Acceleration in a non-inertial reference frome - derevation

Question

The general velocity equation for a point B in on body rotating and translating about point A with respect to the inertial reference frame say 'xyzo' can be expressed as,

$\vec{r_{B/o}} = \vec{r_{A/o}} + \vec{r_{B/A}}$

differentiating the above equation with respect to time

$\vec{v_{B/o}} = \vec{v_{A/o}} + \big(\vec{v_{B/A}}\big)_{\omega = 0} + \omega_{/o} \times \vec{r_{B/A}}$

differentiating again we get,

$\vec{a_{B/o}} = \vec{a_{A/o}} + \big(\vec{a_{B/A}}\big)_{\omega = 0} + \dfrac{d(\omega_{/o} \times \vec{r_{B/A}})}{dt}$

$\vec{a_{B/o}} = \vec{a_{A/o}} + \big(\vec{a_{B/A}}\big)_{\omega = 0} + \dfrac{d(\omega_{/o})}{dt} \times \vec{r_{B/A}} + \omega_{/o} \times \dfrac{r_{B/A}}{dt}$

$\vec{a_{B/o}} = \vec{a_{A/o}} + \big(\vec{a_{B/A}}\big)_{\omega = 0} + \dot{\omega_{/o}} \times \vec{r_{B/A}} + \omega_{/o} \times \big({\vec{v_{B/A}}}_{\omega_{/o} = 0} + \omega_{/o} \times r_{B/A}\big)$

which when expanded gives

$\vec{a_{B/o}} = \vec{a_{A/o}} + \big(\vec{a_{B/A}}\big) + {\alpha} \times \vec{r_{B/A}} + \omega_{/o} \times {\vec{v_{B/A}}}_{\omega_{/o} = 0} + \omega_{/o} \times \big(\omega_{/o} \times r_{B/A}\big)$

The general acceleration equation is given by

$\vec{a_{B/o}} = \vec{a_{A/o}} + \big(\vec{a_{B/A}}\big) + {\alpha} \times \vec{r_{B/A}} + 2(\omega_{/o} \times {\vec{v_{B/A}}}_{\omega_{/o} = 0}) + \omega_{/o} \times \big(\omega_{/o} \times r_{B/A}\big)$

where did the $2(\omega_{/o} \times {\vec{v_{B/A}}}_{\omega_{/o} = 0}) + \omega_{/o}$, 2 come from in the above derivation?

Answer 1

You forgot one term when you derived $\left(\vec{v_{B/ A}}\right)_{\omega_0}$ with respect to the inertial reference frame. The correct would be: $$\frac{d}{dt}\left(\vec{v_{B/ A}}\right)_{\omega_0}=\left(\vec{a_{B/A}}\right)_{\omega_0}+\omega_{/0}\times\left(\vec{v_{B/A}}\right)_{\omega_0}$$ in same way as you did for deriving $\vec{r_{B/A}}$.

Adding something from the comments:

I will call $\mathbf{i},\mathbf{j}$ the vectors of the intertial reference frame and $\hat{\mathbf{i}},\hat{\mathbf{j}}$ the ones of the moving one

(and I'm going to be a bit loose with the notation)

Basically, for every vector you can imagine you have that: $$\mathbf{r}_B = \mathbf{r}_A+\left(\mathbf{r}_B-\mathbf{r}_A\right)$$ or using your notation: $$\mathbf{r}_B = \mathbf{r}_A+\mathbf{r}_{B/A}$$

In principle you could derive $\mathbf{r}_A$ and $\mathbf{r}_{B/A}$ in the same way, but probably $\mathbf{r}_{B/A}$ can be written much more easy in an basis moving with A, i.e. $\hat{\mathbf{i}},\hat{\mathbf{j}}$.

So we have that:

$$\mathbf{r}_A=r_{Ax}(t)\mathbf{i}+r_{Ay}(t)\mathbf{j}$$ and its derivative $$\frac{d\mathbf{r}_A}{dt}=v_{Ax}(t)\mathbf{i}+v_{Ay}(t)\mathbf{j}=\mathbf{v}_{A/o}$$

