The distance of the centre of mass from end A | vedclass

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(C) The mass of a small element $dx$ at distance $x$ is given by $dm = \rho \cdot dx = \rho_{0} \left(1 - \frac{x^{2}}{L^{2}}\right) dx$.The position

Vedclass · Accepted Answer

$8$

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(C) The mass of a small element $dx$ at distance $x$ is given by $dm = \rho \cdot dx = \rho_{0} \left(1 - \frac{x^{2}}{L^{2}}\right) dx$.The position of the centre of mass $X_{cm}$ is given by the formula:$X_{cm} = \frac{\int x \, dm}{\int dm}$First,calculate the total mass $M = \int_{0}^{L} \rho_{0} \left(1 - \frac{x^{2}}{L^{2}}\right) dx = \rho_{0} \left[ x - \frac{x^{3}}{3L^{2}} \right]_{0}^{L} = \rho_{0} \left( L - \frac{L}{3} \right) = \frac{2}{3} \rho_{0} L$.Next,calculate the integral $\int x \, dm = \int_{0}^{L} x \cdot \rho_{0} \left(1 - \frac{x^{2}}{L^{2}}\right) dx = \rho_{0} \int_{0}^{L} \left( x - \frac{x^{3}}{L^{2}} \right) dx = \rho_{0} \left[ \frac{x^{2}}{2} - \frac{x^{4}}{4L^{2}} \right]_{0}^{L} = \rho_{0} \left( \frac{L^{2}}{2} - \frac{L^{2}}{4} \right) = \frac{1}{4} \rho_{0} L^{2}$.Now,calculate $X_{cm} = \frac{\frac{1}{4} \rho_{0} L^{2}}{\frac{2}{3} \rho_{0} L} = \frac{1}{4} \cdot \frac{3}{2} L = \frac{3L}{8}$.Comparing this with $\frac{3L}{\alpha}$,we get $\alpha = 8$.

Column-$I$	Column-$II$
$(1)$ $\frac{{{m_1}{m_2}}}{{{m_1} + {m_2}}}$	$(a)$ Reduced mass of a two-particle system
$(2)$ $\frac{{{r_1} + {r_2}}}{2}$	$(b)$ Position vector of the center of mass for a system of two equal masses

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$(1)$ $\frac{{{m_1}{m_2}}}{{{m_1} + {m_2}}}$ $(a)$ Reduced mass of a two-particle system
$(2)$ $\frac{{{r_1} + {r_2}}}{2}$ $(b)$ Position vector of the center of mass for a system of two equal masses

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