Savchenko Solutions

Solutions of Savchenko Problems in Physics
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    <h3 id="back-link"><a href="/#3.1">$\leftarrow$Back</a></h3>

    <h3> Statement </h3>
    <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.

    <h3>Solution</h3>
    <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
$$h=R'-\sqrt{{R'}^2-x^2}=R'\left(1-\sqrt{1-\frac{x^2}{{R'}^2}}\right)\quad(1)$$
Where $R'$ is the distance from the center of curvature to the center of mass of the ball.
$$R'=R-r$$
Considering the smallness of $x$
$$x\ll R'\Leftrightarrow \frac{x}{R'}\ll 1$$
So we can use approximation $(1+\xi)^a\approx 1+\xi a$, where $\xi \ll 1$
$$\left(1-\frac{x^2}{{R'}^2}\right)^{1/2}=1-\frac{x^2}{2{R'}^2}$$
Substitute into $(1)$
$$h=R'\cdot \frac{x^2}{2{R'}^2}=\frac{x^2}{2R'}$$
Substitute into the formula for potential energy
$$U=mgh=mg\frac{x^2}{2R'}$$
Substituting the value for $R'$, we obtain the final expression
$$\boxed{U=\frac{mgx^2}{2(R-r)}}$$

Aliaksandr Melnichenka

    <h4>Answer</h4>
    <p>
         $$U=\frac{mgx^2}{2(R-r)}$$
    </p>


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