Question

A thin rod of total length $l$ and of mass $m$ has variable linear density given by $\rho ={\rho }_0[1+(x/l)]$, where $x$ is the distance from one end. Calculate the moment of inertia (in $kg{m}^2$ ) about an axis perpendicular to the length of the rod and passing midway, at $x=l/2$. (Take ${\rho }_0=1kg/m,l=2m$ )

Answer 1

Answer: 1

$dI=\int dm{\left(x-\frac{\ell }{2}\right)}^2$
$=r_0{\rho }_0\underset{\ell /2}{\overset{\ell /2}{\int }}\left(1+\frac{x}{\ell }\right){\left(1-\frac{x}{\ell }\right)}^{2}dx$
$={\rho }_0\underset{0}{\overset{\ell }{\int }}\left(x^2+\frac{{\ell }^2}{4}-\ell x+\frac{x^3}{\ell }+\frac{x{\ell }^2}{4}-x^2\right)dx$
$={\rho }_0\left[\frac{{\ell }^3}{4}-\frac{3{\ell }^3}{8}+\frac{{\ell }^3}{4}\right]$
$=\frac{{\rho }_0{\ell }^3}{8}=1kg{m}^2$