Kinetic Energy of a Rotating Cone 
A hollow cone $(M, R)$ of semi-angle $30$ degrees is rolling without slipping on a fixed cone (same radius and semi-angle) as shown. If the angular speed of rotation is $\text{2 rad/s}$, find energy due to motion of the cone.


My Approach:
After analysing a bit I realised that the cone is not only rotating about its own axis but all points are undergoing circular motion with respect to the vertical axis. To find the kinetic energy  I started off with finding velocity of centre of mass of rotating cone. This can be easily found out by using cross product of $R$ and $\omega$ (angular velocity of rotation about vertical axis, $\omega = \text{2 rad/s})$ and $R$ is distance of centre of mass from vertical axis which can be found out by geometry as $R=\frac23H\sin60$.
Now translational kinetic energy can easily be given as $\frac12mv^2$. My main trouble started with calculation of rotational kinetic energy. I first thought of considering the rotation of particles about the centre of mass via an axis passing through the apex and centre of mass of rotating cone. Being in the centre of mass frame I analysed that each point on the rotating cone takes the same amount of time to complete one revolution around centre of mass as that is required by centre of mass to complete a circle about vertical axis - this is because the cones are symmetric so the two angular velocities turn out to be same.
Now since the body is also rotating about the vertical axis, I think we need to find the rotational energy associated with this axis also. I thought of trying to calculate moment of inertia of the cone about the vertical axis but it seems very difficult. Am I going conceptually wrong somewhere? Should I try another approach? It would be great if someone could help me out with this one.
 A: 
The cone is parametrized by the radius R and the semi- angle $~\alpha$
with
$$\alpha=\arctan\left(\frac{R}{H}\right)$$
you can obtain the height $~H$
we put the coordinate system at the center of mass
the energy is
$$E=\frac 12\mathbf \omega \cdot  \mathbf I\cdot \mathbf\omega+M\,g\,h$$
with:
$$\mathbf\omega= \left[ \begin {array}{ccc} 1&0&0\\ 0&\sin \left( 
\alpha \right) &-\cos \left( \alpha \right) \\ 0&
\cos \left( \alpha \right) &\sin \left( \alpha \right) \end {array}
 \right] \,\left[ \begin {array}{c} 0\\ 0\\ 
\Omega\end {array} \right] 
=\Omega\,\left[ \begin {array}{c} 0\\  -\cos \left( \alpha
 \right) \\ \sin \left( \alpha \right) \end {array}
 \right] 
$$
with the parallel axis transformation the inertia $~\mathbf I~$ at the center of mass is:
$$\mathbf I=\left[ \begin {array}{ccc} 1/20\,M \left( 3\,{R}^{2}+2\,{H}^{2}
 \right) &0&0\\  0&1/20\,M \left( 3\,{R}^{2}+2\,{H}^{
2} \right) &0\\  0&0&3/10\,M{R}^{2}\end {array}
 \right] 
+M\,\left[ \begin {array}{ccc} 0&-r_{{z}}&r_{{y}}\\  r_{
{z}}&0&-r_{{x}}\\ -r_{{y}}&r_{{x}}&0\end {array}
 \right] 
\,\left[ \begin {array}{ccc} 0&-r_{{z}}&r_{{y}}\\  r_{
{z}}&0&-r_{{x}}\\ -r_{{y}}&r_{{x}}&0\end {array}
 \right]$$
with
$$ \left[ \begin {array}{c} r_{{x}}\\  r_{{y}}
\\  r_{{z}}\end {array} \right] =\left[ \begin {array}{c} 0\\  0\\  
1/4\,H\end {array} \right] 
$$
the height of the potential enegrgy is:
$$h^2=\left(\frac H4\right)^2+R^2$$
$\Rightarrow$
\begin{align*}
 &E= \left( -{\frac {3}{20}}\, \left( \cos \left( \alpha \right)  \right)
^{2}{R}^{2}+{\frac {3}{80}}\, \left( \cos \left( \alpha \right)
 \right) ^{2}{H}^{2}+3/10\,{R}^{2} \right) M{\Omega}^{2}\\&+\frac 14\,Mg\sqrt
{{H}^{2}+16\,{R}^{2}} 
 \end{align*}
those results are for solid  cone .
