每日彙整: 2018-04-10

樹莓派、樹莓派之學習、樹莓派之教育

STEM 隨筆︰古典力學︰運動學【二．六．五】

當我們閱讀下面

Rigid body

If a mechanical system is constrained to move parallel to a fixed plane, then the rotation of a body in the system occurs around an axis k̂ perpendicular to this plane. In this case, the moment of inertia of the mass in this system is a scalar known as the polar moment of inertia. The definition of the polar moment of inertia can be obtained by considering momentum, kinetic energy and Newton’s laws for the planar movement of a rigid system of particles.[14][17][23][24]

If a system of n particles, Pi, i = 1, …, n, are assembled into a rigid body, then the momentum of the system can be written in terms of positions relative to a reference point R, and absolute velocities vi

\displaystyle {\begin{aligned}\Delta \mathbf {r} _{i}&=\mathbf {r} _{i}-\mathbf {R} ,\\\mathbf {v} _{i}&={\boldsymbol {\omega }}\times \left(\mathbf {r} _{i}-\mathbf {R} \right)+\mathbf {V} ={\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i}+\mathbf {V} ,\end{aligned}}

where ω is the angular velocity of the system and V is the velocity of R.

For planar movement the angular velocity vector is directed along the unit vector k which is perpendicular to the plane of movement. Introduce the unit vectors ei from the reference point R to a point ri , and the unit vector t̂i = k̂ × êi so

\displaystyle {\begin{aligned}\mathbf {\hat {e}} _{i}&={\frac {\Delta \mathbf {r} _{i}}{\Delta r_{i}}},\quad \mathbf {\hat {k}} ={\frac {\boldsymbol {\omega }}{\omega }},\quad \mathbf {\hat {t}} _{i}=\mathbf {\hat {k}} \times \mathbf {\hat {e}} _{i},\\\mathbf {v} _{i}&={\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i}+\mathbf {V} =\omega \mathbf {\hat {k}} \times \Delta r_{i}\mathbf {\hat {e}} _{i}+\mathbf {V} =\omega \,\Delta r_{i}\mathbf {\hat {t}} _{i}+\mathbf {V} \end{aligned}}

This defines the relative position vector and the velocity vector for the rigid system of the particles moving in a plane.

Note on the cross product: When a body moves parallel to a ground plane, the trajectories of all the points in the body lie in planes parallel to this ground plane. This means that any rotation that the body undergoes must be around an axis perpendicular to this plane. Planar movement is often presented as projected onto this ground plane so that the axis of rotation appears as a point. In this case, the angular velocity and angular acceleration of the body are scalars and the fact that they are vectors along the rotation axis is ignored. This is usually preferred for introductions to the topic. But in the case of moment of inertia, the combination of mass and geometry benefits from the geometric properties of the cross product. For this reason, in this section on planar movement the angular velocity and accelerations of the body are vectors perpendicular to the ground plane, and the cross product operations are the same as used for the study of spatial rigid body movement.

 

文本，是否會懷疑︰這等同於

剛體

在物理學裏，理想剛體（rigid body）是一種有限尺寸，可以忽略形變的固體。不論是否感受到外力，在剛體內部，質點與質點之間的距離都不會改變。這種理想模型適用條件是，運動過程比固體中的彈性波的傳播要緩慢得多。根據相對論，這種物體不可能實際存在，但物體通常可以假定為完美剛體，前提是必須滿足運動速度遠小於光速的條件。

 

的『定義』嗎？

雖未明言，該文本實以任意『質點』 P_i 之『速度』都可用

\mathbf {v} _{i} ={\boldsymbol {\omega }}\times \left(\mathbf {r} _{i}-\mathbf {R} \right)+\mathbf {V}

表示以描述『剛體』。

因此任兩不同『質點』 P_j, \ P_k 滿足

(\mathbf{r}_j - \mathbf{r}_k) \cdot ( \mathbf{v}_j - \mathbf{v}_k) =0 。

也就是說︰

\frac{d}{dt} \left( (\mathbf{r}_j - \mathbf{r}_k) \cdot (\mathbf{r}_j - \mathbf{r}_k) \right) =0 ，

