Non-Newtonian description of an accelerated object

[VicenteMMOS]

Hello, everyone!

In the first few chapters of Physics 1, there is the description of motion, and the computation of the acceleration vector as being the sum of the rate of change of velocity, with the division of linear velocity squared by the instant radius of curvature:

a = dv/dt + v^2/r , tangential + centripetal acceleration components

The above equation is calculated with an inertial reference frame. Next, I went to the chapter describing the three-dimensional movement of an object, but with a constant angular speed, omega. Here is the description of that problem, just for context:

• inertial reference frame Oxyz, with versors i, j, k
• non inertial reference frame Ox'y'z', with versors i', j', k'
• constant angular velocity omega (W), in a random fixed direction

so here is how the chapter goes:

1. di'/dt = W ^ i' (vector product)
2. dj'/dt = W ^ j'
3. dk'/dt = W ^ k'

let r(t) be the position vector of the point of interest on the surface of the rotating object:

r(t) = x.i + y.j + z.k = x'.i' + y'.j' + z'.k' = r'(t) (the position vector as measured in the non-inertial reference frame)​

​

let's calculate the time derivative of the above equation​

​

dr(t)/dt = dx/dt.i + dy/dt.j + dz/dt.k = dx'/dt.i' + dy'/dt.j' + dz'/dt.k' + x'.di'/dt + y'.dj'/dt + z'.dk'/dt​

​

where dx'/dt.i' + dy'/dt.j' + dz'/dt.k' = [dr(t)/dt]' calculated in the non-inertial reference frame, v'(t)​

​

so, [dr(t)/dt] = [dr(t)/dt]' + x'.W ^ i' + y'.W ^ j' + z'.W ^ k'​

​

[dr(t)/dt] = [dr(t)/dt]' + (x'.W ^ i' + y'.W ^ j' + z'.W ^ k')​

[dr(t)/dt] = [dr(t)/dt]' + W ^ (x'.i' + y'.j' + z'.k')​

[dr(t)/dt] = [dr(t)/dt]' + W ^ r'(t)​

​

the inertial velocity is the sum of the non-inertial velocity and the vector product of the angular velocity vector and the non-inertial position vector. So far, so good. Now, in my deranged idea, I though of calculating the equations of motion of a body under any accelaration vector, not only under a constant angular velocity. I'm not sure my way of thinking is correct here, I'll publish my calculations, here they go:​

​

d2r(t)/dt2 = d2x/dt2.i + d2y/dt2.j + d2z/dt2.k = (take a deep breath...)​

= (d2x'/dt2.i' + dx'/dt.di'/dt) + (d2y'/dt2.j' + dy'/dt.dj'/dt) + (d2z'/dt2.k' + dz'/dt.dk'/dt) + (dx'/dt.di'/dt + x'.d2i'/dt2) + (dy'/dt.dj'/dt + y'.d2j'/dt2) + (dz'/dt.dk'/dt + z'.d2k'/dt2)​

= (d2x'/dt2.i' + d2y'/dt2.j' + d2z'/dt2.k') +2.(dx'/dt.di'/dt + dy'/dt.dj'/dt + dz'/dt.dk'/dt) + (x'.d2i'/dt2 + y'.d2j'/dt2 + z'.d2k'/dt2)​

​

now, expanding on the equations described in points (1), (2) and (3):​

​

di'/dt = W ^ i' -> di'/dt = [Wx; Wy; Wz] ^ i' -> di'/dt = Wx.i'​

​

so, d2i'/dt2 = dWx/dt.i' + Wx.di'/dt = dWx/dt.i' + Wx.(W ^ i') = dWx/dt.i' + Wx.(Wx.i') = dWx/dt.i' + Wx^2.i' = (dWx/dt + Wx^2).i'​

​

​

dj'/dt = W ^ j' -> dj'/dt = [Wx; Wy; Wz] ^ j' -> dj'/dt = Wy.j'​

​

so, d2j'/dt2 = dWy/dt.j' + Wy.dj'/dt = dWy/dt.j' + Wy.(W ^ j') = dWy/dt.j' + Wy.(Wy.j') = dWy/dt.j' + Wy^2.j' = (dWy/dt + Wy^2).j'​

​

​

dk'/dt = W ^ k' -> dk'/dt = [Wx; Wy; Wz] ^ k' -> dk'/dt = Wz.k'​

​

so, d2z'/dt2 = dWz/dt.k' + Wz.dk'/dt = dWz/dt.k' + Wz.(W ^ k') = dWz/dt.k' + Wz.(Wz.k') = dWz/dt.k' + Wz^2.k' = (dWz/dt + Wz^2).k'​

​

therefore,​

d2r(t)/dt2 = (d2x'/dt2.i' + d2y'/dt2.j' + d2z'/dt2.k') + 2.(dx'/dt.Wx.i' + dy'/dt.Wy.j' + dz'/dt.Wz.k') + [x'.(dWx/dt + Wx^2).i' + y'.(dWy/dt + Wy^2).j' + z'.(dWz/dt + Wz^2).k']​

​

d2r(t)/dt2 = i'.[d2x'/dt2 + 2.dx'/dt.Wx + x'.(dWx/dt + Wx^2)] + j'.[d2y'/dt2 + 2.dy'/dt.Wy + y'.(dWy/dt + Wy^2)] + k'.[d2z'/dt2.k' + 2.dz'/dt.Wz + z'.(dWz/dt + Wz^2)]​

