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This paper analyzes in detail the moment M \boldsymbol{M} projected on the moving coordinate systemM and moment of momentumH \boldsymbol{H}The relationship between H.
1. Moment M \boldsymbol{M}M and moment of momentumH \boldsymbol{H}What is H ?
Moment M \boldsymbol{M}M is the total external moment of the rigid body relative to the center of mass of the rigid body, which is related to the force and moment arm that each mass element receives without passing through the center of mass; the
moment of momentumH \boldsymbol{H}H is the total external moment of momentum of the rigid body relative to the center of mass of the rigid body, which is related to the linear velocity and moment arm of each mass element.
2. Moment M \boldsymbol{M}M and moment of momentumH \boldsymbol{H}What is the relation of H ?
1. Coordinate system
Static coordinate system S \mathcal{S}S : Assumed to be an inertial coordinate system;
The moving coordinate system D \mathcal{D}D : The origin coincides with the center of mass of the rigid body and is fixed to the rigid body.
Set the moving coordinate system D \mathcal DD' sxxx、yyy、 z z The unit vectors on the z- axis arei \boldsymbol{i}i、 j \boldsymbol{j} j、 k \boldsymbol{k} k , to describe these three vectors, it needs to be projected on a certain coordinate system. When its projection is in the static coordinate systemS \mathcal SOn S , the mathematical representation isi S \boldsymbol{i}^{\mathcal S}iS、 j S \boldsymbol{j}^{\mathcal S} jS、 k S \boldsymbol{k}^{\mathcal S} kS. _ When its projection is in the moving coordinate systemS \mathcal SOn S , the mathematical representation isi D \boldsymbol{i}^{\mathcal D}iD、 j D \boldsymbol{j}^{\mathcal D} jD、k D \boldsymbol{k}^{\mathcal D}kD。
2. Analysis
Suppose the mass element inside the rigid body is dm \text{d} md m , the radius of the mass center of the rigid body pointing to this mass element isrdm \boldsymbol{r}_{_{\text{d} m}}rdm, then HS \boldsymbol{H}^\mathcal{S}HS 可以表示为
H S = ∫ m r d m S × d r d m S d t d m \boldsymbol{H}^\mathcal{S}=\int_m \boldsymbol{r}^\mathcal{S}_{_{\text{d} m}} \times\frac{\text{d} \boldsymbol{r}^\mathcal{S}_{_{\text{d} m}}}{\text{d}t} \text{d} m HS=∫mrdmS×dtdrdmSdm
已知 H S = h x i S + h y j S + h z k S , H D = h x i D + h y j D + h z k D = [ h x , h y , h z ] T \boldsymbol{H}^\mathcal{S}=h_x \boldsymbol{i}^\mathcal{S}+h_y \boldsymbol{j}^\mathcal{S}+h_z \boldsymbol{k}^\mathcal{S},\quad \boldsymbol{H}^\mathcal{D}=h_x \boldsymbol{i}^\mathcal{D}+h_y \boldsymbol{j}^\mathcal{D}+h_z \boldsymbol{k}^\mathcal{D}=[h_x,h_y,h_z]^T HS=hxiS+hyjS+hzkS,HD=hxiD+hyjD+hzkD=[hx,hy,hz]T r S = r x i S + r y j S + r z k S \boldsymbol{r}^\mathcal{S}=r_x \boldsymbol{i}^\mathcal{S}+r_y \boldsymbol{j}^\mathcal{S}+r_z \boldsymbol{k}^\mathcal{S} rS=rxiS+ryjS+rzkS M S = m x i S + m y j S + m z k S \boldsymbol{M}^\mathcal{S}=m_x \boldsymbol{i}^\mathcal{S}+m_y \boldsymbol{j}^\mathcal{S}+m_z \boldsymbol{k}^\mathcal{S} MS=mxiS+myjS+mzkS ω S = ω xi S + ω yj S + ω zk S \boldsymbol{\omega}^\mathcal{S}=\omega_x \boldsymbol{i}^\mathcal{S}+\omega_y \boldsymbol{j}^ \mathcal{S}+\omega_z \boldsymbol{k}^\mathcal{S}ohS=ohxiS+ohyjS+ohzkS where,ω S \boldsymbol{\omega}^\mathcal{S}ohS represents the action scaleD \mathcal{D}D is relative to the static coordinate systemS \mathcal{S}S is in the static coordinate systemS \mathcal{S}Projection in S.
