Question 1.
The moment of inertia \(dI_{Oz}\) of a plate of size \(l\times l\times dz\) about axis \(Oz\) and mass \(dm= \rho l^2 dz\) is given by \[ \boxed{dI_{Oz}= \frac{1}{6} l^2 dm = \frac{1}{6} \rho l^4 dz} \]
Question 2.1 The body admits the \(Oxz\) as plane of symmetry: this implies that the moments \(I_{Gxy}= I_{Gyz}=0\). It also admits the \(Oyz\) as plane of symmetry: this implies that the moments \(I_{Gxy}= I_{Gxz}=0\). We conclude that the inertia matrix about \(G\) of this body on basis \((\boldsymbol{\hat{x}},\boldsymbol{\hat{y}},\boldsymbol{\hat{z}})\) is necessarily diagonal. Also we have \(I_{Gx}=I_{Gy}\) since the axes \(Gx\) and \(Gy\) play the same role geometrically and from the point of view of mass distribution.
Conclusion: \[ \boxed{ [I_G] = \begin{pmatrix} I_{Gx} & 0 & 0 \\ 0 & I_{Gy} & 0 \\ 0 & 0 & I_{Gz} \end{pmatrix}_b } \] on basis \(b= (\boldsymbol{\hat{x}},\boldsymbol{\hat{y}},\boldsymbol{\hat{z}})\). Note that the same matrix is obtained on any basis \((-,-,\boldsymbol{\hat{z}})\).
Question 2.2 Using the result of question 1, we can find the moment of inertia \(I_{Oz}\) of body 1 about axis \(Oz\) as the body can be considered as a distribution of plates of the type described in question 1 with sidelength \(l(z) = a(1-z/h)\) for \(0\leq z\leq h\): \[ I_{Gz}= I_{Oz}= \int (x^2+y^2) dm = \int_0^h dJ_{Oz} = \int_0^h \frac{1}{6} \rho l(z)^4 dz \] Here we set \(z=0\) at \(O\) (the moments \(I_{Oz}\) and \(I_{Gz}\) are identical). Using \(l(z)= a(1-z/h)\), we find \[ I_{Oz}= \int_0^h \frac{1}{6} \rho a^4 (1- z/h)^4 dz = \frac{1}{6} a^4 h \rho \int_0^1 (1- u)^4 du =\frac{1}{30} a^4 h \rho \] since \(\int_0^1 (1-u)^4 = \int_0^1 u^4 du= 1/5\). Since \(\rho = 3m/a^2 h\), we find \[ \boxed{I_{Gz}=I_{Oz}=\frac{1}{10} m a^2} \]
Question 2.3 We need to find \(J= \int z^2 dm\). We can consider a slab of thickness \(dz\) and mass \(dm= \rho [l(z)]^2 dz\) parallel to plane \(Oxy\). Now \(l(z)= a(1- \frac{z+h/4}{h})\) (we have \(z=0\) at \(G\)) for $-h/4 z 3h/4: \[ J= \rho \int_{-h/4}^{3h/4} z^2 [l(z)]^2 dz= a^2 h^3 \rho\int_{-1/4}^{3/4} u^2 (\frac{3}{4}-u)^2 du= \frac{1}{80}\rho a^2 h^3 \] With \(\rho = 3m/a^2 h\), we find \[ \boxed{J= \frac{3}{80}m h^2} \]
Since \(I_{Gx}=I_{Gy}\) we find \(\int x^2 dm= \int y^2 dm\) and \(I_{Gz}= 2 \int y^2 dm\). We conclude that \[ I_{Gx}= \int (y^2+z^2) dm= \frac{1}{2}I_{Gz}+ J = \frac{1}{20} m a^2 +\frac{3}{80}m h^2 \] or \[ \boxed{I_{Gx}= \frac{1}{80} m ( 4 a^2 +3 h^2)} \]
Question 2.3 We use the parallel axis theorem: \[ [I_{C}]_b = [I_G]_b + \begin{pmatrix} 9mh^2/16 & 0 & 0 \\ 0 & 9mh^2/16 & 0 \\ 0 & 0 & 0 \end{pmatrix}_b = \frac{1}{20} m \begin{pmatrix} a^2+ 12h^2 & 0 & 0 \\ 0 & a^2+12 h^2 & 0 \\ 0 & 0 & 2a^2 \end{pmatrix}_b \]
Question 3
We use the angles \((\psi,\theta,\phi)\) as the parameters which define the orientation of basis \(b\). See the sketch below. Point \(C\) is fixed. The orientation of \(\boldsymbol{\hat{z}}\) is defined by angles \(\psi\) and \(\theta\). However angle is assumed fixed: \(\dot{\theta}=0\). We can find \(d\boldsymbol{\hat{z}}/dt\): \[ \frac{d\boldsymbol{\hat{z}}}{dt}= \dot{\psi}\boldsymbol{\hat{z}}_0 \times \boldsymbol{\hat{z}}= \psi \sin\theta\boldsymbol{\hat{v}} \] (see the rotation diagrams below). This gives the velocity of \(G\) (\(\boldsymbol{r}_{CG}= (3h/4) \boldsymbol{\hat{z}}\)) \[ \boldsymbol{v}_G = \frac{3}{4}h \dot{\psi}\sin\theta\boldsymbol{\hat{v}} \] The linear momentum of the body is \(m\boldsymbol{v}_G\). The angular velocity of the body is given by \[ \boldsymbol{\omega}= \dot{\psi}\boldsymbol{\hat{z}}_0 + \dot{\phi}\boldsymbol{\hat{z}} \] Then the angular momentum about \(C\) is \[ \boldsymbol{H}_C = {\cal I}_C \boldsymbol{\omega}= \frac{1}{20} m \begin{pmatrix} a^2+ 12h^2 & 0 & 0 \\ 0 & a^2+12 h^2 & 0 \\ 0 & 0 & 2a^2 \end{pmatrix}_{(\boldsymbol{\hat{w}},\boldsymbol{\hat{v}},\boldsymbol{\hat{z}})} \begin{pmatrix} -\dot{\psi}\sin\theta \\ 0 \\ \dot{\phi}+\dot{\psi}\cos\theta \end{pmatrix}_{b'} \] giving \[ \boxed{ \boldsymbol{H}_C = -\frac{1}{20} m (a^2+ 12h^2)\dot{\psi}\sin\theta\boldsymbol{\hat{w}} + \frac{1}{10} m a^2 (\dot{\phi}+\dot{\psi}\cos\theta)\boldsymbol{\hat{z}} } \] We took advantage of the fact that the matrix \([I_C]_b\) is invariant under a rotation about \(\boldsymbol{\hat{z}}\). We used basis \(b'=(\boldsymbol{\hat{w}},\boldsymbol{\hat{v}},\boldsymbol{\hat{z}})\).
© 2025 R. Valéry Roy.