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The distance of centre of mass from end $\mathrm{A}$ of a one dimensional rod (AB) having mass density $\mathrm{\varrho=\varrho_{0}\left(1-\frac{x^{2}}{L^{2}}\right) \mathrm{kg} / \mathrm{m}}$ and length $\mathrm{L}$ (in meter) is $\mathrm{\frac{3 L}{\alpha} \mathrm{m}}$ . The value of $\alpha$ is____________ . (where $\mathrm{x}$ is the distance from end $\mathrm{A}$ )
Option: 1
8

Option: 2
-

Option: 3
-

Option: 4
-

Answers (1)

$\mathrm{ \rho =\rho _{0}\left(1-\frac{x^2}{L^2}\right)}$

$\mathrm{d m=\rho \: d x}$

$\mathrm{d m=\rho _0\left(1-\frac{x^2}{L^2}\right) d x}$

$\mathrm{x_{c m}=\frac{3 L}{\alpha}=\frac{\int_0^L x d m}{\int_{0}^{L} d m}}$

$\mathrm{=\frac{\left.\rho _{o} \left(\left[\frac{x^2}{2}\right]_0^L-\left[\frac{x^4}{4L^{2}}\right]^L_{0}\right]\right)}{ {\rho _{o}}\left([x] ^{L}_{0}-\left(x^3 / 3 L^2\right)^L_{0}]\right.}}$

$\mathrm{\frac{3L}{\alpha }=\frac{\rho _{o}\left [ \left ( \frac{L_{2}}{2} \right )-\left ( \frac{L^{2}}{4} \right ) \right ]}{\rho _{o}\left ( L-\frac{L}{3} \right )}}$

$\mathrm{=\frac{\frac{L^{2}}{4}}{\frac{2L}{3}}}$

$\mathrm{\frac{3L}{\alpha }=\frac{3L}{8}}$

$\therefore \alpha =8$

The value of $\alpha$ is $8$

Posted by

Sumit Saini

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