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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 |
| $2.3.30.$ Determine the force exerted on the vertical wall by the falling dumbbell when the dumbbell axis is at an angle $\alpha$ with the horizon. The dumbbell starts moving from a vertical position without initial velocity. Mass of each dumbbell ball $m$. | | $2.3.30.$ Determine the force exerted on the vertical wall by the falling dumbbell when the dumbbell axis is at an angle $\alpha$ with the horizon. The dumbbell starts moving from a vertical position without initial velocity. Mass of each dumbbell ball $m$. |
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| </p> | | </p> |
| <center> | | <center> |
| <figure> | | <figure> |
| <img src="statement.png" | | <img src="statement.png" |
| loading="lazy" width="130" /> | | loading="lazy" width="130" /> |
| <figcaption> | | <figcaption> |
| For problem $2.3.30$ | | For problem $2.3.30$ |
| </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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| </p> | | </p> |
| <center> | | <center> |
| <figure> | | <figure> |
| <img src="draw.jpg" | | <img src="draw.jpg" |
| loading="lazy" width="280" /> | | loading="lazy" width="280" /> |
| <figcaption> | | <figcaption> |
| Forces acting on the system | | Forces acting on the system |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| <p> | | <p> |
| | | |
| Along the axis $OZ_\parallel$ "The Two-Ball System": | | Along the axis $OZ_\parallel$ "The Two-Ball System": |
| </p><p> | | </p><p> |
| Centripetal acceleration associated with circular rotation: | | Centripetal acceleration associated with circular rotation: |
| $$a=\frac{v^2}{R}$$ | | $$a=\frac{v^2}{R}$$ |
| From the Newton's second law: | | From the Newton's second law: |
| $$\frac{mv^2}{R}=2mg\sin\alpha-N_x\cos\alpha-N_y\sin\alpha$$ | | $$\frac{mv^2}{R}=2mg\sin\alpha-N_x\cos\alpha-N_y\sin\alpha$$ |
| Law of Conservation of Energy | | Law of Conservation of Energy |
| $$mgR=mgh+\frac{mv^2}{2}$$ | | $$mgR=mgh+\frac{mv^2}{2}$$ |
| $$h=R\sin\alpha$$ | | $$h=R\sin\alpha$$ |
| $$\frac{mv^2}{R}=2mg(1-\sin\alpha)$$ | | $$\frac{mv^2}{R}=2mg(1-\sin\alpha)$$ |
| $$N_y\sin\alpha+N_x\cos\alpha=(4\sin\alpha-2)mg$$ | | $$N_y\sin\alpha+N_x\cos\alpha=(4\sin\alpha-2)mg$$ |
| Along the axis $OZ_\perp$ "The Ball System": | | Along the axis $OZ_\perp$ "The Ball System": |
| $$N_y\cos\alpha=N_x\sin\alpha+mg\cos\alpha$$ | | $$N_y\cos\alpha=N_x\sin\alpha+mg\cos\alpha$$ |
| $$N_y=mg+N_x\tan\alpha$$ | | $$N_y=mg+N_x\tan\alpha$$ |
| $N_y$ found from the axis $OZ_\perp$ we substitute into $N_y$ found from the axis $OZ_\parallel$ | | $N_y$ found from the axis $OZ_\perp$ we substitute into $N_y$ found from the axis $OZ_\parallel$ |
| $$mg\sin\alpha\cos\alpha+N_x\sin^2\alpha+N_x\cos^2\alpha=2mg(2\sin\alpha-2)\cos\alpha$$ | | $$mg\sin\alpha\cos\alpha+N_x\sin^2\alpha+N_x\cos^2\alpha=2mg(2\sin\alpha-2)\cos\alpha$$ |
| $$\frac{N_x}{mg}=4\sin\alpha\cos\alpha-2\cos\alpha-\sin\alpha\cos\alpha$$ | | $$\frac{N_x}{mg}=4\sin\alpha\cos\alpha-2\cos\alpha-\sin\alpha\cos\alpha$$ |
| $$\boxed{N_x=(3\sin\alpha-2)\cos\alpha \cdot mg}$$ | | $$\boxed{N_x=(3\sin\alpha-2)\cos\alpha \cdot mg}$$ |
| When $N_x \leq0$, the shoe will break away from the vertical wall and will no longer act on it: | | When $N_x \leq0$, the shoe will break away from the vertical wall and will no longer act on it: |
| $$N_x=0$$ | | $$N_x=0$$ |
| Condition of detachment: | | Condition of detachment: |
| $$(3\sin\alpha-2)\cos\alpha \leq0$$ | | $$(3\sin\alpha-2)\cos\alpha \leq0$$ |
| $$3\sin\alpha-2\leq0 \Leftrightarrow \boxed{\sin\alpha\leqslant2/3}$$ | | $$3\sin\alpha-2\leq0 \Leftrightarrow \boxed{\sin\alpha\leqslant2/3}$$ |
| </p> | | </p> |
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| <h4>Answer</h4> | | <h4>Answer</h4> |
| <p> | | <p> |
| $$F=mg\cos\alpha(3\sin\alpha-2)\mathrm{~with~}\sin\alpha\geqslant2/3;$$ | | $$F=mg\cos\alpha(3\sin\alpha-2)\mathrm{~with~}\sin\alpha\geqslant2/3;$$ |
| $$F=0\mathrm{~with~}\sin\alpha\leqslant2/3.$$ | | $$F=0\mathrm{~with~}\sin\alpha\leqslant2/3.$$ |
| </p> | | </p> |
| <p style="text-align: right; font-style: italic; font-size: 14;"> | | <p style="text-align: right; font-style: italic; font-size: 14;"> |
| Almaskhan Arsen<br> | | Almaskhan Arsen<br> |
| </p> | | </p> |
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