A rod of length 3, m and its mass per unit le | vedclass

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Question

Mass per unit length, i.e. linear mass density $\mu \propto \mathrm{x} \Rightarrow \mu=\lambda \mathrm{x}, \lambda:$ constant

$\mathrm{dm}=\mu \mat

Vedclass · Accepted Answer

$2$

Vedclass · Answer

Mass per unit length, i.e. linear mass density $\mu \propto \mathrm{x} \Rightarrow \mu=\lambda \mathrm{x}, \lambda:$ constant

$\mathrm{dm}=\mu \mathrm{dx}=\lambda \mathrm{x} \mathrm{dx}$

$x_{c. m .}=\frac{\int x d m}{\int d m}=\frac{\int_{0}^{3} x \lambda x d x}{\int_{0}^{3} \lambda x d x}$

$=\frac{\int_{0}^{3} x^{2} d x}{\int_{0}^{3} x d x}=\frac{\left|\frac{x^{3}}{3}\right|_{0}^{3}}{\left|\frac{x^{2}}{2}\right|_{0}^{3}}=\frac{\frac{3^{3}}{3}}{\frac{3^{2}}{2}}=2 m$

A rod of length $3\, m$ and its mass per unit length is directly proportional to the distance $x$ from its one end. The center of gravity of the rod from that end will be at ........ $ m$

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