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) •
Question 01 (e)

Multiple Choice Question: Select ONE correct answer (i), (ii), (iii), (iv), or (v).
(e) Consider the mass density function
 $d(x)=4x^{2}$
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) $\displaystyle {}{\frac {16}{3}}$

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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

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