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Homework Statement:

A cone rolls without slipping on a table. The halfangle at the vertex is ##\alpha##,
and the axis has height ##h##, base radius ##r##, and slant height ##s##. If it takes ##T## seconds for the cone to complete one revolution with the vertex fixed. What is the velocity of the center of base ##P## in the lab frame?
Relevant Equations:
 Relative motion, spinning tangent velocity $$\vec v_t = \vec \omega \times \vec r$$
Let the vertex of the cone be ##O##, the contact point on the cone all the way to the right be ##D## touching ground. Then ##v_{\text{D relative to the table}} = v_{D/table} =0## since it rolls without slipping.
Due to relative motion $$\vec v_{P/table} = \vec v_{P/D} + \vec v_{D/table} = \vec v_{P/D} + 0 = \vec v_{P/D} $$
One way to solve for the angular velocity of ##D## relative to ##P## is by thinking about the number of revolutions ##D## completes relative to ##P##. If we take a bird view, it can be seen the number of revolutions can be found from
$$N = \frac{2 \pi s}{2 \pi r} = \frac{s}{r}$$
Therefore the angular velocity of the base spinning is $$\omega = \frac{N 2 \pi}{T} = \frac{2 \pi s}{T r}$$
Therefore the velocity of P relative to the table is
$$v_{P/table} = v_{D/P} = \vec \omega \times \vec r = \omega r = \frac{2 \pi s}{T}$$
This answer should be wrong since that's the velocity of a point along the bigger circle with circumference ##2 \pi s## but ##P##'s trajectory projects to a smaller circle. However what's wrong with the arguments used in the calculation? Every argument seems deceivingly correct. Thanks,