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| <h2>Solutions of Savchenko Problems in Physics</h2> | | <h2>Solutions of Savchenko Problems in Physics</h2> |
| <p class="author"> | | <p class="author"> |
| Aliaksandr Melnichenka <br/> | | Aliaksandr Melnichenka <br/> |
| October 2023 | | October 2023 |
| </p> | | </p> |
| </header> | | </header> |
| | | |
| <h3 id="back-link"><a href="../../#2.7">$\leftarrow$Back</a></h3> | | <h3 id="back-link"><a href="../../#2.7">$\leftarrow$Back</a></h3> |
| | | |
| <h3> Statement </h3> | | <h3> Statement </h3> |
| <p> | | <p> |
| $2.7.6.$ 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? | | $2.7.6.$ 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? |
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| <center> | | <center> |
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| <img src="statement.png" | | <img src="statement.png" |
| loading="lazy" width="180" /> | | loading="lazy" width="180" /> |
| <figcaption> | | <figcaption> |
| For problem $2.7.6$ | | For problem $2.7.6$ |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| <p> | | <p> |
| </p> | | </p> |
| | | |
| <h3>Solution</h3> | | <h3>Solution</h3> |
| <p> | | <p> |
| | | |
| </p> | | </p> |
| <center> | | <center> |
| <figure> | | <figure> |
| <img src="draw1.png" | | <img src="draw1.png" |
| loading="lazy" width="230" /> | | loading="lazy" width="230" /> |
| <figcaption> | | <figcaption> |
| Forces acting on the cylinder | | Forces acting on the cylinder |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| <p> | | <p> |
| | | |
| Conditions of equilibrium: | | Conditions of equilibrium: |
| $$\left\{\begin{matrix} | | $$\left\{\begin{matrix} |
| O'X\text{: }F_{fr2} = N_1 \\ | | O'X\text{: }F_{fr2} = N_1 \\ |
| O'Y\text{: }F_{fr1}+N_2 = mg \\ | | O'Y\text{: }F_{fr1}+N_2 = mg \\ |
| \end{matrix}\right.$$ | | \end{matrix}\right.$$ |
| From where | | From where |
| $$\left\{\begin{matrix} | | $$\left\{\begin{matrix} |
| \mu N_2=N_1 \\ | | \mu N_2=N_1 \\ |
| \mu N_1+N_2=mg \\ | | \mu N_1+N_2=mg \\ |
| \end{matrix}\right.$$ | | \end{matrix}\right.$$ |
| Let's find friction force | | Let's find friction force |
| $$F_{fr}=F_{fr1}+F_{fr2}$$ | | $$F_{fr}=F_{fr1}+F_{fr2}$$ |
| $$F_{fr}=\mu (N_1+N_2)=\mu N_2(1+\mu)$$ | | $$F_{fr}=\mu (N_1+N_2)=\mu N_2(1+\mu)$$ |
| | | |
| Considering, $mg = N_2(1+\mu ^2)$: | | Considering, $mg = N_2(1+\mu ^2)$: |
| $$N=\frac{mg}{1+\mu ^2}$$ | | $$N=\frac{mg}{1+\mu ^2}$$ |
| | | |
| According to rotational motion dynamics, the sum of moments of external forces can be written as | | According to rotational motion dynamics, the sum of moments of external forces can be written as |
| $$M=I\varepsilon = mR^2=F_{fr} R$$ | | $$M=I\varepsilon = mR^2=F_{fr} R$$ |
| From where, absolute value of angular acceleration is found as | | From where, absolute value of angular acceleration is found as |
| $$\varepsilon = \frac{M}{I} = \frac{F_{fr}}{mR}$$ | | $$\varepsilon = \frac{M}{I} = \frac{F_{fr}}{mR}$$ |
| $$\varepsilon = \frac{\mu(1+\mu)N_2 }{mR}$$ | | $$\varepsilon = \frac{\mu(1+\mu)N_2 }{mR}$$ |
| $$\varepsilon = \frac{\mu(1+\mu)}{1+\mu ^2} \frac{g}{R}$$ | | $$\varepsilon = \frac{\mu(1+\mu)}{1+\mu ^2} \frac{g}{R}$$ |
| The angle by which the cylinder will rotate to the moment of total stop is | | The angle by which the cylinder will rotate to the moment of total stop is |
| $$\varphi=\frac{\omega ^2t}{2\varepsilon}$$ | | $$\varphi=\frac{\omega ^2t}{2\varepsilon}$$ |
| Until the total stop, the cylinder will make $N$ total turnovers | | Until the total stop, the cylinder will make $N$ total turnovers |
| $$\varphi=2\pi N$$ | | $$\varphi=2\pi N$$ |
| From where | | From where |
| $$N=\frac{\omega ^2t}{4\pi \varepsilon}$$ | | $$N=\frac{\omega ^2t}{4\pi \varepsilon}$$ |
| $$\boxed{N=\frac{1+\mu^{2}}{4\pi \mu(1+\mu)}\frac{\omega^{2}R}{g}}$$ | | $$\boxed{N=\frac{1+\mu^{2}}{4\pi \mu(1+\mu)}\frac{\omega^{2}R}{g}}$$ |
| </p> | | </p> |
| | | |
| <h4>Answer</h4> | | <h4>Answer</h4> |
