Physics - Center of Mass Question with Solution | TestHub

The position of the center of mass is given by $x_{CM}=\frac{\int xdm}{\int dm}$. Here, $dm=\lambda (x)dx=A(1+\frac{x}{L})dx$. The total mass $M={\int }_0^{L}A(1+\frac{x}{L})dx=A[x+\frac{x^2}{2L}{]}_0^L=A(L+\frac{L}{2})=\frac{3AL}{2}$. The numerator is ${\int }_0^{L}xA(1+\frac{x}{L})dx=A{\int }_0^L(x+\frac{x^2}{L})dx=A[\frac{x^2}{2}+\frac{x^3}{3L}{]}_0^L=A(\frac{L^2}{2}+\frac{L^2}{3})=\frac{5A{L}^2}{6}$. Thus, $x_{CM}=\frac{5A{L}^2/6}{3AL/2}=\frac{5L}{9}$.