The reel of tape is played back during time t at the film drawing speed v. The initial radius of the reel (with film) is R, and the final radius (without film) is r. What is the thickness of the film?

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Solutions of Savchenko Problems in Physics
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    <h3 id="back-link"><a href="/#1.5">$\leftarrow$Back</a></h3>

    <h3> Statement </h3>
    <p>
        $1.5.20.$ The reel of tape is played back during time $t$ at the film drawing speed $v$. The initial radius of the reel (with film) is $R$, and the final radius (without film) is $r$. What is the thickness of the film?
    </p>

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

Film wound on a reel takes up an area

$$S = \pi (R^2 - r^2)$$

If the film is completely unwound at a speed of $v$ in time $t$, then the length of the film

$$l = vt$$

In cross-section, the film is a rectangle whose area is $S$ and length is $l$. Then the width of this rectangle (it is also the thickness of the film)

$$\fbox{$d = \frac{S}{l} = \frac{\pi (R^2 - r^2)}{vt}$}$$

    <h4>Answer</h4>
    <p>
        $$d = \frac{\pi (R^2 - r^2)}{vt}$$
    </p>


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