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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="../../#4.2">$\leftarrow$Back</a></h3> | | <h3 id="back-link"><a href="../../#4.2">$\leftarrow$Back</a></h3> |
| | | |
| <h3> Statement </h3> | | <h3> Statement </h3> |
| <p> | | <p> |
| $4.2.26.$ A vertical cylindrical vessel of radius R partially filled with liquid rotates with the liquid around its axis. A balloon of radius $r$ is tied to the side wall of the vessel on a thread of length $l$; during rotation the thread forms an angle $\alpha$ with the wall. Determine the angular velocity of rotation of the vessel. | | $4.2.26.$ A vertical cylindrical vessel of radius R partially filled with liquid rotates with the liquid around its axis. A balloon of radius $r$ is tied to the side wall of the vessel on a thread of length $l$; during rotation the thread forms an angle $\alpha$ with the wall. Determine the angular velocity of rotation of the vessel. |
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| <img src="4.2.26.png" | | <img src="4.2.26.png" |
| loading="lazy" width="200" /> | | loading="lazy" width="200" /> |
| <figcaption> | | <figcaption> |
| For problem $4.2.26$ | | For problem $4.2.26$ |
| </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="forces.png" | | <img src="forces.png" |
| loading="lazy" width="240" /> | | loading="lazy" width="240" /> |
| <figcaption> | | <figcaption> |
| Forces acting on the balloon | | Forces acting on the balloon |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| <p> | | <p> |
| | | |
| Since the ball does not float, the vertical Archimedean force is completely compensated by the vertical component of the thread tension force | | Since the ball does not float, the vertical Archimedean force is completely compensated by the vertical component of the thread tension force |
| $$T\cos\alpha=\frac{4\pi r^3}{3}\rho g$$ | | $$T\cos\alpha=\frac{4\pi r^3}{3}\rho g$$ |
| From where we could find the thread tension force | | From where we could find the thread tension force |
| $$T=\frac{4\pi r^3}{3\cos\alpha}\rho g\quad (1)$$ | | $$T=\frac{4\pi r^3}{3\cos\alpha}\rho g\quad (1)$$ |
| Alternatively, we could write an equilibrium condition for the <b>Horizontal Archimedean force</b>, which is compensated by the horizontal force of the thread tension | | Alternatively, we could write an equilibrium condition for the <b>Horizontal Archimedean force</b>, which is compensated by the horizontal force of the thread tension |
| $$T\sin\alpha = \rho V\omega^2x\quad (2)$$ | | $$T\sin\alpha = \rho V\omega^2x\quad (2)$$ |
| It can be seen from the figure that the ball's center of mass is at a distance $l+r$ from the point of attachment of the thread: | | It can be seen from the figure that the ball's center of mass is at a distance $l+r$ from the point of attachment of the thread: |
| $$x=R-(l+r)\sin\alpha$$ | | $$x=R-(l+r)\sin\alpha$$ |
| After substituting into $(2)$ we obtain | | After substituting into $(2)$ we obtain |
| $$\rho \frac{4\pi r^3}{3}\omega^2\left(R-(l+r)\sin\alpha\right) = T\sin\alpha \quad (3)$$ | | $$\rho \frac{4\pi r^3}{3}\omega^2\left(R-(l+r)\sin\alpha\right) = T\sin\alpha \quad (3)$$ |
| Substituting $(1)$ into $(3)$ | | Substituting $(1)$ into $(3)$ |
| <div class="scroll-wrapper">$$\rho \frac{4\pi r^3}{3}\omega^2\left(R-(l+r)\sin\alpha\right) = \frac{4\pi r^3}{3\cos\alpha}\rho g\sin\alpha$$</div> | | <div class="scroll-wrapper">$$\rho \frac{4\pi r^3}{3}\omega^2\left(R-(l+r)\sin\alpha\right) = \frac{4\pi r^3}{3\cos\alpha}\rho g\sin\alpha$$</div> |
| From where we find the angular velocity | | From where we find the angular velocity |
| $$\boxed{\omega = \sqrt{\frac{g\tan\alpha}{R-(l+r)\sin\alpha}}}$$ | | $$\boxed{\omega = \sqrt{\frac{g\tan\alpha}{R-(l+r)\sin\alpha}}}$$ |
| </p> | | </p> |
| | | |
| <h4>Answer</h4> | | <h4>Answer</h4> |
| <p> | | <p> |
