The balloon deforms when it weakly hits the wall, as shown in the figure. The maximum deformation of the balloon x is much smaller than its radius R. Ignoring the change in excess pressure \Delta p of air in the balloon and the elasticity of the shell, estimate the time of impact with the wall. The mass of the ball m.

Solutions of Savchenko Problems in Physics

Aliaksandr Melnichenka
October 2023

    <h3 id="back-link"><a href="/#3.3">$\leftarrow$Back</a></h3>

    <h3> Statement </h3>
    <p>
        $3.3.8^*.$ The balloon deforms when it weakly hits the wall, as shown in the figure. The maximum deformation of the balloon $x$ is much smaller than its radius $R$. Ignoring the change in excess pressure $\Delta p$ of air in the balloon and the elasticity of the shell, estimate the time of impact with the wall. The mass of the ball $m$.

    <h3>Solution</h3>
    <p>
        

$$y=R-x$$
$$r^2=R^2-(R-x)^2$$
$$S=\pi r^2=\pi (R^2-(R-x)^2)$$
Considering $x \ll R$
$$S=\pi r^2=\pi (R^2-(R-x)^2)=2\pi Rx$$
Newton's second law in differential form:
$$-F=m\ddot{x}$$
$$m\ddot{x}+F=0 \Rightarrow m\ddot{x}+2\pi Rx \cdot \Delta p=0$$
$$\ddot{x}+\frac{2\pi R\Delta p}{m}x=0$$
Recalling the differential equation of harmonic motion
$$\ddot{x}+\omega ^2x=0$$
Find the cyclic frequency $\omega$ of oscillation
$$\omega=\sqrt{\frac{2\pi R\Delta p}{m}}$$
Since we're interested in half of the full period, estimated time of impact with the wall could be found as
$$t=\frac{1}{2}\frac{2\pi}{\omega}=\frac{\pi}{\omega}$$
$$\boxed{t=\pi\sqrt{\frac{m}{2\pi R\Delta p}}}$$

    <h4>Answer</h4>
    <p>
        $$t=\pi\sqrt{m/(2\pi R\Delta p)}$$
    </p>


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