D5. Mass distribution

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The Vector Theorems relate the external interaction torsor on a system ([math]\displaystyle{ \sum\overline{\mathbf{F}}_\mathrm{ext} }[/math], [math]\displaystyle{ \sum\overline{\mathbf{M}}_\mathrm{ext}(\Qs) }[/math]) to the change in time of vectors that depend on how the mass is distributed in the system (mass geometry) and on its motion. In the LMT, this vector is the linear momentum of the system, while in the AMT it its angular momentum (or kinetic momentum). This unit provides the tools necessary to describe the mass geometry of a rigid body and to calculate these two vectors.

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D5.1 Center of masses

The centre of mass of a mechanical system is a point whose kinematics is a weighted kinematics of all the elements of the system that have mass. In this course it is represented by the letter G.

Figure D5.1 Centre of mass of a system with constant mass

system of particles: [math]\displaystyle{ \vec{\Os_R \Gs} = \sum_p \frac{\ms_P}{\Ms}\vec{\Os_R \Ps} }[/math] continuous system: [math]\displaystyle{ \vec{\Os_R \Gs} = \sum_{i=1}^N \bigl( \frac{1}{\Ms_i} \int_{S_i} dm(\Ps)\vec{\Os_R \Ps} \bigr) }[/math]

In a homogeneous rigid body S, the location of G is easy when the rigid body has important symmetries (Figure D5.2).

Figure D5.2 Centre of mass of rigid bodies with important symmetries

When this is not the case, integration must be carried out to obtain it. If M is the total mass of the rigid body: [math]\displaystyle{ \overline{\Os_\Rs\Gs}=\frac{1}{\mathrm{M}} \int_\mathrm{S}\mathrm{dm}(\Ps)\overline{\Os_\Rs\Ps} }[/math]

The Table shows the centre of mass of the most common geometries. From this information and for rigid bodies S formed by several of these elements [math]\displaystyle{ \mathrm{S}_\is }[/math], the position of the centre of mass can be found as a weighted average of the position of each [math]\displaystyle{ \mathrm{G}_\is }[/math].

✏️ Example D5.1: shell

The rigid body S is made up of a cylindrical shell and a spherical semi-shell, both homogeneous and with the same surface density [math]\displaystyle{ \sigma }[/math]. For symmetry reasons, the [math]\displaystyle{ (\xs,\ys) }[/math] coordinates of the centre of mass G are zero: [math]\displaystyle{ (\xs_\mathrm{G},\ys_\mathrm{G})=(0,0) }[/math]. The z coordinate of the cylindrical shell is [math]\displaystyle{ \zs_\mathrm{Gcil}=\Rs/2 }[/math]. That of the spherical semi-shell can be found from the Table:

[math]\displaystyle{ \zs_\mathrm{Gsph}=\Rs+(\Rs/2)=3\Rs/2 }[/math].

The mass of each element is the product of the surface density by the surface area of the element:

[math]\displaystyle{ \ms_\mathrm{cyl}=\sigma 2 \pi \Rs^2 }[/math] , [math]\displaystyle{ \ms_\mathrm{sph}=\sigma \frac{1}{2} 4 \pi \Rs^2= \sigma 2 \pi \Rs^2 }[/math]
.

Hence: [math]\displaystyle{ \zs_\mathrm{G}=\frac{\ms_\mathrm{sph}\zs_\mathrm{Gsph}+\ms_\mathrm{cyl}\zs_\mathrm{Gcyl}}{\ms_\mathrm{sph}+\ms_\mathrm{cyl}}=\frac{2\pi\Rs^2(3/2)\Rs +2\pi\Rs^2(1/2)\Rs}{2\pi\Rs^2+2\pi\Rs^2}=\Rs }[/math]

✏️ Example D5.2: folded plate

The rigid body S is a folded homogeneous triangular plate with a surface density of [math]\displaystyle{ \sigma }[/math]. The centre of mass can be found as the weighted average of the centre of mass of a square plate with a side length of 6L and two triangular plates with legs 6L:

[math]\displaystyle{ (\xs_1,\ys_1)=(3\Ls,3\Ls) \hspace{3cm} (\xs_2,\ys_2)=(8\Ls,2\Ls) \hspace{3cm} (\xs_3,\ys_3)=(2\Ls,4\Ls) }[/math]

[math]\displaystyle{ \hspace{1cm} \ms_1=(6\Ls)(6\Ls)\sigma=36\Ls^2\sigma \hspace{1.5cm} \ms_2=(1/2)(6\Ls)(6\Ls)\sigma=18\Ls^2\sigma \hspace{1cm} \ms_3=(1/2)(6\Ls)(6\Ls)\sigma=18 \Ls ^2 \sigma }[/math]

Therefore: [math]\displaystyle{ (\xs_\mathrm{G},\ys_\mathrm{G})=\frac{36\sigma(3,3)\Ls^3+ 18\sigma (8,2)\Ls^3 + 18\sigma (2,4)\Ls^3}{36\Ls^2\sigma+18\Ls^2\sigma+ 18 \Ls^2 \sigma}=(4,3)\Ls. }[/math]

✏️ Example D5.3: cylinder with a hole

The rigid body is a homogeneous perforated cylinder of density [math]\displaystyle{ \rho }[/math], and can be considered as a superposition of a rigid cylinder, with height 4L and radius 2r, and a cylinder of negative mass, with height 2L and radius r. For symmetry reasons, [math]\displaystyle{ (\xs_\mathrm{G},\ys_\mathrm{G})=(0,0) }[/math]. The z coordinate can be found as a weighted average of the z coordinates of two cylinders: Mass of the rigid cylinder and mass of the hole:

[math]\displaystyle{ \ms_\mathrm{rigid}=\mathrm{V}_\mathrm{rigid}\rho=\pi(2\rs)^24\Ls\rho=16\pi\Ls \rs^2\rho \quad , \quad \ms_\mathrm{hole}=\mathrm{V}_\mathrm{hole} \rho=\pi\rs^22 \Ls\ \rho=2\pi\Ls \rs^2\rho }[/math]

