Rolling Cone

Let the slant height of the cone and the radius of the base of the cone be $l$ and $r$ respectively.

The circumference of the big circle is $C = 2 \pi l$ , and the circumference of the base of the cone is $c = 2 \pi r$ .

The center of mass of the cone takes time $t = \dfrac{2\pi}{\Omega}$ to complete one revolution about the vertex of the cone. During this time, the point of contact between the cone and the ground moves a distance of $2 \pi l$ . This means that the circular base makes $\frac{C}{c} = \frac{l}{r}$ revolutions, or rotates by $2 \pi \frac lr$ radians.

This implies that the circular base is rotating with angular velocity $\dfrac{2 \pi \frac{l}{r} }{\frac{2\pi}{\Omega}} = \dfrac{l \Omega}{r}$ . However, we know that this is equal to $\omega$ . Thus, we can equate them:

$\omega = \frac{l \Omega}{r} \implies \frac{\Omega}{\omega} = \frac{r}{l} = \cos \beta$

The no sliding condition states that the line of contact points is instantaneously at rest, this means that any point in the line of contact rotates with the same velocity around the axis of symetry as it does around a vertical axis that passes through the cone´s vertex but in opossite directions. Considering the contat point that is part of the cone´s base (an arbitrary point on the line of contact could be chosen but I chose the one at the base for the sake of clarity) and is a distant "R" from the vertex. The center of the base moves in a circle around the vertical axis that pases through the vertex, just as the center of mass, because it´s a rigid body. So, our contact point also, instantaneously, has the same angular velocity with respect to the vertical axis as the center of the base does, because our contact point is part of the base (and the rest of the cone) and its instantaneous speed around the axis passing through the vertex is given by: $v=\Omega R$ . Now, we "ride" a reference frame that rotates with angular velocity $\Omega$ around the vertex (as the center of mass does) and, by doing so, we don´t appreciate the angular velocity $\Omega$ and is the axis of symetry that is instantaneously at rest, that is the cone rotates with an angular velocity $\omega$ around this axis and, with respect to it, our contact point has a speed: $v= \omega R cos\beta$ . Since both speeds are equal, se finally get that $\boxed{\frac{\Omega}{\omega}=cos\beta}$