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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 |
| $1.2.11.$ a. In a conical vessel, the water level rises at a constant rate $v_0$. How does the rate of water entering a vessel through a tube of section $s$ depend on time? At time zero, the vessel is empty. | | $1.2.11.$ a. In a conical vessel, the water level rises at a constant rate $v_0$. How does the rate of water entering a vessel through a tube of section $s$ depend on time? At time zero, the vessel is empty. |
| </p> | | </p> |
| | | |
| <p> | | <p> |
| b. A jet of oil hitting the surface of the water spreads over it in a round spot of thickness $h$. How does the speed of movement of the spot boundary depend on time, if the volume of oil $q$ enters per unit of time? At the initial time, the spot radius is zero. | | b. A jet of oil hitting the surface of the water spreads over it in a round spot of thickness $h$. How does the speed of movement of the spot boundary depend on time, if the volume of oil $q$ enters per unit of time? At the initial time, the spot radius is zero. |
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| <figure> | | <figure> |
| <img src="statement.png" | | <img src="statement.png" |
| loading="lazy" width="160" /> | | loading="lazy" width="160" /> |
| <figcaption> | | <figcaption> |
| For problem $1.2.11$ | | For problem $1.2.11$ |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| <p> | | <p> |
| </p> | | </p> |
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| <h3>Solution</h3> | | <h3>Solution</h3> |
| <p> | | <p> |
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| <center> | | <center> |
| <figure> | | <figure> |
| <img src="drawing1.png" | | <img src="drawing1.png" |
| loading="lazy" width="200" /> | | loading="lazy" width="200" /> |
| <figcaption> | | <figcaption> |
| Cone vessel | | Cone vessel |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| | | |
| $a)$ By time $t$, the water level will be $v_0t$. And the rate of change of volume will be equal to: | | $a)$ By time $t$, the water level will be $v_0t$. And the rate of change of volume will be equal to: |
| $$\frac{dV}{dt} = \frac{\pi r(t)^2 dx}{dt}$$ | | $$\frac{dV}{dt} = \frac{\pi r(t)^2 dx}{dt}$$ |
| Where $dx$ is the change in water level: | | Where $dx$ is the change in water level: |
| | | |
| $$\frac{dV}{dt} = \pi r(t)^2 v_0$$ | | $$\frac{dV}{dt} = \pi r(t)^2 v_0$$ |
| | | |
| From Geometry, | | From Geometry, |
| $$r(t) = v_0 t \cdot tg(\alpha)$$ | | $$r(t) = v_0 t \cdot tg(\alpha)$$ |
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| By definition, the velocity of incoming water is equal to | | By definition, the velocity of incoming water is equal to |
| $$v = \frac{dV}{sdt}$$ | | $$v = \frac{dV}{sdt}$$ |
| | | |
| Substituting the previous expressions: | | Substituting the previous expressions: |
| $$v = {\pi v_0^3 t^2 \cdot tg^2(\alpha)}/s$$ | | $$v = {\pi v_0^3 t^2 \cdot tg^2(\alpha)}/s$$ |
| | | |
| $b)$ For a small time interval $dt$, the volume changes by | | $b)$ For a small time interval $dt$, the volume changes by |
| $$dV = q dt$$ | | $$dV = q dt$$ |
| | | |
| Also the volume increment can be written as | | Also the volume increment can be written as |
| $$dV = 2\pi r dr \cdot h $$ | | $$dV = 2\pi r dr \cdot h $$ |
| | | |
| Thus: | | Thus: |
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| $$ q dt = 2\pi r dr \cdot h$$ | | $$ q dt = 2\pi r dr \cdot h$$ |
| | | |
| Considering $v = \frac{dr}{dt}$, | | Considering $v = \frac{dr}{dt}$, |
| $$ \fbox{$v = \frac{q}{2\pi r h}$}$$ | | $$ \fbox{$v = \frac{q}{2\pi r h}$}$$ |
| </p> | | </p> |
| | | |
| <h4>Answer</h4> | | <h4>Answer</h4> |
| <p> | | <p> |
| $$v=\frac{\pi v_0^3t^2\operatorname{tg}^2\alpha}{s}$$ | | $$v=\frac{\pi v_0^3t^2\operatorname{tg}^2\alpha}{s}$$ |
| $$v=\frac{1}{2}\sqrt{\frac{q}{\pi ht}}$$ | | $$v=\frac{1}{2}\sqrt{\frac{q}{\pi ht}}$$ |
| </p> | | </p> |
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