距離不變也！

反向推演，難到不然乎？？

故知積累『質點系統』之『角動量』

\mathbf {L_i} = m_{i}\,\Delta \mathbf {r} _{i}\times \mathbf {v} _{i}

，自得出『慣性張量』的『定義』呦！！

Motion in space of a rigid body, and the inertia matrix

The scalar moments of inertia appear as elements in a matrix when a system of particles is assembled into a rigid body that moves in three-dimensional space. This inertia matrix appears in the calculation of the angular momentum, kinetic energy and resultant torque of the rigid system of particles.[3][4][5][6][25]

For analysis of a spinning top, see Precession § Classical (Newtonian), and Euler’s equations (rigid body dynamics).

Let the system of particles Pi, i = 1, …, n be located at the coordinates ri with velocities vi relative to a fixed reference frame. For a (possibly moving) reference point R, the relative positions are

\displaystyle \Delta \mathbf {r} _{i}=\mathbf {r} _{i}-\mathbf {R}

and the (absolute) velocities are
\displaystyle \mathbf {v} _{i}={\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i}+\mathbf {V} _{\mathbf {R} }
where ω is the angular velocity of the system, and VR is the velocity of R.

Angular momentum

Note that the cross product can be equivalently written as matrix multiplication by combining the first operand and the operator into a, skew-symmetric, matrix, [b], constructed from the components of b = (bx, by, bz):

\displaystyle {\begin{aligned}\mathbf {b} \times \mathbf {y} &\equiv \lbrack \mathbf {b} \rbrack \mathbf {y} \\\lbrack \mathbf {b} \rbrack &\equiv {\begin{bmatrix}0&-b_{z}&b_{y}\\b_{z}&0&-b_{x}\\-b_{y}&b_{x}&0\end{bmatrix}}.\end{aligned}}

The inertia matrix is constructed by considering the angular momentum, with the reference point R of the body chosen to be the centre of mass C:[3][6]
\displaystyle {\begin{aligned}\mathbf {L} &=\left(\sum _{i=1}^{n}m_{i}\,\Delta \mathbf {r} _{i}\times \mathbf {v} _{i}\right)\\&=\left(\sum _{i=1}^{n}m_{i}\,\Delta \mathbf {r} _{i}\times \left({\boldsymbol {\omega }}\times \Delta \mathbf {r} _{i}+\mathbf {V} _{\mathbf {R} }\right)\right)\\&=\left(-\sum _{i=1}^{n}m_{i}\,\Delta \mathbf {r} _{i}\times \left(\Delta \mathbf {r} _{i}\times {\boldsymbol {\omega }}\right)\right)+\left(\sum _{i=1}^{n}m_{i}\,\Delta \mathbf {r} _{i}\times \mathbf {V} _{\mathbf {R} }\right),\end{aligned}}
where the terms containing VR (= C) sum to zero by the definition of centre of mass.

Then, the skew-symmetric matrix [Δri] obtained from the relative position vector Δri = ri − C, can be used to define,

\displaystyle \mathbf {L} =\left(-\sum _{i=1}^{n}m_{i}\left[\Delta \mathbf {r} _{i}\right]^{2}\right){\boldsymbol {\omega }}=\mathbf {I} _{\mathbf {C} }{\boldsymbol {\omega }},

where IC defined by
\displaystyle \mathbf {I} _{\mathbf {C} }=-\left(\sum _{i=1}^{n}m_{i}\left[\Delta \mathbf {r} _{i}\right]^{2}\right),
is the symmetric inertia matrix of the rigid system of particles measured relative to the centre of mass C.

 

自可解讀

Inertia tensor

The inertia matrix is often described as the inertia tensor, which consists of the same moments of inertia and products of inertia about the three coordinate axes.[6][23] The inertia tensor is constructed from the nine component tensors, (the symbol \displaystyle \otimes is the tensor product)

\displaystyle \mathbf {e} _{i}\otimes \mathbf {e} _{j},\quad i,j=1,2,3,

where ei, i = 1, 2, 3 are the three orthogonal unit vectors defining the inertial frame in which the body moves. Using this basis the inertia tensor is given by
\displaystyle \mathbf {I} =\left(\sum _{i=1}^{3}\sum _{j=1}^{3}I_{ij}\mathbf {e} _{i}\otimes \mathbf {e} _{j}\right).
This tensor is of degree two because the component tensors are each constructed from two basis vectors. In this form the inertia tensor is also called the inertia binor.

