The linear mass density (lambda) of a rod of | vedclass

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Question

(b)

$\lambda=\alpha+\beta x$

$d m=(\alpha+\beta x) d x$

$x_{c m}=\frac{\int \limits_0^L x(\alpha+\beta x) d x}{\int \limits_0^L(\alpha+\beta

Vedclass · Accepted Answer

$\frac{(3 \alpha+2 \beta L) L}{3(2 \alpha+\beta L)}$

Vedclass · Answer

(b)

$\lambda=\alpha+\beta x$

$d m=(\alpha+\beta x) d x$

$x_{c m}=\frac{\int \limits_0^L x(\alpha+\beta x) d x}{\int \limits_0^L(\alpha+\beta x) d x}$

$=\frac{\alpha \int \limits_0^L x d x+\beta \int \limits_0^L x^2 d x}{\alpha \int \limits_0^L d x+\beta \int \limits_0^L x d x}$

$x_{c m}=\frac{\frac{\alpha L^2}{2}+\frac{\beta L^3}{3}}{\alpha L+\frac{\beta L^2}{2}}$

$x_{cm} = \frac{{L(3A + 2BL)}}{{3(2A + BL)}}$

The linear mass density $(\lambda)$ of a rod of length $L$ kept along $x$-axis varies as $\lambda=\alpha+\beta x$; where $\alpha$ and $\beta$ are positive constants. The centre of mass of the rod is at ..........

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