Question

Where is the center of mass, $x_{cm}$ ?

• A. $\frac{3l}{7}$
• B. $\frac{6l}{7}$
• C. $\frac{4l}{5}$
• D. $\frac{7l}{9}$

Answer 1

Answer: (C)

$\frac{4l}{5}$

Mass of the rod, $m={\int }_0^l\rho dx=p_0{\int }_0^l\frac{x^3}{l^3}=\frac{{\rho }_{0}l}{4}\Rightarrow \frac{{\rho }_0}{m}=\frac{4}{l}$

Center of mass, $x_{cm}=\frac{1}{m}{\int }_0^l\rho xdx=\frac{{\rho }_0}{m}{\int }_0^l\frac{x^4}{l^3}=\frac{{\rho }_0}{m}\frac{l^5}{5l^3}=\frac{4}{l}\times \frac{l^2}{5}=\frac{4l}{5}$