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In Taylor's Classical Mechanics, it is said that the equations are generally difficult to use because the components $M_1,M_2$ and $M_3$ of the applied torque as seen in the rotating body frame are complicated functions of time. My question is why would the torques $M_1, M_2$ and $M_3$ be in the rotating body reference frame? The Euler equations are derived using: $$ \frac{dL}{dt}_{lab} = M$$ and then putting in: $$ \frac{dL}{dt}_{lab} = \frac{dL}{dt}_{body} + \omega \times L$$ so shouldn't the torque be with respect to the lab frame? What am I missing here?

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My question is why would the torques $M_1, M_2$ and $M_3$ and be in the rotating body reference frame?

They are not. As you noted further, they are the applied torque with respect to the fixed space/lab frame.

These set of equations describe the dynamics as seen in frame fixed in the body or the rotating frame. Unless you have a symmetrical object (such as sphere), in general, it is difficult to keep track principal axes as they vary with time. The principal axis, will not remain principal axis,and the inertia ”seen” in this coordinate system will vary with time.

Since the axes are fixed to the body, we are committed to follow the body as it rotates in order to use these equations and for obtaining the solutions. We need to come up with equations of motion relative to a nonrotating space frame, and before we can do that we need a set of coordinates that specifiy the orientation of our body relative to such a frame -- Euler angles.

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  • $\begingroup$ Thank you, but it is still not clear. $\omega$ for example, is used in the $\omega \times L$ term as $\omega$ in the body frame, but in the $L$ term it is used with respect to the principal axes. Isn't $\omega$ different as viewed in the body and rotating frames? The same with the torque. In the textbook it is said the components $M_1,M_2$ and $M_3$ of the applied torque as seen in the rotating body frame are hard to find, but you said they are in the space frame. $\endgroup$
    – Darkenin
    Aug 24 '20 at 8:41
  • $\begingroup$ 1. All the vectors are resolved with respect to principal axes. The subscripts $1, 2$ and $3$ correspond to the three orthogonal principal axes. 2. $\omega$ will be different in the space frame and body frame. Body frame and rotating frame are the same. 3. The torque is also resolved into three components along the three principal axes which are time-dependent. That can be done right? Any complete basis can be used to resolve a vector. $\endgroup$ Aug 24 '20 at 9:01
  • $\begingroup$ That wasn't my question, let me explain more clearly. We first say $\frac{dL}{dt} = M$, all in space frame. Then we make a substitution and in the end it is said the $M$ is in the body frame, but the equation originally referred to the space frame. $\endgroup$
    – Darkenin
    Aug 24 '20 at 13:13
  • $\begingroup$ We are not saying $M$ is in body frame, we are just saying it has a complicated and time-dependent form when seen in body frame. That substitution is to connect the rotating frame of reference with the fixed/space/lab frame. The equation $\frac{dL}{dt} = M$ holds only in fixed frame. But we know the angular velocities along principal axes which is attached to the body (rotating frame). $\endgroup$ Aug 24 '20 at 13:24
  • $\begingroup$ It is referred to as being expressed in the rotating body reference frame everywhere I see it. Is it a subtlety of change in coordinate system vs change in reference frame? $\endgroup$
    – Darkenin
    Aug 24 '20 at 13:50

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