Science:Math Exam Resources/Courses/MATH103/April 2005/Question 01 (e)

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MATH103 April 2005

  •  Q1 (a)  •  Q1 (b)  •  Q1 (c)  •  Q1 (d)  •  Q1 (e)  •  Q2  •  Q3 (a)  •  Q3 (b)  •  Q3 (c)  •  Q4  •  Q5 (a)  •  Q5 (b)  •  Q5 (c)  •  Q6 (a)  •  Q6 (b)  •  Q6 (c)  •  Q7 (a)  •  Q7 (b)  •  Q7 (c)  •  Q8 (a)  •  Q8 (b)  •  Q8 (c)  •  

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Question 01 (e)

Multiple Choice Question: Select ONE correct answer (i), (ii), (iii), (iv), or (v). (e) Consider the mass density function $d(x)=4-x^{2}$ over the interval $0\leq x\leq 2$. The centre of mass of this mass distribution is (i)   $\displaystyle {}1$     (ii)   $\displaystyle {}4$     (iii)   $\displaystyle {}{\frac {4}{3}}$     (iv)   $\displaystyle {}{\frac {3}{4}}$     (v)   $\displaystyle {}{\frac {16}{3}}$

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 formula for the centre of mass of a continuous distribution over the interval $0\leq x\leq L$ is ${\overline {x}}={\frac {\int _{0}^{L}x\rho (x)dx}{\int _{0}^{L}\rho (x)dx}}$ where $\rho (x)$ is the mass density function.

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Solution

Given the mass density function $d(x)=4-x^{2}$ over the interval $0\leq x\leq 2$, we use the formula for the centre of mass to get ${\begin{aligned}{\overline {x}}&={\frac {\int _{0}^{L}x\rho (x)dx}{\int _{0}^{L}\rho (x)dx}}\\&={\frac {\int _{0}^{2}x(4-x^{2})dx}{\int _{0}^{2}(4-x^{2})dx}}\\&={\frac {\int _{0}^{2}(4x-x^{3})dx}{\int _{0}^{2}(4-x^{2})dx}}\\&={\frac {I_{1}}{I_{2}}}\end{aligned}}$ The integral on the top is calculated as ${\begin{aligned}I_{1}&=\int _{0}^{2}(4x-x^{3})dx\\&=\left[2x^{2}-{\frac {x^{4}}{4}}\right]_{0}^{2}\\&=\left[\left(2(2)^{2}-{\frac {(2)^{4}}{4}}\right)-\left(2(0)^{2}-{\frac {(0)^{4}}{4}}\right)\right]\\&=4\end{aligned}}$ and the integral on the bottom reduces to ${\begin{aligned}I_{2}&=\int _{0}^{2}(4-x^{2})dx\\&=\left[4x-{\frac {x^{3}}{3}}\right]_{0}^{2}\\&=\left[\left(4(2)-{\frac {(2)^{3}}{3}}\right)-\left(4(0)-{\frac {(0)^{3}}{3}}\right)\right]\\&={\frac {16}{3}}\end{aligned}}$ Putting these together we finally obtain ${\overline {x}}={\frac {I_{1}}{I_{2}}}={\frac {4}{\frac {16}{3}}}={\frac {3}{4}}$ Therefore, the correct answer is (iv).

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