Inertia tensor for spinning top

Question

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 $$\begin{pmatrix} \mathbf{e_1'}\\ \mathbf{e_2'}\\ \mathbf{e_3'} \end{pmatrix}=\mathbf{(R_{\psi}R_{\theta}R_{\phi})^T}\begin{pmatrix} \mathbf{e_1}\\ \mathbf{e_2}\\ \mathbf{e_3} \end{pmatrix}.$$

So

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

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.

Answer 1

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 $\mathbf{I}_{zz}$:

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

Answer 2

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}