A rod of length L has non-uniform linear mass | vedclass

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(D) The centre of mass $x_{cm}$ is given by the formula $x_{cm} = \frac{\int x dm}{\int dm}$.Given the linear mass density $\rho(x) = \lambda(x) = a +

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$\frac{3}{4}\left(\frac{2 a+b}{3 a+b}\right) L$

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(D) The centre of mass $x_{cm}$ is given by the formula $x_{cm} = \frac{\int x dm}{\int dm}$.Given the linear mass density $\rho(x) = \lambda(x) = a + b\left(\frac{x}{L}\right)^2$,the mass element is $dm = \lambda(x) dx = \left(a + \frac{b x^2}{L^2}\right) dx$.The total mass $M = \int_0^L dm = \int_0^L \left(a + \frac{b x^2}{L^2}\right) dx = \left[ ax + \frac{b x^3}{3 L^2} \right]_0^L = aL + \frac{bL}{3} = L\left(a + \frac{b}{3}\right) = L\left(\frac{3a+b}{3}\right)$.The moment of mass about the origin is $\int_0^L x dm = \int_0^L x \left(a + \frac{b x^2}{L^2}\right) dx = \int_0^L \left(ax + \frac{b x^3}{L^2}\right) dx = \left[ \frac{a x^2}{2} + \frac{b x^4}{4 L^2} \right]_0^L = \frac{a L^2}{2} + \frac{b L^2}{4} = L^2\left(\frac{2a+b}{4}\right)$.Therefore,$x_{cm} = \frac{L^2\left(\frac{2a+b}{4}\right)}{L\left(\frac{3a+b}{3}\right)} = \frac{3}{4} \left(\frac{2a+b}{3a+b}\right) L$.

$A$ rod of length $L$ has non-uniform linear mass density given by $\rho(x)=a+b\left(\frac{x}{L}\right)^{2}$,where $a$ and $b$ are constants and $0 \leq x \leq L$. The value of $x$ for the centre of mass of the rod is at

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