Question 50 of 130

The distance of centre of mass from end $A$ of a one dimensional rod $\left(AB\right)$ having mass density $\rho ={\rho }_{0}\left(1-\dfrac{{x}^{2}}{{L}^{2}}\right)kg{m}^{-1}$ and length $L$ (in meter) is $\dfrac{3L}{\alpha }m$. The value of $\alpha$ is _____ (where $x$ is the distance form end $A$)

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Answer: 8

Given here:
Mass density, $\rho ={\rho }_{0}\left(1-\dfrac{{x}^{2}}{{L}^{2}}\right)kg{m}^{-1}$
Let us suppose there is an element of length $dx$ at a distance $x$ from left must end.

Mass of this element is $dm=\rho dx={\rho }_{0}\left(1-\dfrac{{x}^{2}}{{L}^{2}}\right)$
The centre of mass of this element is ${X}_{cm}=\dfrac{\int xdm}{\int dm}$
$=\dfrac{{\rho }_{0}{\int }_{0}^{l}\left(x-\dfrac{{x}^{2}}{{L}^{2}}\right)xdx}{{\int }_{0}^{l}{\rho }_{0}\left(1-\dfrac{{x}^{2}}{{L}^{2}}\right)dx}$
$=\dfrac{\dfrac{{L}^{2}}{2}-\dfrac{{L}^{4}}{4{L}^{2}}}{L-\dfrac{{L}^{3}}{3{L}^{2}}}=\dfrac{\dfrac{{L}^{2}}{4}}{\dfrac{2L}{3}}=\dfrac{3L}{8}$
Hence, the value of $\alpha =8$.