A spherical buoy of radius R is attached to the bottom of the reservoir. The water level in the reservoir rises at a speed u. What is the speed of movement of the boundary of the flooded part of the buoy on its surface at the moment when the water level is h above the center of the buoy?

Solutions of Savchenko Problems in Physics

Aliaksandr Melnichenka
October 2023

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    <h3> Statement </h3>
    <p>
        $1.5.19.$ A spherical buoy of radius $R$ is attached to the bottom of the reservoir. The water level in the reservoir rises at a speed $u$. What is the speed of movement of the boundary of the flooded part of the buoy on its surface at the moment when the water level is $h$ above the center of the buoy?

    <h3>Solution</h3>
    <p>
        <center>

By time $t$ the water level will be $h$

In a short period of time $dt$ the water layer will rise by

$$dh = u \\, dt$$

In turn, the boundary of the flooded part will also rise by

$$dh = v\\, dt \cos \alpha$$

Let's equate both equations

$$v \cos \alpha = u$$

$$v = \frac{u}{\cos \alpha}$$

$$\fbox{$v = u \frac{R}{\sqrt{R^2 - h^2}}$}$$

    <h4>Answer</h4>
    <p>
        $$v  = u \frac{R}{\sqrt{R^2 - h^2}}$$
    </p>


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