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.

enter image description here

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.


1 Answer 1


enter image description here

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

$$E=\frac 12\mathbf \omega \cdot \mathbf I\cdot \mathbf\omega+M\,g\,h$$


$$\mathbf\omega= \left[ \begin {array}{ccc} 1&0&0\\ 0&\sin \left( \alpha \right) &-\cos \left( \alpha \right) \\ 0& \cos \left( \alpha \right) &\sin \left( \alpha \right) \end {array} \right] \,\left[ \begin {array}{c} 0\\ 0\\ \Omega\end {array} \right] =\Omega\,\left[ \begin {array}{c} 0\\ -\cos \left( \alpha \right) \\ \sin \left( \alpha \right) \end {array} \right] $$

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: $$h^2=\left(\frac H4\right)^2+R^2$$


\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 .

  • $\begingroup$ The rotational kinetic energy of a solid object rotating around a non-principal axis (such as this one is) needs to take into account the projections of $\omega$ along all three non-principal axes, not just one of the principal axes: $T = \frac{1}{2} \left( I_1 \omega_1^2 + I_2 \omega_2^2 + I_3 \omega_3^2\right)$. $\endgroup$ Commented Feb 16, 2021 at 15:01
  • $\begingroup$ Also, the angle between the $z$ and $z'$ axes is $2 \alpha$ according to your notation, not $\alpha$. $\endgroup$ Commented Feb 16, 2021 at 15:11
  • $\begingroup$ O.k. Thank you I will correct it $\endgroup$
    – Eli
    Commented Feb 16, 2021 at 15:51
  • $\begingroup$ Actually I am in high school and this was a problem in my test. I think we haven't been taught the matrices method for the calculation of angular velocity and moment of inertia. Is there another method to solve this question? $\endgroup$
    – Aasneh
    Commented Feb 16, 2021 at 18:18
  • $\begingroup$ sorry for that, perhaps you can ask your teacher how solve this question with out using matrices ? $\endgroup$
    – Eli
    Commented Feb 17, 2021 at 7:11

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