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In "Analytical Mechanics" by N. A. Lemos, in page 99 the author determines the time derivative relation between an inertial frame $\Sigma$ an an non-inertial frame $\Sigma'$ fixed in a rigid body with angular velocity $\mathbf{\omega}$ around its origin $O$, such that $$ \left(\frac{d}{d t}\right)_{\text {inertial}}=\left(\frac{d}{d t}\right)_{\text {body}}+\omega \times$$ Also in this book, on page 100, the author is trying to prove the uniqueness of the angular velocity of he body, and he considers two frames $\Sigma$ and $\Sigma'$, the latter with angular velocity $\omega _1$, such that an arbitrary point $P$ of the body can be represented by the vector $\mathbf{r}$ and also be represented by the sum of vectors $\mathbf{r_1}$ and $\mathbf{R}$, where $\mathbf{R}$ is the $\Sigma'$ origin position with respect to $\Sigma$ and $\mathbf{r_1}$ is the point $P$ position with respect to $\Sigma'$'s origin, such that $\mathbf{r} = \mathbf{R} + \mathbf{r_1}$. The author states that

$$ \left(\frac{d \mathbf{r}}{d t}\right)_{\Sigma}=\left(\frac{d \mathbf{R}}{d t}\right)_{\Sigma}+\left(\frac{d \mathbf{r_1}}{d t}\right)_{\Sigma}=\left(\frac{d \mathbf{R}}{d t}\right)_{\Sigma}+\omega_{1} \times \mathbf{r_1}$$

which is obviously correct in my conception, but when I try to apply the time derivative relation for non inertial systems, I obtain

$$\left(\frac{d \mathbf{r}}{d t}\right)_{\Sigma}=\left(\frac{d( \mathbf{R} + \mathbf{r_1})}{d t}\right)_{\Sigma'}+ \mathbf{\omega_1} \times (\mathbf{R} + \mathbf{r_1}) = \left(\frac{d \mathbf{\mathbf{R}}}{d t}\right)_{\Sigma'} + \omega_1 \times (\mathbf{\mathbf{R} + \mathbf{r_1}}) $$

which is clearly different from the last equation. Where is my mistake?

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you can get the correct answer if you use this notations:

  • $(\vec a)_B$ mean that the components of the vector a are given in B- frame
  • $(\vec {\dot a})_B$ mean that the time derivative of the vector components
  • ${_B^O}\,S$ is the transformation matrix between B-frame and O-frame (initial frame)
  • ${_O^B}\,S\,{_B^O}\,S=I_3$ unity matrix

thus

  • ${_B^O}\,S(\vec {\dot a})_B= (\vec {\dot a}_B)_O$, the time derivative is taken in B frame but the components of the result are in O frame

your problem:

The components of the vector r are given in O frame and you want to take the time derivative in B frame, so first transform the components to B frame

$$\vec{r}_B={_O^B}\,S\,(\vec{r})_O$$

the time derivative is:

$$\vec{\dot r}_B={_O^B}\,S\,(\vec{\dot r})_O+ {_O^B}\,\dot S\,(\vec{ r})_O\tag 1$$

with : $${_O^B}\,\dot S={_O^B}\,S\,\tilde{\omega}_O$$

and $$\vec{\omega}\times \vec r=\tilde{\omega}\,\vec r$$

thus :

$$\vec{\dot r}_B={_O^B}\,S\,(\vec{\dot r})_O+ {_O^B}\,S\, (\vec \omega_O\times\,\vec{ r}_O)\tag 2$$

multiply equation (2) from the left with ${_B^O}\,S$

$${_B^O}\,S\,\vec{\dot r}_B=\vec{\dot r}_O+ \vec\omega_O\times\,\vec{ r}_O$$

$$\boxed{(\vec{\dot r}_B)_O=\vec{\dot r}_O+ \vec\omega_O\times\,\vec{ r}_O}$$

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