[math]\displaystyle{ \zs_\mathrm{G}=\frac{\ms_\mathrm{rigid}\zs_\mathrm{rigid}- \ms_ \mathrm{hole} \zs_\mathrm{hole}}{\ms_\mathrm{mrigid} -\ms_\mathrm{hole}}= \frac{16\pi\Ls\rs^2\rho \cdot 2\Ls - 2\pi\Ls \rs^2 \rho \cdot 3\Ls}{16\pi\Ls\rs^2\rho-2\pi\Ls\rs^2\rho}=\frac{13}{7}\Ls. }[/math]

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D5.2 Inertia tensor

The calculation of the angular momentum of a rigid body S at a point Q of that rigid body can be done easily from a positive definite symmetric matrix [math]\displaystyle{ \mathrm{II}(\Qs) }[/math], called the inertia tensor of S at point Q, and its angular velocity [math]\displaystyle{ \velang{S}{RTQ} }[/math] (which is equal to [math]\displaystyle{ \velang{S}{Gal} }[/math] since the reference frame RTQ has a translational motion relative to a Galilean one):

[math]\displaystyle{ \overline{\mathbf{H}}_{\text {RTQ }}(\mathbf{Q})=\int_{\mathrm{s}} \overline{\mathbf{Q P}} \times \mathrm{dm}(\mathbf{P}) \overline{\mathbf{v}}_{\text {RTQ }}(\mathbf{P}) \equiv \mathrm{II}(\mathbf{Q}) \bar{\Omega}_{\text {RTQ }}^{\mathrm{s}}=\mathrm{II}(\mathbf{Q}) \bar{\Omega}_{\text {Gal }}^{\mathrm{s}} . }[/math]

The relationship between angular momentum [math]\displaystyle{ \overline{\mathbf{H}}_\mathrm{RTQ}(\Qs) }[/math] and angular velocity [math]\displaystyle{ \velang{S}{Gal} }[/math] is not a simple proportionality, since [math]\displaystyle{ \mathrm{II}(\Qs) }[/math] is a matrix. For that reason, these two vectors are not parallel in general (Figure D5.3).

Figure D5.3 Angular momentum and angular velocity of a rigid body are not parallel in general

The [math]\displaystyle{ \mathrm{II}(\Qs) }[/math] elements in a vector basis B (with axes 123) have to do with the mass distribution of S around some coordinate axes [math]\displaystyle{ (\xs_1,\xs_2,\xs_3) }[/math] with origin in Q (Figure D5.4):

Figure D5.4 Inertia tensor of a rigid body

[math]\displaystyle{ I_{ii} = \sum_S [x_j^2(\Ps)+x_k^2(\Ps)]dm(\Ps) = \sum_S \delta_i^2(\Ps) dm(\Ps) \geq 0 }[/math] [math]\displaystyle{ (\delta_i(\Ps) = \text{distance from } (\Ps) \text{to axis i}) }[/math] [math]\displaystyle{ I_{ij} = I_{ji} = - \sum_S x_i(\Ps)x_j(\Ps)dm(\Ps), \text{ either sign} }[/math]

The elements on the diagonal ([math]\displaystyle{ \mathrm{I}_\mathrm{ii} }[/math]) are called moments of inertia, and can never be negative. Those outside the diagonal ([math]\displaystyle{ \mathrm{I}_\mathrm{ii} }[/math]) are the products of inertia, and can have either sign.

If the B vector basis has a constant orientation relative to S, the [math]\displaystyle{ \mathrm{I}_\mathrm{ii} }[/math] elements are constant. In this course, we always work with inertia tensors with constant elements.

💭Proof➕

When the angular momentum of a rigid body S is calculated at a point Q of that rigid body, rigid body kinematics can be applied to relate the velocity of all the points of S with that of point Q:

[math]\displaystyle{ \overline{\mathbf{H}}_{\text {RTQ }}(\mathbf{Q})=\int_{\mathrm{s}} \overline{\mathbf{Q P}} \times \mathrm{dm}(\mathbf{P}) \vel{P}{RTQ} }[/math]

[math]\displaystyle{ \Ps, \Qs \in \mathrm{S} \quad \Rightarrow \quad \vel{P}{RTQ}=\vel{Q}{RTQ}+ \velang{S}{RTQ} \times \QPvec = \velang{S}{RTQ} \times \QPvec = -\QPvec \times \velang{S}{RTQ} }[/math]

[math]\displaystyle{ \overline{\mathbf{H}}_{\text {RTQ }}(\mathbf{Q})=\int_{\mathrm{s}} \QPvec \times (\QPvec \times \velang{S}{RTQ})\mathrm{dm}(\Ps) }[/math]

If vector [math]\displaystyle{ \QPvec }[/math] is projected on a vector basis B with axes (1,2,3):

[math]\displaystyle{ \braq{\QPvec \times \velang{S}{RTQ}}{B}=\vector{\mathrm{QP}_1}{\mathrm{QP}_2}{\mathrm{QP}_3} \times \vector{\Omega_1}{\Omega_2}{\Omega_3} \equiv \vector{\xs_1}{\xs_2}{\xs_3} \times \vector{\Omega_1}{\Omega_2}{\Omega_3}= \vector{\xs_2\Omega_3-\xs_3\Omega_2}{\xs_3\Omega_1-\xs_1\Omega_3}{\xs_1\Omega_2-\xs_2\Omega_1}=\matriz{0}{-\xs_3}{\xs_2}{\xs_3}{0}{-\xs_1}{-\xs_2}{\xs_1}{0} \vector{\Omega_1}{\Omega_2}{\Omega_3} , }[/math]

[math]\displaystyle{ \braq{\QPvec \times (\QPvec \times \velang{S}{RTQ})}{B}= \matriz{0}{-\xs_3}{\xs_2}{\xs_3}{0}{-\xs_1}{-\xs_2}{\xs_1}{0} \matriz{0}{-\xs_3}{\xs_2}{\xs_3}{0}{-\xs_1}{-\xs_2}{\xs_1}{0} \vector{\Omega_1}{\Omega_2}{\Omega_3} = - \matriz{\xs_2^2 + \xs_3^2}{-\xs_1\xs_2}{-\xs_1\xs_3}{-\xs_1\xs_2}{\xs_1^2 + \xs_3^2}{-\xs_2\xs_3}{-\xs_1\xs_3}{-\xs_2\xs_3}{\xs_1^2 + \xs_2^2} \vector{\Omega_1}{\Omega_2}{\Omega_3}. }[/math]

