{L}^2/2+B{L}^3/3}{AL+B{L}^2/2}=\frac{3A{L}^2+2B{L}^3}{6}\times \frac{2}{2AL+B{L}^2}$

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

$=\frac{L(3A+2BL)}{3(2A+BL)}$