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| <h3 id="back-link"><a href="../../#3.2">$\leftarrow$Back</a></h3> | | <h3 id="back-link"><a href="../../#3.2">$\leftarrow$Back</a></h3> |
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| <h3> Statement </h3> | | <h3> Statement </h3> |
| <p> | | <p> |
| $3.2.13.$ A straight tunnel is dug in the Earth that does not pass through its center. Determine the time of movement of a train with engines turned off through such a tunnel if the influence of the Earth's rotation on the movement of the train and friction are neglected. | | $3.2.13.$ A straight tunnel is dug in the Earth that does not pass through its center. Determine the time of movement of a train with engines turned off through such a tunnel if the influence of the Earth's rotation on the movement of the train and friction are neglected. |
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| For problem $3.2.13$ | | For problem $3.2.13$ |
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| <h3>Solution</h3> | | <h3>Solution</h3> |
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| For problem $3.2.13$ | | For problem $3.2.13$ |
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| <p> | | <p> |
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| Similarly to <a href="../3.2.12">3.2.12</a>, we find the external force $F$ as</p><p> | | Similarly to <a href="../3.2.12">3.2.12</a>, we find the external force $F$ as</p><p> |
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| The body, at a distance $x$ from the core, will be affected by the gravitational force of attraction caused by the inner layers of the planet with density $\rho$, forming a sphere of radius $x$. The mass of this part of the earth | | The body, at a distance $x$ from the core, will be affected by the gravitational force of attraction caused by the inner layers of the planet with density $\rho$, forming a sphere of radius $x$. The mass of this part of the earth |
| $$ M_\oplus = \frac{4}{3} \rho\pi x^3 $$ | | $$ M_\oplus = \frac{4}{3} \rho\pi x^3 $$ |
| Gravitational force acting on a rock at depth $x$ | | Gravitational force acting on a rock at depth $x$ |
| $$ F_G = \frac{GmM_\oplus}{x^2}=mg\frac{x}{R} $$ | | $$ F_G = \frac{GmM_\oplus}{x^2}=mg\frac{x}{R} $$ |
| In this case, only the horizontal component of this force will create a moment. Whence, the resulting external force is equal to | | In this case, only the horizontal component of this force will create a moment. Whence, the resulting external force is equal to |
| $$ F=mg\cos\alpha=mg\frac{x}{R} $$ | | $$ F=mg\cos\alpha=mg\frac{x}{R} $$ |
| From here we find the angular frequency of oscillations | | From here we find the angular frequency of oscillations |
| $$ \omega=\sqrt{\frac{g}{R}} $$ | | $$ \omega=\sqrt{\frac{g}{R}} $$ |
| The period of oscillation in a given system | | The period of oscillation in a given system |
| $$ T=2\pi\sqrt{\frac{R}{g}} $$ | | $$ T=2\pi\sqrt{\frac{R}{g}} $$ |
| Since we are interested in the flight time only in one direction, we take half of this period. | | Since we are interested in the flight time only in one direction, we take half of this period. |
| $$ \boxed{t=\frac{T}{2}=\pi\sqrt{\frac{R}{g}}\approx42\text{ min}} $$ | | $$ \boxed{t=\frac{T}{2}=\pi\sqrt{\frac{R}{g}}\approx42\text{ min}} $$ |
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| <p style="text-align: right; font-style: italic; font-size: 14;"> | | <p style="text-align: right; font-style: italic; font-size: 14;"> |
| Dzikan Mikita<br> | | Dzikan Mikita<br> |
| Aliaksandr Kanashenka<br> | | Aliaksandr Kanashenka<br> |
| </p> | | </p> |
| | | |
| <h4>Answer</h4> | | <h4>Answer</h4> |
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| $$t=42\text{ min}$$ | | $$t=42\text{ min}$$ |
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