The linear mass density of a rod of length L& | vedclass

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Question

$x_{\mathrm{cm}}=\frac{\int(\mathrm{dm}) \mathrm{r}}{\int \mathrm{dm}}=\frac{\int(\lambda \mathrm{d} \mathrm{x}) \mathrm{x}}{\int \lambda \mathrm{d} \

Vedclass · Accepted Answer

$\frac{3L}{4}$

Vedclass · Answer

$x_{\mathrm{cm}}=\frac{\int(\mathrm{dm}) \mathrm{r}}{\int \mathrm{dm}}=\frac{\int(\lambda \mathrm{d} \mathrm{x}) \mathrm{x}}{\int \lambda \mathrm{d} \mathrm{x}}=\frac{\int\left(\mathrm{kx}^{2} \mathrm{dx}\right) \mathrm{x}}{\int \mathrm{kx}^{2} \mathrm{d} \mathrm{x}}$

$=\frac{\int_{0}^{L} x^{3} d x}{\int_{0}^{L} x^{2} d x}=\frac{3 L}{4}$

The linear mass density of a rod of length $L$ varies as $\lambda = kx^2$, where $k$ is a constant and $x$ is the distance from one end. The position of centre of mass of the rod is

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