Finally: [math]\displaystyle{ \braq{\overline{\mathbf{H}}_{\text {RTQ }}(\Qs)}{B}=\matriz{\int_{\mathrm{s}}(\xs_2^2 + \xs_3^2)\mathrm{dm}(\Ps)}{-\int_{\mathrm{s}}(\xs_1\xs_2)\mathrm{dm}(\Ps)}{-\int_{\mathrm{s}}(\xs_1\xs_3)\mathrm{dm}(\Ps)}{-\int_{\mathrm{s}}(\xs_1\xs_2)\mathrm{dm}(\Ps)}{\int_{\mathrm{s}}(\xs_1^2 + \xs_3^2)\mathrm{dm}(\Ps)}{-\int_{\mathrm{s}}(\xs_2\xs_3)\mathrm{dm}(\Ps)} {-\int_{\mathrm{s}}(\xs_1\xs_3)\mathrm{dm}(\Ps)} {-\int_{\mathrm{s}}(\xs_2\xs_3\mathrm{dm}(\Ps)}{\int_{\mathrm{s}}(\xs_1^2 + \xs_2^2)\mathrm{dm}(\Ps)} \vector{\Omega_1}{\Omega_2}{\Omega_3} \equiv \Bigr[\mathrm{II}(\Qs) \Bigr]_\mathrm{B} }[/math]

✏️ Example D5.4: discrete rigid body

The rigid body is made up of six particles with mass m connected by massless rigid bars. Since it is a discrete mass distribution, no integral needs to be performed to calculate the inertia tensor. The inertia moments of that tensor at point O and the vector basis 123 are:

[math]\displaystyle{ \mathrm{I}_{11}=6\ms\Ls^2 \left\{\begin{array}{l} \bullet \text {contribution of particles on axis} 2: 2 \cdot \ms\Ls^2 \\ \bullet \text {contribution of particles on axis} 3: 2 \cdot \ms\Ls^2 \\ \bullet \text {contribution of particles in quadrants} 13: 2 \cdot \ms\Ls^2 \end{array}\right. }[/math]

[math]\displaystyle{ \mathrm{I}_{22}=6\ms\Ls^2 \left\{\begin{array}{l} \bullet \text {contribution of particles on axis} 2: 0\\ \bullet \text {contribution of particles on axis} 3: 2 \cdot \ms\Ls^2 \\ \bullet \text {contribution of particles in quadrants} 13: 2 \cdot \ms(\sqrt{2}\Ls)^2 \end{array}\right. }[/math]

[math]\displaystyle{ \mathrm{I}_{33}=4\ms\Ls^2 \left\{\begin{array}{l} \bullet \text {contribution of particles on axis} 2: 2 \cdot \ms\Ls^2 \\ \bullet \text {contribution of particles on axis} 3: 0 \\ \bullet \text {contribution of particles in quadrants} 13: 2 \cdot \ms\Ls^2 \end{array}\right. }[/math]

The inertia products are:

[math]\displaystyle{ \mathrm{I}_{12}=0 \left\{\begin{array}{l} \bullet \text {contribution of particles on axis} 2 \text{and} 3: 0 \text{ (since} \xs_1=0)\\ \bullet \text {contribution of particles in quadrants} 13: 0 \text{ (since } \xs_2=0) \end{array}\right. }[/math]

[math]\displaystyle{ \mathrm{I}_{13}=2\ms\Ls^2 \left\{\begin{array}{l} \bullet \text {contribution of particles on axis} 2 \text{ and } 3 : \xs_1=0 \Rightarrow 0 \\ \bullet \text {contribution of particles in quadrants}1^+3^-: \xs_1=\Ls, \xs_3=-\Ls \Rightarrow \ms\Ls^2 \\ \bullet \text {contribution of particles in quadrants}1^-3^+: \xs_1=-\Ls, \xs_3=\Ls \Rightarrow \ms\Ls^2 \end{array}\right. }[/math]

[math]\displaystyle{ \mathrm{I}_{23}=0 \left\{\begin{array}{l} \bullet \text {contribution of particles on axis} 2: 0 \text{ (since} \xs_3=0)\\ \bullet \text {contribution of particles on axis} 3: 0 \text{ (since} \xs_2=0)\\ \bullet \text {contribution of particles in quadrants} 13: 0 \text{ (since} \xs_2=0)\\ \end{array}\right. }[/math]

Finally:[math]\displaystyle{ \Bigr[\mathrm{II}(\Os) \Bigr]_\mathrm{B}= \matriz{6}{0}{2}{0}{6}{0}{2}{0}{4} \ms \Ls^2 }[/math]

The Table summarizes the moments and products of inertia of continuous rigid bodies (not made up of particles, as in EXAMPLE D5.3 ) with simple geometry.

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D5.3 Some relevant properties of the inertia tensor

The [math]\displaystyle{ \mathrm{II}(\Qs) }[/math] matrix elements depend on the vector basis being used:[math]\displaystyle{ \Bigr[\mathrm{II}(\Qs) \Bigr]_{\mathrm{B}1} \neq \Bigr[\mathrm{II}(\Qs) \Bigr]_{\mathrm{B}2} }[/math]. If it is the eigenbasis EB of the matrix (the one with the axes in the direction of the eigenvectors), it is diagonal:

[math]\displaystyle{ \Bigr[\mathrm{II}(\Qs) \Bigr]_\mathrm{EB}= \matriz{\mathrm{I}_{11}}{0}{0}{0}{\mathrm{I}_{22}}{0}{0}{0}{\mathrm{I}_{33}}. }[/math]

The directions of the EB vector through [math]\displaystyle{ \Qs }[/math] are called principal directions of inertia for point [math]\displaystyle{ \Qs }[/math] (PDI for[math]\displaystyle{ \Qs }[/math]) or principal axes of inertia (the word “axis” implies a direction through a particular point), and the corresponding inertia moments are the principal moments for point[math]\displaystyle{ \Qs }[/math]. If the angular velocity [math]\displaystyle{ \velang{S}{Gal} }[/math] is parallel to one of the principal axes, the angular momentum [math]\displaystyle{ \overline{\mathbf{H}}_\mathrm{RTQ}(\Qs) }[/math] and the angular velocity [math]\displaystyle{ \velang{S}{Gal} }[/math] are parallel (Figure D5.5).

