$2.7.6.$ A thin-walled cylinder of radius $R$ was spun up to an angular velocity $\omega$ and placed in a corner, as shown in the figure. The coefficient of friction between the corner walls and the cylinder is $\mu$. How many revolutions will the cylinder make before it comes to a complete stop?
Conditions of equilibrium: $$\left\{\begin{matrix} O'X\text{: }F_{fr2} = N_1 \\ O'Y\text{: }F_{fr1}+N_2 = mg \\ \end{matrix}\right.$$ From where $$\left\{\begin{matrix} \mu N_2=N_1 \\ \mu N_1+N_2=mg \\ \end{matrix}\right.$$ Let's find friction force $$F_{fr}=F_{fr1}+F_{fr2}$$ $$F_{fr}=\mu (N_1+N_2)=\mu N_2(1+\mu)$$ Considering, $mg = N_2(1+\mu ^2)$: $$N=\frac{mg}{1+\mu ^2}$$ According to rotational motion dynamics, the sum of moments of external forces can be written as $$M=I\varepsilon = mR^2=F_{fr} R$$ From where, absolute value of angular acceleration is found as $$\varepsilon = \frac{M}{I} = \frac{F_{fr}}{mR}$$ $$\varepsilon = \frac{\mu(1+\mu)N_2 }{mR}$$ $$\varepsilon = \frac{\mu(1+\mu)}{1+\mu ^2} \frac{g}{R}$$ The angle by which the cylinder will rotate to the moment of total stop is $$\varphi=\frac{\omega ^2t}{2\varepsilon}$$ Until the total stop, the cylinder will make $N$ total turnovers $$\varphi=2\pi N$$ From where $$N=\frac{\omega ^2t}{4\pi \varepsilon}$$ $$\boxed{N=\frac{1+\mu^{2}}{4\pi \mu(1+\mu)}\frac{\omega^{2}R}{g}}$$
$$N=\omega^{2}R(1+\mu^{2})/[4\pi g\mu(1+\mu)]$$