However, for the case of $\mathbf{r}_{B/A}$ you have that $$\mathbf{r}_{B/A}=r_{B/Ax}(t)\hat{\mathbf{i}}(t)+r_{B/Ay}(t)\hat{\mathbf{j}}(t)$$ You see the difference?, $\hat{\mathbf{i}},\hat{\mathbf{j}}$ also depend on the time because the move along with A, so now if you derive them you have two products: $$\frac{d\mathbf{r}_{B/A}}{dt}=v_{B/Ax}(t)\hat{\mathbf{i}}(t)+v_{B/Ay}(t)\hat{\mathbf{j}}(t)+r_{B/Ax}(t)\frac{d\hat{\mathbf{i}}(t)}{dt}+r_{B/Ay}(t)\frac{d\hat{\mathbf{j}}(t)}{dt}$$ that is $$\frac{d\mathbf{r}_{B/A}}{dt}=\mathbf{v}_{B/A}+\omega\times\mathbf{r}_{B/A}$$ where $$\mathbf{v}_{B/A}=v_{B/Ax}(t)\hat{\mathbf{i}}(t)+v_{B/Ay}(t)\hat{\mathbf{j}}(t)$$ And HERE you can see that $\mathbf{v}_{B/A}$ is again writen in terms of $\hat{\mathbf{i}},\hat{\mathbf{j}}$ son when you derive it for the aceleration you will have: $$\frac{d\mathbf{v}_{B/A}}{dt}=a_{B/Ax}(t)\hat{\mathbf{i}}(t)+a_{B/Ay}(t)\hat{\mathbf{j}}(t)+v_{B/Ax}(t)\frac{d\hat{\mathbf{i}}(t)}{dt}+v_{B/Ay}(t)\frac{d\hat{\mathbf{j}}(t)}{dt}$$ that is $$\frac{d\mathbf{v}_{B/A}}{dt}=\mathbf{a}_{B/A}+\omega\times\mathbf{v}_{B/A}$$

Answer 2

Coriolis and Centrifugal Accelerations

The components of the vector to center of mass (vector R) are given in body coordinates system (index b)

To write the equations of motion we have to calculate the components of vector R in space coordinate system (Index 0) \begin{align*} &\vec{R}_o =S\,\vec{R}_b\\\\ &\text{S is the transformation matrix (rotation matrix) }\\&\text{ between body system and space system}\\ &\text{S can construct from the euler angles }\quad S=S(\alpha(t)\,,\beta(t)\,,\gamma(t))\\\\ &\Rightarrow\\ &\text{Velocity:}\qquad \vec{\dot{R}}_o =\dot{S}\,\vec{R}_b+S\,\dot{R}_b \\\\ &\text{Acceleration:}\qquad \vec{\ddot{R}}_o =\ddot{S}\vec{R}_b+2\,\dot{S}\,\dot{R}_b+S\,\ddot{R}_b\\\\ &\text{With:}\\\\ &\dot{S}=S\,\tilde{\omega}\,\quad,\quad \ddot{S}=\dot{S}\,\tilde{\omega} +S\,\tilde{\dot{\omega}}\\ &\Rightarrow\\\\ &\vec{\ddot{R}}_o =S\,\left[\tilde{\omega}^2\,\vec{R}_b+\tilde{\dot{\omega}}\,\vec{R}_b +2\,\tilde{\omega}\,\dot{\vec{R}}_b+\ddot{\vec{R}}_b\right]\\ &\text{If we multiply this equation from the left with $S^T (\quad S^T\,S=I_3)\quad$ we get}\\ &\text{the acceleration in components of the body system }\\\\ &\vec{\ddot{R}}_b := \tilde{\omega}^2\,\vec{R}_b+\tilde{\dot{\omega}}\,\vec{R}_b +2\,\tilde{\omega}\,\dot{\vec{R}}_b+\ddot{\vec{R}}_b= \vec{\omega}\times\,\left(\vec{\omega}\,\times\,\vec{R}_b\right)+\vec{\dot{\omega}}\times\,\vec{R}_b +2\left(\vec{\omega}\times \,\dot{\vec{R}}_b\right)+\ddot{\vec{R}}_b \end{align*} \begin{align*} &\text{The equations of motion are:} \\\\ &m\,\ddot{R}_o=\sum \,f_o\qquad \Rightarrow\quad\,m\,S^T \,\ddot{R}_o=S^T\,\sum \,f_o=\sum\,f_b\\ &\Rightarrow\\ &m\,\ddot{\vec{R}}_b=\sum\,f_b-m\,\left[\vec{\omega}\times\,\left(\vec{\omega}\,\times\,\vec{R}_b\right)+\vec{\dot{\omega}}\times\,\vec{R}_b +2\left(\vec{\omega}\times \,\dot{\vec{R}}_b\right)\right] \end{align*}

Remarks: \begin{align*} &\text{I use for cross product this notation:} &\vec{\omega}\times \vec{r}=\tilde{\omega}\,\vec{r}= \begin{bmatrix} 0 & -\omega_z & \omega_y \\ \omega_z & 0 & -\omega_x \\ -\omega_y & \omega_x & 0 \\ \end{bmatrix} \begin{bmatrix} r_x \\ r_y \\ r_z \\ \end{bmatrix} \end{align*}