wzy75 said:
Isn't the absolute angular velocity of the second frame (please see the figure) w1+w2?
Yes and no. Those two vectors ω
1 and ω
2 are not expressed in the same reference frames. One of them has to be transformed so that they can be added. Angular velocities are typically expressed in rotating frame coordinates. Three reasons:
- The inertia tensor of a rigid body is constant in body frame coordinates. It is a nasty, time varying beast when expressed in inertial coordinates.
- One standard way to measure the angular velocity of a rotating body is to attach a set gyros to that body. Those measurements will naturally be in rotating frame coordinates.
- Last but not least, expressing angular velocity in rotating frame coordinates is the convention.
So, rather than saying
<br />
\boldsymbol{\omega} = \boldsymbol{\omega}_1 + \boldsymbol{\omega}_2<br />
it is better to say
<br />
\boldsymbol{\omega} =<br />
\boldsymbol T_{1:2}\,\boldsymbol{\omega}_1 +<br />
\boldsymbol{\omega}_2<br />
Here \boldsymbol T_{1:2} is the transformation matrix from frame 1 to frame 2. Differentiating the above yields
<br />
\dot{\boldsymbol{\omega}} =<br />
\boldsymbol T_{1:2}\,\dot{\boldsymbol{\omega}}_1 +<br />
\dot {\boldsymbol T}_{1:2}\,\boldsymbol{\omega}_1 +<br />
\dot{\boldsymbol{\omega}}_2<br />If you want to go against convention and represent ω
1 in inertial coordinates and ω
2 in frame 1 rotating coordinates, the angular velocity of frame 2 with respect to inertial is instead
<br />
\boldsymbol{\omega} =<br />
\boldsymbol{\omega}_1 +<br />
\boldsymbol T_{1:I}\,\boldsymbol{\omega}_2<br />
Differentiating will once again yield a term that involves the time derivative of a transformation matrix. So, what is the time derivative of a transformation matrix? You already have a big clue as to what this is in your expression (1),
<br />
\dot{\mathbf v} = \dot{\mathbf v}' + \boldsymbol{\omega}\times\mathbf v<br />
At the undergraduate level the derivation of the above is typically involves some rather kludgy hand waving. Let's do this with less hand waving (there still will be some hand waving at the end). Suppose
q is some vector quantity. This can be expressed in two different coordinate systems, call them
A and
B. The relationship between these different representations of this vector is
<br />
\mathbf q_A = \boldsymbol T_{B:A}\, \mathbf q_B<br />
Using the chain rule to differentiating with respect to time,
<br />
\dot{\mathbf q}_A =<br />
\boldsymbol T_{B:A}\, \dot{\mathbf q}_B +<br />
\dot {\boldsymbol T}_{B:A}\, \mathbf q_B<br />
The components of the time derivatives \dot{\mathbf q}_A and \dot{\mathbf q}_B are the component-by-component time derivatives of the vector
q expressed in frames
A and
B. In other words, these are time derivatives of the vector
q as observed in and expressed in frames
A and
B respectively. All we are left with is the time derivative of the transformation matrix.
The derivation of that is a two or three page mathout (being blinded by too much whiteness is called
whiteout; being blinded by too much math is called
mathout). So, without derivation,
<br />
\begin{aligned}<br />
\dot T_{B:A}<br />
&= \boldsymbol T_{B:A} \boldsymbol S(\boldsymbol{\omega}_{A:B,B}) \\<br />
&= \boldsymbol S(\boldsymbol{\omega}_{A:B,A}) \boldsymbol T_{B:A} \\<br />
\dot T_{A:B}<br />
&= -\,\boldsymbol T_{A:B} \boldsymbol S(\boldsymbol{\omega}_{A:B,A}) \\<br />
&= -\boldsymbol S(\boldsymbol{\omega}_{A:B,B}) \boldsymbol T_{A:B}<br />
\end{aligned}<br />
Here,
S(ω) is the skew symmetric cross product matrix generated from
ω. This matrix let's one write the cross product in matrix form:
<br />
\boldsymbol{\omega} \times \mathbf q =<br />
\boldsymbol S(\boldsymbol{\omega}) \mathbf q<br />
I was very explicit with coordinate systems in the above. For example, \boldsymbol{\omega}_{A:B,A} is the angular velocity of frame B with respect to frame A, expressed in frame A coordinates.
Going against convention and expressing angular velocity in parent frame coordinates leads to (being sloppy with coordinates here)
<br />
\dot{\boldsymbol{\omega}} =<br />
\dot{\boldsymbol{\omega}}_1 +<br />
\dot{\boldsymbol{\omega}}_2 +<br />
\boldsymbol{\omega}_1 \times \boldsymbol{\omega}_2<br />
This is equivalent your equation (3). Going with convention will lead to
<br />
\dot{\boldsymbol{\omega}} =<br />
\dot{\boldsymbol{\omega}}_1 +<br />
\dot{\boldsymbol{\omega}}_2 -<br />
\boldsymbol{\omega}_2 \times \boldsymbol{\omega}_1<br />
This is what your equation (4) should have been. Note that this result is consistent with the above.
