Science:Math Exam Resources/Courses/MATH103/April 2016/Question 04 (b)/Solution 1

To find the centre of mass of the bar, we first need to compute its mass and the moment.

From Question 04 (a), we know that the mass of the bar is $m=2.$

On the other hand, the moment $M$ of the bar is

$M=\int _{0}^{3}x\rho (x)\,dx=\int _{0}^{3}{\frac {x}{\sqrt {x+1}}}\,dx.$

By the change of variable $u=x+1,du=dx,$ we have

${\begin{aligned}\int _{0}^{3}{\frac {x}{\sqrt {x+1}}}\,dx=\int _{1}^{4}{\frac {u-1}{\sqrt {u}}}\,du&=\int _{1}^{4}\left({\frac {u}{\sqrt {u}}}-{\frac {1}{\sqrt {u}}}\right)du\\&=\int _{1}^{4}u^{\frac {1}{2}}\,du-\int _{1}^{4}u^{-{\frac {1}{2}}}\,du\\&=\left.{\frac {2}{3}}u^{\frac {3}{2}}\right|_{1}^{4}-\left.2{\sqrt {u}}\right|_{1}^{4}\\&={\frac {2}{3}}(8-1)-2(2-1)={\frac {8}{3}}.\end{aligned}}$

Therefore, by the formula for the centre of mass ${\bar {x}}$ in the hint, we have

${\bar {x}}={\frac {M}{m}}={\frac {8/3}{2}}=\color {blue}{\frac {4}{3}}.$