A uniform thin rod AB of length L has linear mass density μ x = a + b x L , where x is measured from A. If…

1. $2a=b$
2. $a=2b$
3. $a=b$
4. $3a=2b$

Solution

Consider an infinitisemal small element of length $dx$ and mars density $\lambda$

$x_{cm}=\frac{\int \left(dm\right)x}{\int dm}=\frac{\int \left(\lambda dx\right)x}{\int \lambda dx}$

$=\frac{{\int }_0^L\left(a+\frac{b}{L}x\right)xdx}{{\int }_0^L\left(a+\frac{b}{L}x\right)dx}$

$=\frac{{\int }_0^{L}axdx+{\int }_0^L\frac{b}{L}{x}^{2}dx}{{\int }_0^{L}adx+{\int }_0^L\frac{b}{L}xdx}$

$=\frac{a{\left[\frac{x^2}{2}\right]}_0^L+\frac{b}{L} {\left[\frac{x^3}{3}\right]}_0^L}{a{\left[x\right]}_0^L+\frac{b}{L}{\left[\frac{x^2}{2}\right]}_0^L}$

$=\frac{\frac{a}{2}\left(L^2-0^2\right)+\frac{b}{3L} \left(L^3-0^3\right)}{a\left(L-0\right)+\frac{b}{2L}\left(L^2-0^2\right)}$

$=\frac{\frac{a}{2}{L}^2+\frac{b{L}^2}{3}}{aL+\frac{bL}{2}}=\frac{\frac{a}{2}L+\frac{bL}{3}}{a+\frac{b}{2}}$

$\because x_{cm}=\frac{7L}{12}=\frac{\left(\frac{a}{2}+\frac{b}{3}\right)L}{a+\frac{b}{2}}\Rightarrow \frac{7}{12}= \frac{\frac{3a+2b}{6}}{\frac{2a+b}{2}}$

$\Rightarrow \frac{7}{12}=\frac{3a+2b}{3}\times \frac{1}{\left(2a+b\right)}\Rightarrow 7\left(2a+b\right)=4\left(3a+2b\right)$

$\Rightarrow 14a+7b=12a+8b\Rightarrow 2a=b$

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