For a rigid system of particles Pk, k = 1, …, N each of mass mk with position coordinates rk = (xk, yk, zk), the inertia tensor is given by

\displaystyle \mathbf {I} =\sum _{k=1}^{N}m_{k}\left(\left(\mathbf {r} _{k}\cdot \mathbf {r} _{k}\right)\mathbf {E} -\mathbf {r} _{k}\otimes \mathbf {r} _{k}\right),

where E is the identity tensor
\displaystyle \mathbf {E} =\mathbf {e} _{1}\otimes \mathbf {e} _{1}+\mathbf {e} _{2}\otimes \mathbf {e} _{2}+\mathbf {e} _{3}\otimes \mathbf {e} _{3}.
In this case, the components of the inertia tensor are given by
\displaystyle {\begin{aligned}I_{11}=I_{xx}&=\left(\sum _{k=1}^{N}m_{k}\left(y_{k}^{2}+z_{k}^{2}\right)\right),\\I_{22}=I_{yy}&=\left(\sum _{k=1}^{N}m_{k}\left(x_{k}^{2}+z_{k}^{2}\right)\right),\\I_{33}=I_{zz}&=\left(\sum _{k=1}^{N}m_{k}\left(x_{k}^{2}+y_{k}^{2}\right)\right),\\I_{12}=I_{21}=I_{xy}&=-\left(\sum _{k=1}^{N}m_{k}x_{k}y_{k}\right),\\I_{13}=I_{31}=I_{xz}&=-\left(\sum _{k=1}^{N}m_{k}x_{k}z_{k}\right),\\I_{23}=I_{32}=I_{yz}&=-\left(\sum _{k=1}^{N}m_{k}y_{k}z_{k}\right).\end{aligned}}
The inertia tensor for a continuous body is given by
\displaystyle \mathbf {I} =\iiint \limits _{Q}\rho (\mathbf {r} )\left(\left(\mathbf {r} \cdot \mathbf {r} \right)\mathbf {E} -\mathbf {r} \otimes \mathbf {r} \right)\,\mathrm {d} V,
where r defines the coordinates of a point in the body and ρ(r) is the mass density at that point. The integral is taken over the volume V of the body. The inertia tensor is symmetric because Iij = Iji.

Alternatively it can also be written in terms of the hat operator[clarification needed] as:

\displaystyle \mathbf {I} =\iiint \limits _{Q}\rho (\mathbf {r} )({\hat {r}})^{2}\,\mathrm {d} V,

The inertia tensor can be used in the same way as the inertia matrix to compute the scalar moment of inertia about an arbitrary axis in the direction n,
\displaystyle I_{n}=\mathbf {n} \cdot \mathbf {I} \cdot \mathbf {n} ,
where the dot product is taken with the corresponding elements in the component tensors. A product of inertia term such as I12 is obtained by the computation
\displaystyle I_{12}=\mathbf {e} _{1}\cdot \mathbf {I} \cdot \mathbf {e} _{2},
and can be interpreted as the moment of inertia around the x-axis when the object rotates around the y-axis.

The components of tensors of degree two can be assembled into a matrix. For the inertia tensor this matrix is given by,

\displaystyle \mathbf {I} ={\begin{bmatrix}I_{11}&I_{12}&I_{13}\\I_{21}&I_{22}&I_{23}\\I_{31}&I_{32}&I_{33}\end{bmatrix}}={\begin{bmatrix}I_{xx}&I_{xy}&I_{xz}\\I_{yx}&I_{yy}&I_{yz}\\I_{zx}&I_{zy}&I_{zz}\end{bmatrix}}.

It is common in rigid body mechanics to use notation that explicitly identifies the x, y, and z axes, such as Ixx and Ixy, for the components of the inertia tensor.

 

文本中『一般』之於『限定』哩◎