A rod of length is 3 ;m and its mass acting p | vedclass

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Question

Here $\rho= kx$

where $k$ is a constant

mass of small element of $dx$ length is $dm = kx \cdot dx$

$x_{c m}=\frac{\int x \cdot d m}{\int d m}

Vedclass · Accepted Answer

$2$

Vedclass · Answer

Here $\rho= kx$

where $k$ is a constant

mass of small element of $dx$ length is $dm = kx \cdot dx$

$x_{c m}=\frac{\int x \cdot d m}{\int d m}=\frac{\int_{0}^{3} x(x d x)}{\int_{0}^{3} x \cdot d x}$$=\frac{\left[\frac{x^{3}}{3}\right]_{0}^{3}}{\left[\frac{x^{2}}{2}\right]_{0}^{3}}=\frac{\frac{27}{3}}{\frac{9}{2}}=2$

A rod of length is $3 \;m$ and its mass acting per unit length is directly proportional to distance $x$ from one of its end then its centre of gravity from that end will be at

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