Science:Math Exam Resources/Courses/MATH103/April 2012/Question 02 (c)

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MATH103 April 2012

  •  Q1 (a)  •  Q1 (b)  •  Q1 (c)  •  Q1 (d)  •  Q1 (e)  •  Q1 (f)  •  Q1 (g)  •  Q2 (a)  •  Q2 (b)  •  Q2 (c)  •  Q2 (d)  •  Q2 (e)  •  Q2 (f)  •  Q3 (a)  •  Q3 (b)  •  Q4 (a)  •  Q4 (b)  •  Q4 (c)  •  Q4 (d)  •  Q4 (e)  •  Q4 (f)  •  Q5 (a)  •  Q5 (b)  •  Q5 (c)  •  

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Question 02 (c)
Short answer question. Simplify your answer to the best possible. Show in detail how you arrive at your answer (work will be considered for this problem). Find the center of mass of the distribution $\rho (x)=\cos \left({\frac {x}{2}}\right),\quad 0<x<\pi .$

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
The centre of mass of the shape delimited by the x-axis, the vertical lines x = a, x = b and the curve y = ƒ(x) is given by ${\bar {x}}={\frac {\int _{a}^{b}xf(x)\,dx}{\int _{a}^{b}f(x)\,dx}}$

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Solution

The centre of mass is given by ${\bar {x}}={\frac {I}{M}}$ where ${\begin{aligned}I&=\int _{0}^{\pi }x\cos \left({\frac {x}{2}}\right)\,dx\\M&=\int _{0}^{\pi }\cos \left({\frac {x}{2}}\right)\,dx\end{aligned}}$ We start with the simpler integral M and use the substitution w = x/2. Then dw = dx/2, or equivalently, dx = 2dw. Further, when x = π we have w = π/2 and when x = 0 we have w = 0. The integral M therefore becomes ${\begin{aligned}M&=\int _{0}^{\pi }\cos \left({\frac {x}{2}}\right)\,dx\\&=\int _{0}^{\pi /2}\cos(w)2\,dw\\&=\left.2\sin(w)\right|_{0}^{\pi /2}\\&=2\sin(\pi /2)-2\sin(0)=2.\end{aligned}}$ Next we calculate I. This is done using integration by parts: Choose u = x and $\displaystyle dv=\cos(x/2)dx$. Then du = dx and from the integral we just solve, we know that $v=\displaystyle 2\sin(w)=2\sin(x/2)$. Hence ${\begin{aligned}I&=\int _{0}^{\pi }x\cos \left({\frac {x}{2}}\right)\,dx=\left.2x\sin(x/2)\right|_{0}^{\pi }-\int _{0}^{\pi }2\sin(x/2)\,dx\\&=2\pi -0-\left(-\left.4\cos(x/2)\right|_{0}^{\pi }\right)\\&=2\pi +0-4=2\pi -4.\end{aligned}}$ For the second integral we again used our favourite substitution w = x/2. The center of mass is thus ${\begin{aligned}{\bar {x}}={\frac {I}{M}}={\frac {2\pi -4}{2}}=\pi -2.\end{aligned}}$

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