If the linear density of a rod of length 3m v | vedclass

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Question

Linear density of rod varies with distance

$\frac{d m}{d x}=\lambda$

$x_{c m}=\frac{\int d m 	imes x}{\int d m}$

$=\frac{\int_{0}^{3}(\lambd

Vedclass · Accepted Answer

$12/7\, m$

Vedclass · Answer

Linear density of rod varies with distance

$\frac{d m}{d x}=\lambda$

$x_{c m}=\frac{\int d m 	imes x}{\int d m}$

$=\frac{\int_{0}^{3}(\lambda d x) x}{\int_{0}^{3} \lambda d x}=\frac{\left.\int_{0}^{3}(2+x) x d xight)}{\int_{0}^{3}(2+x) d x}$

$=\frac{\left[x^{2}+\frac{x^{3}}{3}ight]_{0}^{3}}{\left[2 x+\frac{x^{3}}{3}ight]_{0}^{3}}$

$=\frac{9+9}{6+\frac{9}{2}}=\frac{36}{21}=\frac{12}{7} m$

If the linear density of a rod of length $3m$ varies as $\lambda = 2 + x$, then the position of centre of gravity of the rod is :

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