6. A reel of film needs to be rewound from one spool to another. The diameters of the empty spools are the same and equal to $a$. Find the time required for rewinding if the length of the film is $L$, the thickness of the film is small and equal to $S$, and the receiving spool rotates at a constant angular velocity $\omega$.
Answer: $T=\frac{\pi}{S \omega}\left(\sqrt{a^{2}+\frac{4 S L}{\pi}}-a\right)$. Solution. For each revolution of the receiving coil, one turn of the film is wound onto it, meaning the radius increases by $S$. One revolution of the coil takes time $t_{0}=\frac{2 \pi}{\omega}$. Over time $t$, the coil will complete $n=\frac{t}{t_{0}}$ revolutions, so the radius will increase by $\Delta r=S \cdot n$. Therefore, the radius of the receiving coil changes over time according to the law: $r(t)=\frac{a}{2}+\frac{S \omega}{2 \pi} t$. The speed $V$ of the film winding: $V=\omega\left(\frac{a}{2}+\frac{S \omega}{2 \pi} t\right)$.

Thus, the film moves with constant acceleration, and the time $T$ required to pass the entire length of the film $L$ will be determined from the equation: $\frac{L}{\omega}=\frac{S \omega}{4 \pi} T^{2}+\frac{a T}{2}$.

In this case, the equation has one positive root, which is the answer to the problem:

$$
T=\frac{\pi}{S \omega}\left(\sqrt{a^{2}+\frac{4 S L}{\pi}}-a\right)
$$

Answer to variant 162: $T=\frac{\pi}{S \omega}\left(\sqrt{a^{2}+\frac{4 S L}{\pi}}-a\right)$.

Answer to variant 163: $T=\frac{\pi}{S \omega}\left(\sqrt{a^{2}+\frac{4 S L}{\pi}}-a\right)$.

Answer to variant 164: $T=\frac{\pi}{S \omega}\left(\sqrt{a^{2}+\frac{4 S L}{\pi}}-a\right)$.