I was trying to find the equations of motion of a symmetric, solid and fixed spinning top employing Euler's angles and lagrangian mechanics. As for Euler angles I used the following convention (the one in Wolfram's MathWorld): $\psi$ corresponds to the angle of rotation about the spinning top, $\phi$ would be the angle of precession and $\theta$ the nutation one. Its lagrangian should look like $$\mathcal{L}=\mathcal{T}_{\text{rot}}-\mathcal{U}=\dfrac{1}{2}\boldsymbol{\Omega}^\mathbf{T}\mathbf{I}\boldsymbol{\Omega}-mgR_{\text{CM}}\cos{\theta}$$ where $\boldsymbol{\Omega}$ is its total angular velocity, $\mathbf{I}$ the inertia tensor and $R_{\text{CM}}$ the length from the origin to the spinning top's center of mass.

Up until now I managed to find the total angular velocity by expressing the rotation axes ($\boldsymbol{\hat{\psi}},\boldsymbol{\hat{\phi}}, \boldsymbol{\hat{\theta}}$) in canonical basis and the rotated canonical basis, related by $$\begin{pmatrix} \mathbf{e_1'}\\ \mathbf{e_2'}\\ \mathbf{e_3'} \end{pmatrix}=\mathbf{(R_{\psi}R_{\theta}R_{\phi})^T}\begin{pmatrix} \mathbf{e_1}\\ \mathbf{e_2}\\ \mathbf{e_3} \end{pmatrix}.$$


$$\boldsymbol{\Omega}=\dot{\psi}\boldsymbol{\hat{\psi}}+\dot{\phi}\boldsymbol{\hat{\phi}}+\dot{\theta}\boldsymbol{\hat{\theta}}= \begin{pmatrix} \dot{\psi}\sin\theta\sin\phi+\dot{\theta}\cos\phi\\ -\dot{\psi}\sin\theta\cos\phi+\dot{\theta}\sin\phi\\ \dot{\psi}\cos\theta+\dot{\phi} \end{pmatrix}_{\mathbf{e}} = \begin{pmatrix} \dot{\phi}\sin\theta\sin\psi+\dot{\theta}\cos\psi\\ \dot{\phi}\sin\theta\cos\psi-\dot{\theta}\sin\psi\\ \dot{\phi}\cos\theta+\dot{\psi} \end{pmatrix}_{\mathbf{e'}}.$$

Regarding $R_{CM}$, it can be calculated using triple integrals assuming the top has a shape described by $z^2-z^4$ for $z\in [0,1]$ rotated about the $z$ axis ($x^2+y^2=(z^2-z^4)^2$).

Now what I'm left with is calculating the inertia tensor. I'd need help on figuring out which of the two basis is most practical, whether the tensor presents symmetries on the non-diagonal elements and what would its elements look like expressed as integrals.

  • $\begingroup$ This post shows how inertia tensors are calculated. We need to see your own attempts. $\endgroup$
    – Kurt G.
    Aug 12, 2023 at 16:17
  • $\begingroup$ Looks like all off diagonal elements are $0$ because either the sine or cosine in $y$ or $x$ integrated over $[0,2\pi]$ yields $0$. That for the canonical basis equation $x^2+y^2=(z^2-z^4)^2$ in cylindrical coordinates. Also, we get $\mathbf{I}_{zz}=\frac{\rho\pi}{2}\int_0^1(z^2-z^4)^4dz$ and by symmetry $\mathbf{I}_{xx}=\mathbf{I}_{yy}=\rho\pi(\frac{1}{2}\mathbf{I}_{zz}+\int_0^1z^2(z^2-z^4)^2dz).$ Should the limits of integration be the same for all?: $0\leq r\leq z^2-z^4$, $0\leq z\leq 1$ and $0\leq t\leq 2\pi$. $\endgroup$ Aug 12, 2023 at 22:59
  • $\begingroup$ After computing those integrals I got that $$\mathbf{I}=\begin{pmatrix} \dfrac{103m}{221} & 0 & 0\\ 0 & \dfrac{103m}{221} & 0\\ 0 & 0 & \dfrac{56m}{2431} \end{pmatrix}.$$ And so since $\mathbf{I}_{xx}=\mathbf{I}_{yy}$, $$\dfrac{1}{2}\boldsymbol{\Omega^T}\mathbf{I}\boldsymbol{\Omega}=\mathbf{I}_{xx}(\Omega_1^2+\Omega^2_2)+\mathbf{I}_{zz}\Omega^2_3.$$ $\endgroup$ Aug 12, 2023 at 23:26
  • $\begingroup$ Your calculations look wrong. Since the volume element in cylindrical coordinates is $dV=r\,dr\,d\varphi\,dz$ and since we integrate $mr^2$ we get if I am not mistaken \begin{align} \mathbf{I}_{zz}&=\int_0^1\iint\limits_{x^2+y^2\le (z^2-z^4)^2}m\underbrace{(x^2+y^2)}_{\textstyle r^\color{red}{2}}\, dx\,dy\,dz=m\int_0^1\int_0^{(z^2-z^4)}\int_0^{2\pi} r^\color{red}{3} \,dr\,d\varphi\,dz\,,\\ &=2\pi m\int_0^1\frac{(z^2-z^4)^4}{4}\,dz=2\pi m\frac{32}{109395}\,. \end{align} $\endgroup$
    – Kurt G.
    Aug 13, 2023 at 7:44

2 Answers 2


First let's calculate the spinning top's volume: $$\begin{aligned} \mathcal{V}=\iiint_{V: x^2+y^2\leq (z^2-z^4)^2} dV&=\int_0^{2\pi}\int_0^1\int_0^{z^2-z^4} rdrdzdt\\ &=2\pi\int_0^1\dfrac{(z^2-z^4)^2}{2}dz\\ &=\pi\int_0^1(z^4-2z^6-z^8)dz=\dfrac{8\pi}{315}. \end{aligned}$$

