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A right circular cone with base radius $R$ and height $H=3 R \sqrt{7}$ is laid on its side on a plane and rolled so that its vertex remains stationary. How many revolutions will the base make before the cone returns to its original position?
The base circle rolls along the circumference of a circle of radius $L$, where $L$ is the slant height of the cone. Therefore, the number of revolutions is $\frac{2 \pi L}{2 \pi R}=\frac{\sqrt{H^{2}+R^{2}}}{R}=8$.

## Answer

8 revolutions.