A thin bar of length L has a mass per unit le | vedclass

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Question

(C) Let the mass per unit length be $\lambda(x) = \lambda_0 + kx$.Total mass $M = \int_{0}^{L} (\lambda_0 + kx) dx = \lambda_0 L + \frac{kL^2}{2}$.Fro

Vedclass · Accepted Answer

$\frac{2L}{3} - \frac{\lambda_0 L^2}{6M}$

Vedclass · Answer

(C) Let the mass per unit length be $\lambda(x) = \lambda_0 + kx$.Total mass $M = \int_{0}^{L} (\lambda_0 + kx) dx = \lambda_0 L + \frac{kL^2}{2}$.From this,$k = \frac{2(M - \lambda_0 L)}{L^2} = \frac{2M}{L^2} - \frac{2\lambda_0}{L}$.The center of mass $x_{cm}$ is given by $x_{cm} = \frac{1}{M} \int_{0}^{L} x dm = \frac{1}{M} \int_{0}^{L} x (\lambda_0 + kx) dx$.$x_{cm} = \frac{1}{M} [\frac{\lambda_0 x^2}{2} + \frac{kx^3}{3}]_{0}^{L} = \frac{1}{M} (\frac{\lambda_0 L^2}{2} + \frac{kL^3}{3})$.Substituting $k = \frac{2M}{L^2} - \frac{2\lambda_0}{L}$ into the equation:$x_{cm} = \frac{1}{M} [\frac{\lambda_0 L^2}{2} + \frac{L^3}{3} (\frac{2M}{L^2} - \frac{2\lambda_0}{L})] = \frac{1}{M} [\frac{\lambda_0 L^2}{2} + \frac{2ML}{3} - \frac{2\lambda_0 L^2}{3}]$.$x_{cm} = \frac{2L}{3} + \frac{\lambda_0 L^2}{M} (\frac{1}{2} - \frac{2}{3}) = \frac{2L}{3} - \frac{\lambda_0 L^2}{6M}$.

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

$A$ thin bar of length $L$ has a mass per unit length $\lambda$ that increases linearly with distance $x$ from one end. If its total mass is $M$ and its mass per unit length at the lighter end $(x=0)$ is $\lambda_0$,then the distance of the centre of mass from the lighter end is:

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Match Column-$I$ with Column-$II$.
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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