since the bodies moves on an inertial frame and the value of J is fixed in the frame, the J derivative is zero.
I see, I mean the pin is fixed to the frame of the body.
there is the physical interpretation, but it also comes from the geometrical and calculus derivation.
If you write the angular constraint equation, this is.
J . w = b
. Dot product operator
J is the pin
W the angular velocity
b some desired angular velocity function.
Remember we say we solve on the force and torque domain
So, we need to get the angular acceleration from the constraint equation. And that is achieved by taking the time derivative. So, you get
D(j.w) = D(b)
D is the time derivative operator.
We get
D(j) . W + j . D (w) = c
Let us go again over the easy terms.
C is the acceleration that the joint specifies ex (motor, free, fix, etc)
J. D(w) = J . a
That is the angular velocity of the body projected over the pin j.
The last term is where I confused you, It is very simple.
We just need to expand the expression and see what it leads.
To calculate that term, we must know what the value of D(j) is.
And that's where the meaning of the frame of reference is important.
We assume that j is of fix length, so it is invariance in any frame.
But j can change orientation with time, therefore it's time derivative is the
cross product of the pin and the pins angular velocity.
Let us call the pin angular velocity Wp
Now we can write.
D(j). W = (Wp x J) . W
Now we need to know how to get Wp.
And there are at least three common classic cases.
1- inertial frame, one that is not rotating, and not acceleratind
On this frame, the pin angular velocity is the body angular velocity,
simply because the pin is fix to the body.
Wp = w
So, we get
Wp. X J . W = W x J . W
But by the triple product identity
a x b . c= c x a . b
We then get
W x J . W = W x W . J = 0
Because a x a = 0
x is the cross product operator.
2-case. moving on a rotation frame.
Here the pin omega is the angular velocity of the body plus the angular velocity of the frame
wp = w + wf
Wf is the instantaneous angular velocity of the frame and may not be zero
plugging that into the equation and the frame inflict an acceleration of the body.
3-case. the pin is fixed to the frame of reference, example for this are contact joints.
Here D(j) is zero since is constant.
There can be other scenarios like pin that change in length, and things like that but, I stop covering those.