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

MATH103 April 2016

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Question 04 (b)
The density of a metal bar 3 metres long is is given by $\rho (x)={\frac {1}{\sqrt {x+1}}}$ kilograms per metre ($0\leq x\leq 3$). Find the centre of mass of the bar.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Let $m$ be the mass of the bar (obtained in Question 04 (a)) and $M$ its moment. Then the centre of mass of the bar, ${\bar {x}}$, is given by ${\bar {x}}={\frac {M}{m}}.$

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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 ${\bar {x}}={\frac {M}{m}}={\frac {8/3}{2}}=\color {blue}{\frac {4}{3}}.$