The mass per unit length of a rod of length l | vedclass

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Question

$\mathrm{X}_{\mathrm{cm}}=\frac{\int(\mathrm{dm}) \mathrm{x}}{\int \mathrm{dm}}=\frac{\int(\lambda \mathrm{d} \mathrm{x}) \mathrm{x}}{\int \lambda \ma

Vedclass · Accepted Answer

$\frac{2l}{3}$

Vedclass · Answer

$\mathrm{X}_{\mathrm{cm}}=\frac{\int(\mathrm{dm}) \mathrm{x}}{\int \mathrm{dm}}=\frac{\int(\lambda \mathrm{d} \mathrm{x}) \mathrm{x}}{\int \lambda \mathrm{dx}}=\frac{\int_{0}^{\ell}\left(\frac{\mathrm{m}_{0} \mathrm{x}}{\ell} \mathrm{dx}\right) \mathrm{x}}{\int_{0}^{\ell} \frac{\mathrm{m}_{0} \mathrm{x}}{\ell} \mathrm{dx}}$

$=\frac{\int_{0}^{\ell} x^{2} d x}{\int_{0}^{\ell} x d x}=\frac{\left(\frac{x^{3}}{3}\right)_{0}^{\ell}}{\left(\frac{x^{2}}{2}\right)_{0}^{\ell}}=\frac{2}{3} \ell$

The mass per unit length of a rod of length $l$ is given by : $\lambda = \frac{M_0x}{l}$ ,where $M_0$ is a constant and $x$ is the distance from one end of the rod. The position of centre of mass of the rod is

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