개념-inertia tensor

Posted Apr 22, 2024

1 min read

cheet sheet

1.moment of inertia

\(I = i_{xx}cos^2\alpha + i_{yy}cos^2\beta + i_{zz}cos^2\gamma + 2i_{xy}{cos\alpha}{cos\beta} + 2i_{yz}{cos\beta}{cos\gamma} + 2i_{xz}{cos\alpha}{cos\gamma}\) - 식(1)

2.dyadic product

2.1.dyadic product of two vector a and b is denoted by ab (병치)

2.2.dyadic product of two vector a and b in \(\mathbb{R}^3\) euclidean space is as following form.

\[\mathbf {ab} = a_{1}b_{1}\mathbf {ii} +a_{1}b_{2}\mathbf {ij} +a_{1}b_{3}\mathbf {ik}\] \[a_{2}b_{1}\mathbf {ji} +a_{2}b_{2}\mathbf {jj} +a_{2}b_{3}\mathbf {jk}\] \[a_{3}b_{1}\mathbf {ki} +a_{3}b_{2}\mathbf {kj} +a_{3}b_{3}\mathbf {kk}\]

3.moment of inertia tensor

\[\mathbf{I} = \hat{n}^T \mathbf {I} \hat{n}\]

such that

\[\mathbf{I} = {\begin{bmatrix}I_{11}&I_{12}&I_{13} \\ I_{21}&I_{22}&I_{23} \\ I_{31}&I_{32}&I_{33}\end{bmatrix}}\] \[\hat{n} = \hat{i}{cos\alpha} + \hat{j}{cos\beta} + \hat{k}{cos\gamma}\]

4.angular momentum

\[\vec{L} = (\sum (m_ir_i^2) - \sum m_i \vec{r_i} (\vec{r_i})) \cdot \vec{w} = \mathbf{I}\vec{w}\]

여기서, \(\vec{r}_i(\vec{r}_i)\)는 dyadic product이다.

5.kinematics energy

\[T= \frac{1}{2}m_iv_i^2 = \frac{1}{2}\vec{w}\vec{L} = \frac{1}{2}\vec{w}\mathbf{I}\vec{w}^T\]

process

1.moment of inertia

\[\hat{n} = \hat{i}{cos\alpha} + \hat{j}{cos\beta} + \hat{k}{cos\gamma}\] \[|r_i \times \hat{n}|^2 = r_{i\perp}^2 = (y_i^2 + z_i^2)cos^2\alpha + (x_i^2 + z_i^2)cos^2\beta + (x_i^2 + y_i^2)cos^2\gamma - 2x_iy_i{cos\alpha}{cos\beta} 2y_iz_i{cos\beta}{cos\gamma} 2x_iz_i{cos\alpha}{cos\gamma}\]

such that

\(\hat{n}\) unit vector of rotational axis, \(r_i\) is position vector of particle.

\[I = \sum r_{i\perp}^2m_i\]

2.dyadic product

\[\mathbf {ab} =\sum _{j=1}^{N}\sum _{i=1}^{N}a_{i}b_{j}\mathbf {e} _{i}\mathbf {e} _{j}\]

3.moment of inertia tensor

식(1) 참고

4.angular momentum, 5.kinematics energy

reference 참고 (angular momentum, kinematics energy represented as inertia tensor)

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3D Rigid Body Dynamics