Figure D5.5 Angular momentum when the direction of the angular velocity is a PDI

✏️ Example D5.5: discrete rigid body

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Let us consider a general rotation [math]\displaystyle{ \velang{S}{Gal} }[/math] of the discrete solid in example D5.4 . The angular momentum is not parallel to the angular velocity: [math]\displaystyle{ \braq{\overline{\mathbf{H}}_{\text {RTO }}(\Os)}{B}=\matriz{6}{0}{2}{0}{6}{0}{2}{0}{4}\ms \Ls^2 \vector{\Omega_1}{\Omega_2}{\Omega_3}= \vector{6\Omega_1 + 2\Omega_3}{6\Omega_2}{2\Omega_1 + 4\Omega_3}\ms \Ls^2 }[/math] The elements of the [math]\displaystyle{ \Bigr[\mathrm{II}(\Os) \Bigr]_\mathrm{B} }[/math] tensor, however, show that direction 2 is a DPI for point [math]\displaystyle{ \Os }[/math]. Therefore, if the angular velocity [math]\displaystyle{ \velang{S}{Gal} }[/math] has that direction [math]\displaystyle{ \overline{\mathbf{H}}_\mathrm{RTO}(\Os) }[/math] is parallel to [math]\displaystyle{ \velang{S}{Gal} }[/math]:

[math]\displaystyle{ \braq{\velang{S}{RTO}}{B}=\braq{\velang{S}{T}}{B}= \vector{0}{\Omega_2}{0} \quad \Rightarrow \quad \braq{\overline{\mathbf{H}}_{\text{RTO }}(\Os)}{}=\vector{0}{6\Omega_2}{0} \ms \Ls^2 \quad \Rightarrow \quad \overline{\mathbf{H}}_{\text{RTO }}(\Os) \parallel \velang{S}{T} }[/math]

If the [math]\displaystyle{ \velang{S}{Gal} }[/math] direction is that of axes 1 or 3, [math]\displaystyle{ \overline{\mathbf{H}}_{\text{RTO }}(\Os) }[/math] and [math]\displaystyle{ \velang{S}{Gal} }[/math] are not parallel any more:

[math]\displaystyle{ \braq{\velang{S}{RTO}}{B}=\braq{\velang{S}{T}}{B}= \vector{\Omega_1}{0}{0} \quad \Rightarrow \quad \braq{\overline{\mathbf{H}}_{\text{RTO }}(\Os)}{}=\vector{6\Omega_1}{0}{2\Omega_1} \ms \Ls^2 \quad \Rightarrow \quad \overline{\mathbf{H}}_{\text{RTO }}(\Os) }[/math] in quadrant [math]\displaystyle{ 1^+3^- }[/math]

[math]\displaystyle{ \braq{\velang{S}{RTO}}{B}=\braq{\velang{S}{T}}{B}= \vector{0}{0}{\Omega_3} \quad \Rightarrow \quad \braq{\overline{\mathbf{H}}_{\text{RTO }}(\Os)}{}=\vector{2\Omega_3}{0}{4\Omega_3} \ms \Ls^2 \quad \Rightarrow \quad \overline{\mathbf{H}}_{\text{RTO }}(\Os) }[/math] in quadrant [math]\displaystyle{ 1^-3^+ }[/math]

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D5.4 Steiner’s Theorem

When calculating the inertia tensor of a rigid body, it is necessary to make a qualitative assessment before resorting to the Table, since it only contains minimal information (the expression for an entire tensor is never given). This section presents some general properties that facilitate this assessment.

Property 1: In a planar rigid body (Figure D5.6), the direction perpendicular to it (direction k) is always a principal direction of inertia ([math]\displaystyle{ \mathrm{I}_\mathrm{ik}=\mathrm{I}_\mathrm{jk}=0 }[/math]) for any point [math]\displaystyle{ \Qs }[/math], and the corresponding inertia moment is the sum of the other two (Pythagoras theorem):

Figure D5.6 Planar rigid body

[math]\displaystyle{ x_k(\mathbf{P}) = 0 \Rightarrow \begin{cases} I_{ik}(\Qs) = -\int_S x_i(\Ps) x_k(\Ps) \, d m(\Ps) = 0 \\[10pt] I_{jk}(\Qs) = -\int_S x_j(\Ps) x_k(\Ps) \, d m(\Ps) = 0 \end{cases} }[/math] [math]\displaystyle{ \Is_{kk}(\Qs) = \int_S \delta_k^2 (\Ps) dm(\Ps) = \int_S \bigl[ \delta^2_i(\Ps) + \delta^2_j(\Ps) \bigr] dm(\Ps) = \Is_{ii}(\Qs) + \Is_{jj}(\Qs) }[/math]

Property 2:: In a planar rigid body (Figure D5.7), the sign of the contribution of each quadrant ij to the inertia product [math]\displaystyle{ \mathrm{I}_\mathrm{ij}(\Qs) }[/math] is:

Figure D5.7 Planar rigid body

[math]\displaystyle{ \Is_{ij} = -\int_S x_i(\Ps)x_j(\Ps)dm(\Ps) }[/math] quadrants [math]\displaystyle{ \is^+\js^+, \is^-\js^- }[/math]: [math]\displaystyle{ \Is_{ij}(\Qs) \lt 0 }[/math] quadrants [math]\displaystyle{ \is^+\js^-, \is^-\js^+ }[/math]: [math]\displaystyle{ \Is_{ij}(\Qs) \gt 0 }[/math]

Property 3: In any rigid body, if there is symmetry with respect to the plane ij through a point Q (Figure D4.8), the k direction is the principal direction of inertia for any point on that plane:

[math]\displaystyle{ \left.\begin{array}{l} \xs_\is(\Ps)=\xs_\is (\Ps^{\prime)} \\ \xs_\js(\Ps)=\xs_\js (\Ps^{\prime}) \\ \xs_\ks(\Ps)=\xs_\ks (\Ps^{\prime}) \end{array}\right\} \Rightarrow\left\{\begin{array}{l} \mathrm{I}_\mathrm{i k}(\Qs \in \text { plane of symmetry })=-\int_{\mathrm{s}} \xs_\is(\Ps) \xs_\ks(\Ps) \mathrm{dm}(\Ps)=0 \\ \mathrm{I}_\mathrm{j k}(\Qs \in \text { plane of symmetry })=-\int_{\mathrm{s}} \xs_\js(\Ps) \xs_\ks(\Ps) \mathrm{dm}(\Ps)=0 \end{array}\right. }[/math]

