A uniform thin rod AB of length L has linear& | vedclass

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Question

Centre of mass of the rod is given by:

$\begin{array}{l}
{x_{cm}} = \frac{{\int\limits_0^L {(ax + \frac{{b{x^2}}}{L})dx} }}{{\int\limits_0^L {\lef

Vedclass · Accepted Answer

$2a\, = b$

Vedclass · Answer

Centre of mass of the rod is given by:

$\begin{array}{l}
{x_{cm}} = \frac{{\int\limits_0^L {(ax + \frac{{b{x^2}}}{L})dx} }}{{\int\limits_0^L {\left( {a + \frac{{bx}}{L}} ight)dx} }}\
\,\,\,\,\,\,\,\,\, = \frac{{\frac{{a{L^2}}}{2} + \frac{{b{L^2}}}{3}}}{{aL + \frac{{bL}}{2}}} = \frac{{L\left( {\frac{a}{2} + \frac{b}{3}} ight)}}{{a + \frac{b}{2}}}\
Now\,\frac{7}{{12}} = \frac{{\frac{a}{2} + \frac{b}{3}}}{{a + \frac{b}{2}}}
\end{array}$

On solving we get, $b = 2a$

A uniform thin rod $AB$ of length $L$ has linear mass density $\mu \left( x \right) = a + \frac{{bx}}{L}$ , where $x$ is measured from $A$. If the $CM$ of the rod lies at a distance of $\left( {\frac{7}{12}} \right)L$ from $A$, then $a$ and $b$ are related as

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