Therefore the top's density would be $\rho=\dfrac{m}{\mathcal{V}}=\dfrac{315m}{8\pi}.$

As for the moments of inertia, since its products $\mathbf{I}_{ij}$ are $0$ and, by symmetry, $\mathbf{I}_{xx}=\mathbf{I}_{yy}$, it is only necessary to calculate $\mathbf{I}_{xx}$ and $\mathbf{I}_{zz}$:

$$\begin{aligned} \mathbf{I}_{zz}&=\iiint_{V}(x^2+y^2)\rho dV\\ &=\rho\int_0^{2\pi}\int_0^1\int_0^{z^2-z^4} r^2 rdrdzdt\\ &=\dfrac{\pi}{2}\rho\int_0^1 (z^2-z^4)^4dz=\dfrac{\pi}{2}\rho\dfrac{128}{109395}=\dfrac{\pi}{2}\dfrac{315m}{8\pi}\dfrac{128}{109395}=\dfrac{56m}{2431}. \end{aligned}$$ $$\begin{aligned} \mathbf{I}_{xx}&=\iiint_{V}(y^2+z^2)\rho dV\\ &=\rho\iiint_V (r^2\sin^2t+z^2) rdrdzdt\\ &=\rho\iiint_V r^3\sin^2t\ drdzdt+\rho\iiint_Vrz^2drdzdt\\ &=\rho\int_0^{2\pi}\sin^2tdt\int_0^1\int_0^{z^2-z^4} r^3 drdz+2\pi\rho\int_0^1\int_0^{z^2-z^4}rz^2drdzdt\\ &=\dfrac{1}{2}\underbrace{\left(\dfrac{\pi}{2}\rho\int_0^1 (z^2-z^4)^4dz\right)}_{\mathbf{I}_{zz}}+\pi\rho\int_0^1 z^2(z^2-z^4)^2dz\\ &=\dfrac{1}{2}\dfrac{56m}{2431}+\dfrac{8\pi}{693}\rho=\dfrac{56m}{4862}+\dfrac{315m}{693}=\dfrac{103m}{221}. \end{aligned}$$


Your calculations look wrong. Since the volume element in cylindrical coordinates is $dV=r\,dr\,d\varphi\,dz$ and since we integrate (with constant density) $\rho\,r^2=m\,r^2/V$ we get if I am not mistaken \begin{align} \mathbf{I}_{zz}&=\int_0^1\iint\limits_{x^2+y^2\le (z^2-z^4)^2}\rho \,\underbrace{(x^2+y^2)}_{\textstyle r^\color{red}{2}}\, dx\,dy\,dz=\rho \,\int_0^1\int_0^{(z^2-z^4)}\int_0^{2\pi} r^\color{red}{3} \,dr\,d\varphi\,dz\,,\\ &=2\pi \rho \,\int_0^1\frac{(z^2-z^4)^4}{4}\,dz=2\pi \rho \,\frac{32}{109395}\,. \end{align} The calculation of $\mathbf{I}_{xx}$ is trickier. Using wolfram alpha a lot I get \begin{align} \mathbf{I}_{xx}&=\int_0^1\iint\limits_{x^2+y^2\le (z^2-z^4)^2}\rho \,(\color{red}{z^2}+y^2)\, dx\,dy\,dz\\ &=\int_0^1\rho \,z^2\,\underbrace{\pi(z^2-z^4)^2}_{\text{area of disk}}\,dz\,+\,m\int_0^1\int\limits_{-(z^2-z^4)}^{(z^2-z^4)}\int\limits_{-\sqrt{(z^2-z^4)^2-x^2}}^{\sqrt{(z^2-z^4)^2-x^2}}y^2\,dy\,dx\,dz\\ &=\frac{8\pi\,\rho\,}{693}+\rho \,\int_0^1\int\limits_{-(z^2-z^4)}^{(z^2-z^4)}2\frac{((z^2-z^4)^2-x^2)^{3/2}}{3}\,dx\,dz\\ &=\frac{8\pi\,\rho \,}{693}+ \frac{32\pi\,\rho \,}{109395}\,. \end{align}

  • $\begingroup$ What I don't get is what $m$ is, should it be density? If it weren't then the inertia element would have wrong dimensions ($\text{kg m}^5$) instead of ($\text{kg m}^2$). For the first integral, I got your result but $m$ being the density. What I did was finding the top's volume and substituting the density for $m/V$, $m$ being its mass and $V$ being $8\pi/315$. $\endgroup$ Aug 13, 2023 at 10:01
  • $\begingroup$ The same goes for the 2nd integral: if you substitute the density for $m/V$ you get the same as me. $\endgroup$ Aug 13, 2023 at 10:16
  • $\begingroup$ That $m$ is your mass $m$ in your very first equation in OP. The density $\rho=dm/dV$ is uniform. $\mathbf{I}$ has undoubtetly the dimension kg m$^2\,.$ This is a nice example where the powers in the integration limits make us believe that this may not be the case. But this is an artefact. Edit: I simply don't see what this has to do with density versus mass. The density is uniform across $V$. $\endgroup$
    – Kurt G.
    Aug 13, 2023 at 10:17
  • $\begingroup$ Can I post my procedure to clear things out? $\endgroup$ Aug 13, 2023 at 10:22
  • $\begingroup$ You don't need my permission to post. Whatever you find most conventient. $\endgroup$
    – Kurt G.
    Aug 13, 2023 at 10:28

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