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


  • ${_B^O}\,S(\vec {\dot a})_B= (\vec {\dot a}_B)_O$, 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


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 ${_B^O}\,S$

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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.