The distance of centre of mass from end A of a one dimensional rod (AB) having mass density $\rho=\rho_{0}\left(1-\frac{x^{2}}{L^{2}}\right) \mathrm{kg} / \mathrm{m}$ and length L (in meter) is $\frac{3 L}{\alpha} \mathrm{m}$. The value of $\alpha$ is ___________. (where x is the distance from end A)
Answer (integer)
8
Solution
<p>$\rho = {\rho _0}\left( {1 - {{{x^2}} \over {{L^2}}}} \right)$ kg/m</p>
<p>$${x_{cm}} = {{A\int\limits_0^L {{\rho _0}\left( {1 - {{{x^2}} \over {{L^2}}}} \right)x\,dx} } \over {A\int\limits_0^L {{\rho _0}\left( {1 - {{{x^2}} \over {{L^2}}}} \right)\,dx} }}$$</p>
<p>$${x_{cm}} = {{{{{L^2}} \over 2} - {{{L^2}} \over 4}} \over {L - {L \over 3}}} = {{{{{L^2}} \over 4}} \over {{{2L} \over 3}}} = {{3L} \over 8}$$</p>
<p>$\Rightarrow \alpha = 8$</p>
About this question
Subject: Physics · Chapter: Rotational Motion · Topic: Centre of Mass
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