可以得到 M S = d H S d t = h x ˙ i S + h y ˙ j S + h z ˙ k S + h x d i S d t + h y d j S d t + h z d k S d t \boldsymbol{M}^\mathcal{S}=\frac{\text{d}\boldsymbol{H}^\mathcal{S}}{\text{d}t}=\dot{h_x} \boldsymbol{i}^\mathcal{S}+\dot{h_y} \boldsymbol{j}^\mathcal{S}+\dot{h_z} \boldsymbol{k}^\mathcal{S}+h_x \frac{\text{d}\boldsymbol{i}^\mathcal{S}}{\text{d}t}+h_y \frac{\text{d}\boldsymbol{j}^\mathcal{S}}{\text{d}t}+h_z \frac{\text{d}\boldsymbol{k}^\mathcal{S}}{\text{d}t} MS=dtdHS=hx˙iS+hy˙jS+hz˙kS+hxdtof iS+hydtdjS+hzdtdkS
Note that the two derivation writing methods here only distinguish between vectors and scalars, since both are projected on the static coordinate system S \mathcal{S}S , so the two are computationally the same. If it is projected into the motion coordinate system, it cannot be calculated directly.
According to Poisson's formula ,
di S dt = ω S × i S \frac{\text{d}\boldsymbol{i}^\mathcal{S}}{\text{d}t}=\boldsymbol{\omega }^\mathcal{S}\times\boldsymbol{i}^\mathcal{S}dtof iS=ohS×iS d j S d t = ω S × j S \frac{\text{d}\boldsymbol{j}^\mathcal{S}}{\text{d}t}=\boldsymbol{\omega}^\mathcal{S}\times\boldsymbol{j}^\mathcal{S} dtdjS=ohS×jS d k S d t = ω S × k S \frac{\text{d}\boldsymbol{k}^\mathcal{S}}{\text{d}t}=\boldsymbol{\omega}^\mathcal{S}\times\boldsymbol{k}^\mathcal{S} dtdkS=ohS×kS
则有
M S = d H S d t = h x ˙ i S + h y ˙ j S + h z ˙ k S + ω S × H S \boldsymbol{M}^\mathcal{S}=\frac{\text{d}\boldsymbol{H}^\mathcal{S}}{\text{d}t}=\dot{h_x} \boldsymbol{i}^\mathcal{S}+\dot{h_y} \boldsymbol{j}^\mathcal{S}+\dot{h_z} \boldsymbol{k}^\mathcal{S}+\boldsymbol{\omega}^\mathcal{S}\times\boldsymbol{H}^\mathcal{S} MS=dtdHS=hx˙iS+hy˙jS+hz˙kS+ohS×HS
Then projected to the action frame D \mathcal{D}D,则有
MD = hx ˙ i D + hy ˙ j D + hz ˙ k D + ω D × HD \boldsymbol{M}^\mathcal{D}=\dot{h_x} \boldsymbol{i}^\mathcal {D}+\dot{h_y} \boldsymbol{j}^\mathcal{D}+\dot{h_z} \boldsymbol{k}^\mathcal{D}+\boldsymbol{\omega}^\mathcal{D} \times\boldsymbol{H}^\mathcal{D}MD=hx˙iD+hy˙jD+hz˙kD+ohD×HD即MD = H ˙ D + ω D × HD \boldsymbol{M}^\mathcal{D}={\dot{\boldsymbol{H}}^\mathcal{D}}+\boldsymbol{\omega}^\ mathcal{D}\times\boldsymbol{H}^\mathcal{D}MD=H˙D+ohD×HD Note: These are all projected on the action frame.
Summarize
The above is the moment M \boldsymbol{M} in the moving coordinate systemM and moment of momentumH \boldsymbol{H}For the detailed derivation of the relationship between H , the most important point is to distinguish the dynamic and static of the coordinate system.