Figure D5.8 Rigid body with a simmetry plane ij

✏️ Example D5.6: planar rigid body

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The planar rigid body consists of three homogeneous bars of the same material attached to a massless frame. Plane rigid body in the 23 plane: by property 1, direction 1 is a DPI, and [math]\displaystyle{ \mathrm{I}_{11}=\mathrm{I}_{22}+ \mathrm{I}_{33} }[/math]. [math]\displaystyle{ \Bigr[\mathrm{II}(\Os) \Bigr]_\mathrm{B}= \matriz{\mathrm{I}_{22}+ \mathrm{I}_{33}}{0}{0}{0}{\mathrm{I}_{22}}{\mathrm{I}_{23}}{0}{\mathrm{I}_{23}}{\mathrm{I}_{33}} }[/math] by property 3, since the 13 plan is a symmetry plane, direction2 is a DPI: [math]\displaystyle{ \Bigr[\mathrm{II}(\Os) \Bigr]_\mathrm{B}= \matriz{\mathrm{I}_{22}+ \mathrm{I}_{33}}{0}{0}{0}{\mathrm{I}_{22}}{0}{0}{0}{\mathrm{I}_{33}} }[/math]

If we take into account that the central bar does not contribute to the [math]\displaystyle{ \mathrm{I}_{33} }[/math] inertia moment because it is on axis 3 and its distance to it is zero, it is easy to see that [math]\displaystyle{ \mathrm{I}_{22}\gt \mathrm{I}_{33} }[/math].

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Property 4: When a rigid body has two equal principal moments of inertia (according to orthogonal directions) for a point [math]\displaystyle{ \Os }[/math] ([math]\displaystyle{ \Is_\mathrm{ii} (\Os) = \Is_\mathrm{jj}(\Os) \equiv \Is , \Is_\mathrm{ij} (\Os)= 0 }[/math]), its inertia tensor at [math]\displaystyle{ \Os }[/math] is invariant under rotations about the k direction:[math]\displaystyle{ \Bigr[\mathrm{II}(\Os) \Bigr]_\mathrm{ijk}=\Bigr[\mathrm{II}(\Os) \Bigr]_\mathrm{i'j'k} }[/math]. Indeed:

[math]\displaystyle{ \Bigr[\mathrm{II}(\Os) \Bigr]_\mathrm{ijk}= \matriz{\mathrm{I}}{0}{0}{0}{\mathrm{I}}{0}{0}{0}{\mathrm{I}_\mathrm{kk}} }[/math]

To relate [math]\displaystyle{ \Bigr[\mathrm{II}(\Os) \Bigr]_\mathrm{ijk} }[/math] and [math]\displaystyle{ \Bigr[\mathrm{II}(\Os) \Bigr]_\mathrm{i'j'k} }[/math], we only need to transform the upper left quadrant (since the k axis is the same). If [math]\displaystyle{ [\mathrm{S}] }[/math] is the matrix of the change of basis [math]\displaystyle{ (\is,\js) \rightarrow (\is',\js'): }[/math]

[math]\displaystyle{ \left[\begin{array}{l} \text { upper } \\ \text { left } \\ \text { quadrant } \end{array}\right]_{\is' \js'}=[\mathrm{S}]^{-1}\left[\begin{array}{ll} \mathrm{I} & 0 \\ 0 & \mathrm{I} \end{array}\right][\mathrm{S}]=\mathrm{I}[\mathrm{S}]^{-1}\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right][\mathrm{S}]=\mathrm{I}[\mathrm{S}]^{-1}[\mathrm{S}]=\mathrm{I}\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \text {. } }[/math]

The rigid body is a symmetrical rotor at point [math]\displaystyle{ \Os }[/math]. If its angular velocity is contained in the ij plane or has the k direction, the angular momentum at [math]\displaystyle{ \Os ( \overline{\mathbf{H}}_\mathrm{RTO}(\Os)) }[/math] and the angular velocity [math]\displaystyle{ \velang{S}{RTO}(=\velang{S}{Gal}) }[/math] are parallel.

✏️ Example D5.7: symmetrical rotor

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The homogeneous rigid body consists of two identical triangular plates. Planar figure in the 12 plane: by property 1, direction 3 is a DPI, and:[math]\displaystyle{ \mathrm{I}_{33}=\mathrm{I}_{11}+ \mathrm{I}_{22} }[/math]. [math]\displaystyle{ \Bigr[\mathrm{II}(\Os) \Bigr]_\mathrm{B}= \matriz{\mathrm{I}_{11}}{\mathrm{I}_{12}}{0}{\mathrm{I}_{12}}{\mathrm{I}_{22}}{0}{0}{0}{\mathrm{I}_{11}+\mathrm{I}_{22}} }[/math] [math]\displaystyle{ \mathrm{I}_{33}(\Os)=\mathrm{I}_{11}(\Os)+\mathrm{I}_{22}(\Os), \mathrm{I}_{13}(\Os)=\mathrm{I}_{23}(\Os)=0. }[/math]

Concerning the inertia moments about axes 1 and 2, it is easy to see that they are equal: the distances to axis 1 and axis 2 of the dm of the triangle located in the lower right quadrant ([math]\displaystyle{ \mathrm{dm}(\Ps) }[/math]) are equal to the distances to axis 2 and axis 1, respectively, of the dm of the triangle located in the lower left quadrant ([math]\displaystyle{ \mathrm{dm}(\Ps') }[/math]):

[math]\displaystyle{ \left.\begin{array}{l} \delta_1(\Ps)=\delta_2\left(\Ps'\right) \\ \delta_2(\Ps)=\delta_1\left(\Ps'\right) \end{array}\right\} \Rightarrow \Is_{11}(\Os)=\Is_{22}(\Os) \equiv \Is }[/math] On the other hand: [math]\displaystyle{ \left.\begin{array}{l} \xs_1(\Ps)=-\xs_2\left(\Ps'\right) \\ \xs_2(\Ps)=\xs_1\left(\Ps'\right) \end{array}\right\} \Rightarrow \Is_{12}(\Os)=0 }[/math]

Finally:

