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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="../../#2.4">$\leftarrow$Back</a></h3> | | <h3 id="back-link"><a href="../../#2.4">$\leftarrow$Back</a></h3> |
| | | |
| <h3> Statement </h3> | | <h3> Statement </h3> |
| <p> | | <p> |
| $2.4.9^*.$ At the ends of a long thread, weights of mass $m$ each are suspended. The thread is spanned over two light little blocks located at a distance of $2l$ from each other. A $2m$ weight is attached to it in the middle between the blocks, and the system starts moving. Find the speed of loads after a sufficiently long period of time has elapsed. | | $2.4.9^*.$ At the ends of a long thread, weights of mass $m$ each are suspended. The thread is spanned over two light little blocks located at a distance of $2l$ from each other. A $2m$ weight is attached to it in the middle between the blocks, and the system starts moving. Find the speed of loads after a sufficiently long period of time has elapsed. |
| </p> | | </p> |
| <center> | | <center> |
| <figure> | | <figure> |
| <img src="2.4.9.png" | | <img src="2.4.9.png" |
| loading="lazy" width="200" /> | | loading="lazy" width="200" /> |
| <figcaption> | | <figcaption> |
| For problem $2.4.9^*$ | | For problem $2.4.9^*$ |
| </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="2.4.9_1.png" | | <img src="2.4.9_1.png" |
| loading="lazy" width="450" /> | | loading="lazy" width="450" /> |
| <figcaption> | | <figcaption> |
| Changing the position of weights | | Changing the position of weights |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| <p> | | <p> |
| | | |
| From the geometry of the drawing, a trigonometric expression can be obtained by | | From the geometry of the drawing, a trigonometric expression can be obtained by |
| $$\tan\varphi=\frac{l}{h}\quad(1)$$ | | $$\tan\varphi=\frac{l}{h}\quad(1)$$ |
| $$\sin\varphi=\frac{l}{\sqrt{l^2+h^2}};\quad\cos\varphi=\frac{h}{\sqrt{l^2+h^2}}\quad(2)$$ | | $$\sin\varphi=\frac{l}{\sqrt{l^2+h^2}};\quad\cos\varphi=\frac{h}{\sqrt{l^2+h^2}}\quad(2)$$ |
| Over the long period of time the $2m$ weight would go down infinitely ($h\to \infty$) | | Over the long period of time the $2m$ weight would go down infinitely ($h\to \infty$) |
| From $(1)$, we obtain that as $h$ increases $\tan\varphi$ decreases | | From $(1)$, we obtain that as $h$ increases $\tan\varphi$ decreases |
| $$\tan\varphi \to 0;\quad\varphi \to 0$$ | | $$\tan\varphi \to 0;\quad\varphi \to 0$$ |
| For the small angle $\varphi$ we could use an approximation to the second order of magnitude | | For the small angle $\varphi$ we could use an approximation to the second order of magnitude |
| $$\sin\varphi \approx\tan\varphi\approx\varphi;\quad\cos\varphi\approx 1$$ | | $$\sin\varphi \approx\tan\varphi\approx\varphi;\quad\cos\varphi\approx 1$$ |
| From the geometry of the figure | | From the geometry of the figure |
| $$\sqrt{l^2+h^2}=l+\Delta x $$ | | $$\sqrt{l^2+h^2}=l+\Delta x $$ |
| by the Pythagorean theorem, the change in the bottom of the thread connecting $2m$ | | by the Pythagorean theorem, the change in the bottom of the thread connecting $2m$ |
| $$\Delta x = \sqrt{l^2+h^2}-l$$ | | $$\Delta x = \sqrt{l^2+h^2}-l$$ |
| Similarly, from the geometry of the figure</p><p> | | Similarly, from the geometry of the figure</p><p> |
| Considering $h\gg l$, | | Considering $h\gg l$, |
| $$\sqrt{l^2+h^2}\approx l\Rightarrow \boxed{h-\Delta x=l}\quad(2)$$ | | $$\sqrt{l^2+h^2}\approx l\Rightarrow \boxed{h-\Delta x=l}\quad(2)$$ |
| $L_1,~L_2$ — Length of threads at the initial moment</p><p> | | $L_1,~L_2$ — Length of threads at the initial moment</p><p> |
| Conservation of the mechanical energy | | Conservation of the mechanical energy |
| $$2mgL_1-2mgL_2=\frac{2mv_2^2}{2}-\frac{2mv_1^2}{2}$$ | | $$2mgL_1-2mgL_2=\frac{2mv_2^2}{2}-\frac{2mv_1^2}{2}$$ |
| $$v_1^2+v_2^2=2g\left(h-\Delta x\right)\quad(3)$$ | | $$v_1^2+v_2^2=2g\left(h-\Delta x\right)\quad(3)$$ |
| By the Newton's second law for the $m$ weight | | By the Newton's second law for the $m$ weight |
| $$ma=T-mg$$ | | $$ma=T-mg$$ |
| Similarly, for the $2m$ weight | | Similarly, for the $2m$ weight |
| $$2ma^*=2mg-2T\cos\varphi\approx2mg-2T\Leftrightarrow $$ | | $$2ma^*=2mg-2T\cos\varphi\approx2mg-2T\Leftrightarrow $$ |
| $$\Leftrightarrow ma^*=mg-T$$ | | $$\Leftrightarrow ma^*=mg-T$$ |
| As they are connected by the same inextensible thread, their acceleration will be equal | | As they are connected by the same inextensible thread, their acceleration will be equal |
| $$\left|a\right|=\left|a^*\right|\Rightarrow v=v^*$$ | | $$\left|a\right|=\left|a^*\right|\Rightarrow v=v^*$$ |
| Given that the system is being run from a stationary state $(v_0=v_0^*=0)$, let's substitute into $(3)$ | | Given that the system is being run from a stationary state $(v_0=v_0^*=0)$, let's substitute into $(3)$ |
| $$2v^2=2g\left(h-\Delta x\right)$$ | | $$2v^2=2g\left(h-\Delta x\right)$$ |
| Taking into consideration $(2)$ | | Taking into consideration $(2)$ |
| $$v^2=gl\Rightarrow \boxed{v=\sqrt{gl}}$$ | | $$v^2=gl\Rightarrow \boxed{v=\sqrt{gl}}$$ |
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| </p> | | </p> |
| | | |
| <h4>Answer</h4> | | <h4>Answer</h4> |
