Example 2.78. Find the kinetic energy of a circular cylinder of height $h$ with density varying according to the law $\rho(r)=\rho_{0}+k r$, where $0 \leqslant r \leqslant R, r$ - distance from the axis of the cylinder, $R$ - radius of the base, if the cylinder rotates around its axis with a constant angular velocity $\omega$.
Solution. Let's divide the cylinder into concentric layers of thickness $d r$ (where $r$ is the distance from the axis of the cylinder to the inner surface), similar to problem 2.72. Then

$$
d m=2 \pi r h \rho d r=\left(\rho_{0}+k r\right) 2 \pi r h d r
$$

The linear velocity $v$ of the mass $d m$ is related to the angular velocity by $v=r \omega$. The kinetic energy of the hollow cylinder of mass $d m$

$$
d E=\frac{v^{2}(r) d m}{2}=\omega^{2} r^{2}(\rho+k r) \pi r h d r=\omega^{2} \pi h\left(\rho_{0} r^{2}+k r^{3}\right) d r
$$

Integrating both sides of the last equation, we get

$$
\begin{aligned}
E & =\int_{0}^{R} \omega^{2} \pi h\left(\rho_{0} r^{2}+k r^{3}\right) d r=\omega^{2} \pi h\left(\int_{0}^{R} \rho_{0} r^{2} d r+\int_{0}^{R} k r^{3} d r\right)= \\
& =\left.\omega^{2} \pi h\left(\frac{\rho_{0} r^{3}}{3}+\frac{k r^{4}}{4} d r\right)\right|_{0} ^{R}=\omega^{2} \pi h\left(\frac{\rho_{0} R^{3}}{3}+\frac{k R^{4}}{4} d r\right)
\end{aligned}
$$