[math]\displaystyle{ \Bigr[\mathrm{II}(\Os) \Bigr]_\mathrm{B}= \matriz{\mathrm{I}_{11}}{0}{0}{0}{\mathrm{I}_{22}}{0}{0}{0}{\mathrm{I}_{11}+\mathrm{I}_{22}}= \matriz{\mathrm{I}}{0}{0}{0}{\mathrm{I}}{0}{0}{0}{2\mathrm{I}} }[/math]

It is a symmetrical rotor (property 4). Therefore, the inertia tensor at [math]\displaystyle{ \Os }[/math] is invariant under rotation of the vector basis about axis 3: [math]\displaystyle{ \Bigr[\mathrm{II}(\Os) \Bigr]_{1'2'3'}=\Bigr[\mathrm{II}(\Os) \Bigr]_{123} }[/math]

The qualitative aspect of [math]\displaystyle{ \Bigr[\mathrm{II}(\Os) \Bigr]_\mathrm{B} }[/math] shows that [math]\displaystyle{ \overline{\mathbf{H}}_\mathrm{RTO}(\Os) }[/math] is parallel to [math]\displaystyle{ \velang{S}{RTO} }[/math] when that angular velocity is contained in the 12 plane or is parallel to direction 3:

[math]\displaystyle{ \braq{\overline{\mathbf{H}}_{\text {RTO }}(\Os)}{B}=\Bigr[\mathrm{II}(\Os) \Bigr]_\mathrm{B} \vector{\Omega_1}{\Omega_2}{0}=\vector{\Is \Omega_1}{\Is \Omega_2}{0} \quad , \quad \braq{\overline{\mathbf{H}}_{\text {RTO }}(\Os)}{B}=\Bigr[\mathrm{II}(\Os) \Bigr]_\mathrm{B} \vector{0}{0}{\Omega_3}=\vector{0}{0}{2\Is \Omega_3} }[/math].

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Property 5: When a rigid body has three or more equal moments of inertia in the same plane ij for a point [math]\displaystyle{ \Os }[/math], it is also a symmetrical rotor for [math]\displaystyle{ \Os }[/math]:[math]\displaystyle{ \Bigr[\mathrm{II}(\Os) \Bigr]_\mathrm{ijk}=\Bigr[\mathrm{II}(\Os) \Bigr]_\mathrm{i'j'k} }[/math]. The proof is longer than that of property 4, and is omitted.

✏️ Example D5.8: symmetrical rotor

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The rigid body consists of three identical homogeneous hexagonal plates. planar figure in the 12 plane: by property 1, direction 3 is a DPI, and [math]\displaystyle{ \mathrm{I}_{33}=\mathrm{I}_{11}+ \mathrm{I}_{22} }[/math]. [math]\displaystyle{ \Bigr[\mathrm{II}(\Gs) \Bigr]_\mathrm{B}= \matriz{\mathrm{I}_{11}}{\mathrm{I}_{12}}{0}{\mathrm{I}_{12}}{\mathrm{I}_{22}}{0}{0}{0}{\mathrm{I}_{11}+\mathrm{I}_{22}} }[/math]

The rigid body has no planes of symmetry, so it is not easy to see if the inertia product [math]\displaystyle{ \Is_{12} }[/math] is zero or non-zero. It is also not easy to assess which of the two inertia moments [math]\displaystyle{ \Is_{11},\Is_{22} }[/math] is higher. But there are three coplanar axes that generate the same mass distribution on each side, and therefore the inertia moments with respect to those axes are equal.

By property 5, it is a symmetrical rotor. Therefore, all directions in the 12 plane through [math]\displaystyle{ \Gs }[/math] are principal directions with the same inertia moment.

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Property 6: When the three principal inertia moments (about orthogonal directions) of a rigid body are equal for a point [math]\displaystyle{ \Os }[/math], its inertia tensor at [math]\displaystyle{ \Os }[/math] does not depend on the vector basis: [math]\displaystyle{ \Bigr[\mathrm{II}(\Os) \Bigr]_\mathrm{ijk}=\Bigr[\mathrm{II}(\Os) \Bigr]_\mathrm{i'j'k'} }[/math]. .The rigid body is a spherical rotor for point [math]\displaystyle{ \Os }[/math], and the angular momentum at [math]\displaystyle{ \Os }[/math] and the angular velocity are always parallel: [math]\displaystyle{ \overline{\mathbf{H}}_\mathrm{RTO}(\Os) \parallel \velang{S}{RTO}(=\velang{S}{Gal}). }[/math]

✏️ Example D5.9: spherical rotor

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The solid is formed by a homogeneous ring, with mass 2m, and a particle [math]\displaystyle{ \Ps }[/math] with mass m. The bars that join these elements have negligible mass. The inertia tensor at [math]\displaystyle{ \mathbf{C} }[/math] is the sum of two tensors: [math]\displaystyle{ \mathrm{II}(\mathbf{C})=\mathrm{II}^{\mathrm{part}}(\mathbf{C})+\mathrm{II}^{\mathrm{anella}}(\mathbf{C}) }[/math]. That of the particle is straightforward: [math]\displaystyle{ \left.\begin{array}{l} \delta_1(\Ps)=\delta_2\left(\Ps\right), \delta_3(\Ps)=0\\ \xs_1(\Ps)=\xs_2\left(\Ps\right)=0 \end{array}\right\} \Rightarrow \Bigr[\mathrm{II}^\mathrm{part}(\Cs) \Bigr]_\mathrm{B}= \matriz{\mathrm{I}_\mathrm{p}}{0}{0}{0}{\mathrm{I}_\mathrm{p}}{0}{0}{0}{0} }[/math]

Since the ring is a planar rigid body and it is symmetrical about the axes 1 and 2, property 1 and property 3 lead to:

[math]\displaystyle{ \Bigr[\mathrm{II}_\mathrm{anella}(\Cs) \Bigr]_\mathrm{B}= \matriz{\mathrm{I}_\mathrm{a}}{0}{0}{0}{\mathrm{I}_\mathrm{a}}{0}{0}{0}{2\mathrm{I}_\mathrm{a}} }[/math].

