A rod of length 3, m has a mass per unit leng | vedclass

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(B) The mass per unit length,i.e.,the linear mass density $\mu$,is given by $\mu \propto x$,which implies $\mu = \lambda x$,where $\lambda$ is a const

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$2$

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(B) The mass per unit length,i.e.,the linear mass density $\mu$,is given by $\mu \propto x$,which implies $\mu = \lambda x$,where $\lambda$ is a constant.Consider a small element of length $dx$ at a distance $x$ from the end. The mass of this element is $dm = \mu dx = \lambda x dx$.The center of mass $x_{cm}$ is calculated as:$x_{cm} = \frac{\int x dm}{\int dm} = \frac{\int_{0}^{3} x (\lambda x dx)}{\int_{0}^{3} \lambda x dx}$$x_{cm} = \frac{\int_{0}^{3} x^2 dx}{\int_{0}^{3} x dx} = \frac{[x^3/3]_0^3}{[x^2/2]_0^3}$$x_{cm} = \frac{27/3}{9/2} = \frac{9}{4.5} = 2\, m$.

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

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