A disk of mass $M$ and radius $R$ is initially rotating at an angular velocity of $\omega$. While rotating, it is placed on a horizontal surface whose coefficient of friction is $\mu = 0.5$. How long will it take for the disk to stop rotating?
Someone had solved the problem the following way:
$$ \mathrm{df}=\mu \mathrm{mg} \cdot \frac{2 \pi \mathrm{xdx}}{\pi \mathrm{R}^{2}}=\frac{2 \mu \mathrm{mgx}}{\mathrm{R}^{2}} \mathrm{dx}\\ \Rightarrow \mathrm{d} \tau=\mathrm{xdf}=\frac{2 \mu \mathrm{mgx}^{2} \mathrm{dx}}{\mathrm{R}^{2}}\\ \Rightarrow \int \mathrm{d} \tau=\frac{2 \mu \mathrm{mg}}{\mathrm{R}^{2}} \int_{0}^{R} \mathrm{x}^{2} \mathrm{~d} \mathrm{x} \\\text{Again}, \quad \tau=\frac{2}{3} \mu \mathrm{mgR}\\ \tau=\mathrm{I} \alpha=\frac{1}{2} \mathrm{~m} \mathrm{R}^{2} \alpha\\ \Rightarrow \alpha=\frac{4}{3} \frac{\mu \mathrm{g}}{\mathrm{R}}\\ \alpha \mathrm{t}=\omega \Rightarrow \mathrm{t}=\frac{\omega}{\alpha}=\frac{3 \omega \mathrm{R}}{4 \mathrm{\mu g}} $$
Unfortunately, I didn't understand the first line $\text{df} = \mu \text{mg} \cdot \dfrac{2 \pi x \text{dx}}{\pi R^2}$. Could someone please explain it to me? Or, is there any other way to solve the problem?
