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| <span><img src="../../img/book.png"></span><span>Savchenko Solutions</span> | | <span><img src="../../img/book.png"></span><span>Savchenko Solutions</span> |
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| Solutions of Savchenko Problems in Physics <br> | | Solutions of Savchenko Problems in Physics <br> |
| <i><b>knowledge must be free</b></i> | | <i><b>knowledge must be free</b></i> |
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| <h3 id="back-link"><a href="../../#1.3">$\leftarrow$Back</a></h3> | | <h3 id="back-link"><a href="../../#1.3">$\leftarrow$Back</a></h3> |
| | | |
| <h3> Statement </h3> | | <h3> Statement </h3> |
| <p> | | <p> |
| $1.3.3.$ At what angle to the vertical should a smooth chute be directed from point $A$ so that the ball slides down it to the inclined plane in the shortest time? | | $1.3.3.$ At what angle to the vertical should a smooth chute be directed from point $A$ so that the ball slides down it to the inclined plane in the shortest time? |
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| For problem $1.3.3$ | | For problem $1.3.3$ |
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| <h3>Solution</h3> | | <h3>Solution</h3> |
| <p> | | <p> |
| Let's consider the possible directions of the gutter from point $A$. It is clear that by changing the angle of the gutter to the plane, the length of the gutter changes, but the acceleration with which the ball will move along the gutter also changes. | | Let's consider the possible directions of the gutter from point $A$. It is clear that by changing the angle of the gutter to the plane, the length of the gutter changes, but the acceleration with which the ball will move along the gutter also changes. |
| From geometric considerations: | | From geometric considerations: |
| $$ h = x \cdot \tan{\alpha} + x \cdot \tan{\varphi} $$ | | $$ h = x \cdot \tan{\alpha} + x \cdot \tan{\varphi} $$ |
| | | |
| $$ l = \frac{x}{\cos{\alpha}}\; (1) $$ | | $$ l = \frac{x}{\cos{\alpha}}\; (1) $$ |
| The ball will roll with acceleration: | | The ball will roll with acceleration: |
| $a = g \cdot \sin{\alpha}$. | | $a = g \cdot \sin{\alpha}$. |
| We find the time from the condition that the ball has traveled $l$: | | We find the time from the condition that the ball has traveled $l$: |
| $$ l = \frac{a t^2}{2} $$ | | $$ l = \frac{a t^2}{2} $$ |
| | | |
| $$ \frac{x}{\cos{\alpha}} = \frac{g \cdot \sin{\alpha} \cdot t^2}{2} $$ | | $$ \frac{x}{\cos{\alpha}} = \frac{g \cdot \sin{\alpha} \cdot t^2}{2} $$ |
| Therefore, expressing from $(1)$ $x$, we obtain from the last equation: | | Therefore, expressing from $(1)$ $x$, we obtain from the last equation: |
| $$ t^2 = \frac{2h}{g \cdot \sin{\alpha} \cdot \cos{\alpha} (\tan{\alpha} + \tan{\varphi})} $$ | | $$ t^2 = \frac{2h}{g \cdot \sin{\alpha} \cdot \cos{\alpha} (\tan{\alpha} + \tan{\varphi})} $$ |
| The time will be minimal if $\xi=g \cdot \sin{\alpha} \cdot \cos{\alpha} (\tan{\alpha} + \tan{\varphi})$ is maximal. | | The time will be minimal if $\xi=g \cdot \sin{\alpha} \cdot \cos{\alpha} (\tan{\alpha} + \tan{\varphi})$ is maximal. |
| $$ \xi = \frac{\sin{\alpha}}{\cos{\varphi}} \sin{(\alpha + \varphi)} $$ | | $$ \xi = \frac{\sin{\alpha}}{\cos{\varphi}} \sin{(\alpha + \varphi)} $$ |
| | | |
| $$ \xi = \frac{1}{2 \cos{\varphi}} (\cos{\varphi} - \cos{(2\alpha + \varphi)}) $$ | | $$ \xi = \frac{1}{2 \cos{\varphi}} (\cos{\varphi} - \cos{(2\alpha + \varphi)}) $$ |
