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| <h3 id="back-link"><a href="../../#3.1">$\leftarrow$Back</a></h3> | | <h3 id="back-link"><a href="../../#3.1">$\leftarrow$Back</a></h3> |
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| <h3> Statement </h3> | | <h3> Statement </h3> |
| <p> | | <p> |
| $3.1.9.$ A ball of mass $m$ and radius $r$ slides on the surface of a well whose radius of curvature is $R$. Find the dependence of the potential energy of the ball on a small displacement $x$ from the equilibrium position. | | $3.1.9.$ A ball of mass $m$ and radius $r$ slides on the surface of a well whose radius of curvature is $R$. Find the dependence of the potential energy of the ball on a small displacement $x$ from the equilibrium position. |
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| <figcaption> | | <figcaption> |
| For problem $3.1.9$ | | For problem $3.1.9$ |
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| <p> | | <p> |
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| <h3>Solution</h3> | | <h3>Solution</h3> |
| <p> | | <p> |
| As in <a href="../3.1.5">3.1.5</a>, we will propose two possible solutions: through small transformations and through integration</p><p> | | As in <a href="../3.1.5">3.1.5</a>, we will propose two possible solutions: through small transformations and through integration</p><p> |
| When the body deviates horizontally by a small value $x$, its height changes by the value | | When the body deviates horizontally by a small value $x$, its height changes by the value |
| $$ h=R'-\sqrt{{R'}^2-x^2}=R'\left(1-\sqrt{1-\frac{x^2}{{R'}^2}}\right)\quad(1) $$ | | $$ h=R'-\sqrt{{R'}^2-x^2}=R'\left(1-\sqrt{1-\frac{x^2}{{R'}^2}}\right)\quad(1) $$ |
| Where $R'$ is the distance from the center of curvature to the center of mass of the ball. | | Where $R'$ is the distance from the center of curvature to the center of mass of the ball. |
| $$ R'=R-r $$ | | $$ R'=R-r $$ |
| Considering the smallness of $x$ | | Considering the smallness of $x$ |
| $$ x\ll R'\Leftrightarrow \frac{x}{R'}\ll 1 $$ | | $$ x\ll R'\Leftrightarrow \frac{x}{R'}\ll 1 $$ |
| So we can use approximation $(1+\xi)^a\approx 1+\xi a$, where $\xi \ll 1$ | | So we can use approximation $(1+\xi)^a\approx 1+\xi a$, where $\xi \ll 1$ |
| $$ \left(1-\frac{x^2}{{R'}^2}\right)^{1/2}=1-\frac{x^2}{2{R'}^2} $$ | | $$ \left(1-\frac{x^2}{{R'}^2}\right)^{1/2}=1-\frac{x^2}{2{R'}^2} $$ |
| Substitute into $(1)$ | | Substitute into $(1)$ |
| $$ h=R'\cdot \frac{x^2}{2{R'}^2}=\frac{x^2}{2R'} $$ | | $$ h=R'\cdot \frac{x^2}{2{R'}^2}=\frac{x^2}{2R'} $$ |
| Substitute into the formula for potential energy | | Substitute into the formula for potential energy |
| $$ U=mgh=mg\frac{x^2}{2R'} $$ | | $$ U=mgh=mg\frac{x^2}{2R'} $$ |
| Substituting the value for $R'$, we obtain the final expression | | Substituting the value for $R'$, we obtain the final expression |
| $$ \boxed{U=\frac{mgx^2}{2(R-r)}} $$ | | $$ \boxed{U=\frac{mgx^2}{2(R-r)}} $$ |
| </p> | | </p> |
| <p style="text-align: right; font-style: italic; font-size: 14;"> | | <p style="text-align: right; font-style: italic; font-size: 14;"> |
| Aliaksandr Melnichenka<br> | | Aliaksandr Melnichenka<br> |
| </p> | | </p> |
| <h3>Alternative solution</h3> | | <h3>Alternative solution</h3> |
| <p> | | <p> |
| The change in energy with a small change in coordinate can be found by definition | | The change in energy with a small change in coordinate can be found by definition |
| $$ dU=Fdx $$ | | $$ dU=Fdx $$ |
| In this case, we will find the horizontal force that will return the ball to stable equilibrium taking into account the smallness of the angle $\varphi$ | | In this case, we will find the horizontal force that will return the ball to stable equilibrium taking into account the smallness of the angle $\varphi$ |
| $$ F=mg\sin\varphi=mg\frac{x}{R-r} $$ | | $$ F=mg\sin\varphi=mg\frac{x}{R-r} $$ |
| Where we find the potential energy through integration $dU$ | | Where we find the potential energy through integration $dU$ |
| $$ U=\int_{0}^{U}dU=\int_{0}^{x}mg\frac{x}{R-r}dx=\frac{mg}{R-r}\int_{0}^{x}xdx $$ | | $$ U=\int_{0}^{U}dU=\int_{0}^{x}mg\frac{x}{R-r}dx=\frac{mg}{R-r}\int_{0}^{x}xdx $$ |
| This integral is solved as | | This integral is solved as |
| $$ \int_{0}^{x}xdx=\frac{mg}{R-r}\cdot\frac{x^2}{2} $$ | | $$ \int_{0}^{x}xdx=\frac{mg}{R-r}\cdot\frac{x^2}{2} $$ |
| Where does the potential energy of the ball come from? | | Where does the potential energy of the ball come from? |
| $$ \boxed{U=\frac{mgx^2}{2(R-r)}} $$ | | $$ \boxed{U=\frac{mgx^2}{2(R-r)}} $$ |
| </p> | | </p> |
| <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> |
| </p> | | </p> |
| | | |
| <h4>Answer</h4> | | <h4>Answer</h4> |
| <p> | | <p> |
| $$U=\frac{mgx^2}{2(R-r)}$$ | | $$U=\frac{mgx^2}{2(R-r)}$$ |
| </p> | | </p> |
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