Question

A rod of length $L$ and cross section area $A$ has variable density according to the relation $\rho \left(x\right)={\rho }_0+kx$ for $0\le x\le \frac{L}{2}$ and $\rho \left(x\right)=2x^2$ for $\frac{L}{2}\le x\le L$ where ${\rho }_0$ and $k$ are constants. Find the mass of the rod.

• A. $\left(\begin{array}{c}\frac{7L^3}{24}+{\rho }_0\frac{L}{2}+\frac{k{L}^2}{8}\end{array}\right)A$
• B. $\left(\begin{array}{c}\frac{7L^3}{24}+{\rho }_0\frac{L}{2}+\frac{k{L}^2}{4}\end{array}\right)A$
• C. $\left(\begin{array}{c}\frac{14L^3}{24}+{\rho }_0\frac{L}{2}+\frac{k{L}^2}{8}\end{array}\right)A$
• D. $\left(\begin{array}{c}\frac{14L^3}{24}+{\rho }_0\frac{L}{4}+\frac{k{L}^2}{8}\end{array}\right)A$

Answer 1

Answer: (C)

$\left(\begin{array}{c}\frac{14L^3}{24}+{\rho }_0\frac{L}{2}+\frac{k{L}^2}{8}\end{array}\right)A$

$M_1=\underset{0}{\stackrel{L/2}{\int }}A\left(\begin{array}{c}{\rho }_0+kx\end{array}\right)dx=\left(\begin{array}{c}{\rho }_0\frac{L}{2}+\frac{k{L}^2}{8}\end{array}\right)A$

$M_2=\underset{L/2}{\stackrel{L}{\int }}A\left(2x^{2}dx\right)=\frac{2}{3}\left[\begin{array}{c}{L}^3-\frac{L^3}{8}\end{array}\right]=\frac{14L^3}{24}A$

$M_{total}=M_1+M_2=\left(\begin{array}{c}\frac{14L^3}{24}+{\rho }_0\frac{L}{2}+\frac{k{L}^2}{8}\end{array}\right)A$