The rigid body is a symmetrical rotor for point [math]\displaystyle{ \Cs }[/math] since two principal moments are equal:

[math]\displaystyle{ \Bigr[\mathrm{II}(\Cs) \Bigr]_\mathrm{B}= \matriz{\mathrm{I}_\mathrm{p}}{0}{0}{0}{\mathrm{I}_\mathrm{p}}{0}{0}{0}{0} + \matriz{\mathrm{I}_\mathrm{a}}{0}{0}{0}{\mathrm{I}_\mathrm{a}}{0}{0}{0}{2\mathrm{I}_\mathrm{a}}= \matriz{\mathrm{I}_\mathrm{p} + \mathrm{I}_\mathrm{a}}{0}{0}{0}{\mathrm{I}_\mathrm{p} + \mathrm{I}_\mathrm{a}}{0}{0}{0}{2\mathrm{I}_\mathrm{a}}. }[/math]

The quantitative tensor can be found without need of the table:

[math]\displaystyle{ \left.\begin{array}{l} \delta_1(\Ps)=\delta_2\left(\Ps\right)=\Rs\quad \Rightarrow \quad \mathrm{I}_\mathrm{p}=\ms\Rs^2\\ \delta_3(\mathrm{dm} \in \mathrm{anella}) \quad \Rightarrow \quad 2\mathrm{I}_\mathrm{a}=2\ms\Rs^2 \end{array}\right\} \quad \Rightarrow \quad \Bigr[\mathrm{II}(\Cs) \Bigr]_\mathrm{B}= \matriz{2}{0}{0}{0}{2}{0}{0}{0}{2} \ms\Rs^2 }[/math]

It is a spherical rotor, so [math]\displaystyle{ \overline{\mathbf{H}}_ \mathrm{RTC}(\Cs) }[/math] is always parallel to [math]\displaystyle{ \velang{S}{T} }[/math] : [math]\displaystyle{ \overline{\mathbf{H}}_ \mathrm{RTC}(\Cs)=\Is\Is(\Cs) \velang{S}{T}=2\ms\Rs^2\velang{S}{T} }[/math].

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D5.5 Change of vector basis

The inertia tensor of a rigid body in a vector basis B and for a point [math]\displaystyle{ \Ps }[/math] or for a point [math]\displaystyle{ \Qs }[/math] do not have the same expression: [math]\displaystyle{ \Is\Is(\Ps) \neq \Is\Is(\Qs) }[/math] . The relationship between the two can be found by means of Steiner's Theorem, which can be proved from the barycentric decomposition of the angular momentum:

[math]\displaystyle{ \Is\Is(\Qs)=\Is\Is(\Gs)+\Is\Is^\oplus(\Qs) }[/math] ,

where [math]\displaystyle{ \Is\Is^\oplus(\Qs) }[/math] is the inertia tensor of a particle with mass equal to that of the rigid body and located at its centre of mass [math]\displaystyle{ \Gs }[/math].

If the theorem is applied to two different points and the equations are combined, we come out with the relationship between [math]\displaystyle{ \Is\Is(\Ps) }[/math] and [math]\displaystyle{ \Is\Is(\Qs) }[/math] :

[math]\displaystyle{ \left.\begin{array}{l} \Is\Is(\Qs)=\Is\Is(\Gs) + \Is\Is^\oplus(\Qs)\\ \Is\Is(\Ps)=\Is\Is(\Gs) + \Is\Is^\oplus(\Ps) \end{array}\right\} \quad \Rightarrow \quad \Is\Is(\Qs)=\Is\Is(\Ps) - \Is\Is^\oplus(\Ps) - \Is\Is^\oplus(\Qs) }[/math]

✏️ Example D5.10: parallel bars

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The solid is formed by two short and one long homogeneous bar, of the same linear density, and connected to a massless frame. We want to find out the qualitative aspect of the inertia tensor at point [math]\displaystyle{ \Os }[/math]. When analysing the inertia tensor, it is useful to think of the long bar as two short bars. The two bars in the upper quadrants have the same inertia moment about axis 2 and axis 3 [math]\displaystyle{ \left(\Is_{22}^{\mathrm{quad.sup.}}=\Is_{33}^{\mathrm{quad.sup.}} \right) }[/math], but the bars in the lower quadrants are farther from axis 2 than from axis 3 [math]\displaystyle{ \left( \Is_{22}^{\mathrm{quad.inf.}}\gt \Is_{33}^{\mathrm{quad.inf.}} \right) }[/math]. Hence: [math]\displaystyle{ \Is_{22}\gt \Is_{33} }[/math].

Since there is no symmetry about axis 3, it is difficult to guess the sign of the inertia product [math]\displaystyle{ \Is_{23} }[/math] . Therefore:

[math]\displaystyle{ \Bigr[\mathrm{II}(\Os) \Bigr]_\mathrm{B}= \matriz{\Is_{22}+\Is_{33}}{0}{0}{0}{\Is_{22}}{\Is_{23}}{0}{\Is_{23}}{\Is_{33}} }[/math].

The [math]\displaystyle{ \Is_{23} }[/math] sign can be deduced easily if tensor [math]\displaystyle{ \Is\Is(\Os) }[/math] is referred to the tensors of the four identical bars to their centres of mass through Steiner's theorem:

[math]\displaystyle{ \Is\Is(\Os)=\sum_{\is=1}^{4}\Is\Is_\is(\Os)=\sum_{\is=1}^{4} \Bigr[\mathrm{II}_\is(\Gs_\is) + \Is\Is_\is^\oplus(\Os) \Bigr] }[/math]

[math]\displaystyle{ \Bigr[\mathrm{II}_\is(\Gs_\is) \Bigr]_\mathrm{B}= \matriz{2\Is}{0}{0}{0}{\Is}{-|\Is_{23}'|}{0}{-|\Is_{23}'|}{\Is} }[/math] ,

[math]\displaystyle{ \sum_{\is=1}^{4} \Bigr[\Is\Is_\is^\oplus(\Os) \Bigr]_\mathrm{B}=\Bigr[\Is\Is_\mathrm{quad.sup.}^\oplus(\Os) \Bigr]_\mathrm{B} + \Bigr[\Is\Is_\mathrm{quad.inf.}^\oplus(\Os) \Bigr]_\mathrm{B}= 2\ms \left( \frac{\Ls}{2} \right)^2 \matriz{2}{0}{0}{0}{1}{0}{0}{0}{1} + 2\ms \left( \frac{\Ls}{2} \right)^2 \matriz{10}{0}{0}{0}{9}{0}{0}{0}{1} = \ms\Ls^2 \matriz{6}{0}{0}{0}{5}{0}{0}{0}{1} }[/math]

[math]\displaystyle{ \Bigr[\mathrm{II}(\Os) \Bigr]_\mathrm{B}= \matriz{8\Is+6\ms\Ls^2}{0}{0}{0}{4\Is + 5\ms\Ls^2}{-4|\Is_{23}'|}{0}{-4|\Is_{23}'|}{4\Is + \ms\Ls^2} }[/math].

