# Confusion about relation between inertial an non-inertial reference frames in respect tto the motion of a rigid body

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?

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

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 $${_B^O}\,S$$

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

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