A non-uniform rod OM (of length l) is kept al | vedclass

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(D) Let the rod be along the $x$-axis from $r=0$ to $r=l$. The axis $AB$ is at $r=x$. The distance of a small element $dr$ at position $r$ from the ax

Vedclass · Accepted Answer

$\frac{4l}{5}$

Vedclass · Answer

(D) Let the rod be along the $x$-axis from $r=0$ to $r=l$. The axis $AB$ is at $r=x$. The distance of a small element $dr$ at position $r$ from the axis $AB$ is $|r-x|$.The mass of the element is $dm = \lambda dr = \lambda_0 \frac{r^3}{l^3} dr$.The moment of inertia about axis $AB$ is $I_{AB} = \int_0^l (r-x)^2 dm = \int_0^l (r-x)^2 \frac{\lambda_0}{l^3} r^3 dr$.$I_{AB} = \frac{\lambda_0}{l^3} \int_0^l (r^2 - 2rx + x^2) r^3 dr = \frac{\lambda_0}{l^3} \int_0^l (r^5 - 2xr^4 + x^2r^3) dr$.$I_{AB} = \frac{\lambda_0}{l^3} \left[ \frac{r^6}{6} - \frac{2xr^5}{5} + \frac{x^2r^4}{4} \right]_0^l = \frac{\lambda_0}{l^3} \left( \frac{l^6}{6} - \frac{2xl^5}{5} + \frac{x^2l^4}{4} \right) = \lambda_0 \left( \frac{l^3}{6} - \frac{2xl^2}{5} + \frac{x^2l}{4} \right)$.For $I_{AB}$ to be minimum,$\frac{dI_{AB}}{dx} = 0$.$\frac{d}{dx} \left( \lambda_0 (\frac{l^3}{6} - \frac{2xl^2}{5} + \frac{x^2l}{4}) \right) = \lambda_0 (0 - \frac{2l^2}{5} + \frac{2xl}{4}) = 0$.$\frac{xl}{2} = \frac{2l^2}{5} \Rightarrow x = \frac{4l}{5}$.

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