✏️ Example D5.11: planar rigid body, qualitative tensor

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The rigid body consists of two identical homogeneous square plates. We want to find the inertia tensor at point [math]\displaystyle{ \Ps }[/math]. The qualitative aspect of the tensor at its centre of mass [math]\displaystyle{ \Gs }[/math] is straightforward (it is a planar figure and the mass is located in the quadrants that contribute with a positive sign to the inertia product):

[math]\displaystyle{ \Bigr[\mathrm{II}(\Gs) \Bigr]_\mathrm{B}= \matriz{\Is}{+|\Is_{12}|}{0}{+|\Is_{12}|}{\Is}{0}{0}{0}{2\Is} }[/math]

The Table can be used to calculated the inertia tensor of a rectangular plate:

[math]\displaystyle{ \Is\Is(\Gs)=\Is\Is_\mathrm{placa.inf.}(\Gs)+\Is\Is_\mathrm{placa.sup} (\Gs) }[/math]

[math]\displaystyle{ \Bigr[\mathrm{II}(\Gs) \Bigr]= 2 \matriz{\frac{1}{3}\ms (2\Ls)^2}{\frac{1}{4}\ms (2\Ls)^2}{0}{\frac{1}{4}\ms (2\Ls)^2}{\frac{1}{3}\ms (2\Ls)^2}{0}{0}{0}{\frac{2}{3}\ms (2\Ls)^2}= \frac{2}{3} \ms\Ls^2 \matriz{4}{3}{0}{3}{4}{0}{0}{0}{8} }[/math]

Now we must move on to point [math]\displaystyle{ \Ps }[/math] with Steiner's theorem:[math]\displaystyle{ \Is\Is(\Ps)=\Is\Is(\Gs)+\Is\Is^\oplus(\Ps) }[/math].

[math]\displaystyle{ \Bigr[\mathrm{II}^\oplus(\Ps) \Bigr]= \matriz{2\ms (2\Ls)^2}{-2\ms (2\Ls)(-2\Ls)}{0}{-2\ms (2\Ls)(-2\Ls)}{2\ms (2\Ls)^2}{0}{0}{0}{4\ms (2\Ls)^2}= 8 \ms\Ls^2 \matriz{1}{1}{0}{1}{1}{0}{0}{0}{2} }[/math]

[math]\displaystyle{ \Bigr[\mathrm{II}(\Ps) \Bigr]=\frac{2}{3} \ms\Ls^2 \matriz{4}{3}{0}{3}{4}{0}{0}{0}{8}+ 8 \ms\Ls^2 \matriz{1}{1}{0}{1}{1}{0}{0}{0}{2}=\frac{2}{3} \ms\Ls^2 \matriz{16}{15}{0}{15}{16}{0}{0}{0}{32} }[/math]

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D5.6 Change of vector basis

An inertia tensor expressed in a vector basis [math]\displaystyle{ \mathrm{B} }[/math] can be transformed to another vector basis [math]\displaystyle{ \mathrm{B}' }[/math] by means of the change of basis matrix [math]\displaystyle{ \mathrm{S} }[/math], whose columns are the unit vectors of the [math]\displaystyle{ \mathrm{B}'(\overline{\mathbf{e}}_{\is'}) }[/math] vector basis projected onto the [math]\displaystyle{ \mathrm{B} }[/math] one:

[math]\displaystyle{ \Bigr[\mathrm{II}(\Ps) \Bigr]_{\mathrm{B}'}=\Bigr[\mathrm{S} \Bigr]^{-1} \Bigr[\mathrm{II}(\Ps) \Bigr]_\mathrm{B} \Bigr[\mathrm{S}\Bigr] \quad , \quad \Bigr[\mathrm{S} \Bigr] = \Bigr[ \braq{\overline{\mathbf{e}}_{1'}}{B} \quad \braq{\overline{\mathbf{e}}_{2'}}{B} \quad \braq{\overline{\mathbf{e}}_{3'}}{B} \Bigr] }[/math].

It is easy to see that [math]\displaystyle{ \Is_{\is'\js'}(\Ps)=\Bigl\{ \overline{\mathbf{e}}_{\is'} \Bigl\}_{\mathrm{B}}^{\Ts} \Bigr[\mathrm{II}(\Ps) \Bigr]_{\mathrm{B}} \braq{\overline{\mathbf{e}}_{\js'}}{B} }[/math] .

✏️ EXAMPLE D5.12: planar rigid body, change of vector basis

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The circular plate is homogeneous, with mass m and radius r. We want to find the inertia moment about axis p-p' through its centre and forming an angle of [math]\displaystyle{ 45^o }[/math] with the plate axis.
The Table contains information on the plate inertia tensor for the vertical and horizontal axes. From that tensor[math]\displaystyle{ \Is_{\ps\ps'}(\Os) }[/math] is readily obtained:

[math]\displaystyle{ \Bigr[\mathrm{II}(\Os) \Bigr]_\mathrm{B}= \frac{1}{4} \ms\rs^2\matriz{1}{0}{0}{0}{1}{0}{0}{0}{2} }[/math] , [math]\displaystyle{ \Is_{\ps\ps'}(\Os)=\Is_{1'1'}=\Bigl\{ \overline{\mathbf{e}}_{1'} \Bigl\}_{\mathrm{B}}^{\Ts} \Bigr[\mathrm{II}(\Os) \Bigr]_{\mathrm{B}} \braq{\overline{\mathbf{e}}_{1'}}{B} }[/math]

[math]\displaystyle{ \Is_{\ps\ps'}(\Os)=\frac{1}{\sqrt{2}} \{ 0 \quad 1 \quad 1\} \frac{1}{4} \ms \rs^2 \matriz{1}{0}{0}{0}{1}{0}{0}{0}{2} \frac{1}{\sqrt{2}} \vector{0}{1}{1}=\frac{3}{8} \ms\rs^2 }[/math]

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D5.7 Table of centers of mass and inertia tensors of homogeneous rigid bodies

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