Calculate the center of mass, the moment of inertia, and radius of gyration

Consider a rod of length $L$ placed along the $x$ -axis with one at the origin. Give a mass density function

$\rho(x) = x^2 \quad \text{for} \quad 0 \leq x \leq L.$

compute

The center of mass is given by
$\begin{align*} \overline{x} &= \frac{\int_0^L x \rho(x) \, dx}{\int_0^L \rho(x) \, dx} \\ &= \frac{\int_0^L x^3 \,dx}{\int_0^L x^2 \, dx} \\ &= \frac{L^4/4}{L^3/3} \\ &= \frac{3L}{4}. \end{align*}$
The moment of inertia is
$\text{moment of inertia} = \int_0^L x^2 \rho(x) \, dx = \int_0^L x^4 \, dx = \frac{L^5}{5}.$
The radius of gyration is given by
$\begin{align*} r^2 &= \frac{\int_0^L x \rho(x) \, dx}{\int_0^L \rho(x) \, dx} \\ &= \frac{L^5/5}{L^3/3} \\ &= \frac{3L^2}{5} \\ \implies r &= \frac{\sqrt{3}L}{\sqrt{5}} \\ &= \frac{\sqrt{15}L}{5}. \end{align*}$

Stumbling Robot