Ok, after all of this, here's my question, am I going crazy? How many mistakes did I make there in my thoughts? Is this all correct, or am I delusiuonal? I'm going crazy over this, this is my attempt at describing a motion problem that I'd like to implement on a car model, believe it or not. Thank you so much for taking the time to read (and perhaps answer) my post. I appreciate it. I tried to make it as intelligible and brief as possible! =P

Thank you so much again,
Vicente

 

[vanhees71]

Can you type this in readable form? The LaTeX guide is on the left just below the typing area:

https://www.physicsforums.com/help/latexhelp/

 

[VicenteMMOS]

I'm following the guide on LaTeX, but when I click 'Preview', I only see the regular text, not the formatted equations... do I need to install anything? I'm following the guide, but I don't know what I'm missing...

 

[Ibix]

There's a known issue where MathJax doesn't load for preview if there isn't already $\LaTeX$ on the page. Since I've used it in my previous sentence you should be fine now.

In future you can workaround it like this:
1. Type your post including some $\LaTeX$
2. Copy your text to clipboard (just in case)
3. Hit preview (maths won't render)
4. Refresh the page (maths should render)
You should now be able to go back to the editor, make changes, and preview with rendered maths. If your text is gone from the editor, paste it back in.

 

[VicenteMMOS]

I've spent quite some time trying to implement the equations with LaTeX, and I was succeeding, but then at some point, it stopped working for some reason. Is there a cap to how many equations you can put in a single post?

 

[vanhees71]

No, there are some troubles with the rendering for some time (since an "update" of the forum software ;-)). I use Firefox as a browser. There it helps to simply reload the page and use the preview button again to preview. Reloading the page fortunately doesn't erase what you typed before. That's however not working with the Chrome browser.

Sometimes also after posting the math isn't rendered too. Also then the reload button helps.

 

[VicenteMMOS]

Finally, after all these hours, I managed to compile this LaTeX code! =DLet's do this! I'll publish also the text from the original post, for better contextualizing my calculations. Here it goes:

[...] the description of motion, and the computation of the acceleration vector as being the sum of the rate of change of velocity, with the division of linear velocity squared by the instant radius of curvature:

$a = \frac {dv} {dt} + \frac {v^2} {r}$ , tangential + centripetal acceleration componentsThe above equation is calculated with an inertial reference frame. Next, I went to the chapter describing the three-dimensional movement of an object, but with a constant angular speed, $\Omega$. Here is the description of that problem, just for context:

• inertial reference frame $O_{xyz}$, with versors $\hat {\mathbf i}$, $\hat {\mathbf j}$, $\hat {\mathbf k}$
• non inertial reference frame $O_{x'y'z'}$, with versors $\hat {\mathbf i'}$, $\hat {\mathbf j'}$, $\hat {\mathbf k'}$
• constant angular velocity $\Omega$, in a random fixed direction

so here is how the chapter goes:

$$\frac {d\hat {\mathbf i'}} {dt} = \vec \Omega \times \hat {\mathbf i'} (1), ~~ \frac {d\hat {\mathbf j'}} {dt} = \vec \Omega \times \hat {\mathbf j'} (2), ~~ \frac {d\hat {\mathbf k'}} {dt} = \vec \Omega \times \hat {\mathbf k'} (3)$$

let $r(t)$ be the position vector of the point of interest on the surface of the rotating object:

$r(t) = x.\hat {\mathbf i} + y.\hat {\mathbf j} + z.\hat {\mathbf k} = x'.\hat {\mathbf i'} + y'.\hat {\mathbf j'} + z'.\hat {\mathbf k'} = r'(t)$ (the position vector as measured in the non-inertial reference frame)

let's calculate the time derivative of the above equation

$\frac {dr(t)} {dt} = \frac {dx} {dt}.\hat {\mathbf i} + \frac {dy} {dt}.\hat {\mathbf j} + \frac {dz} {dt}.\hat {\mathbf k} = \frac {dx'} {dt}.\hat {\mathbf i'} + \frac {dy'} {dt}.\hat {\mathbf j'} + \frac {dz'} {dt}.\hat {\mathbf k'} + x'.\frac {d{\hat {\mathbf i'}}} {dt} + y'.\frac {d{\hat {\mathbf j'}}} {dt} + z'.\frac {d{\hat {\mathbf k'}}} {dt}$

where $\frac {dx'} {dt}.\hat {\mathbf i'} + \frac {dy'} {dt}.\hat {\mathbf j'} + \frac {dz'} {dt}.\hat {\mathbf k'} = [\frac {dr(t)} {dt}]^{'}$ calculated in the non-inertial reference frame, $v'(t)$

so, $\left [\frac {dr(t)} {dt} \right ] = \left [\frac {dr(t)} {dt} \right ]' + x'.\vec \Omega \times \hat {\mathbf i'} + y'.\vec \Omega \times \hat {\mathbf j'} + z'.\vec \Omega \times \hat {\mathbf k'}$

$\left [\frac {dr(t)} {dt} \right ] = \left [\frac {dr(t)} {dt} \right ]' + (x'.\vec \Omega \times \hat {\mathbf i'} + y'.\vec \Omega \times \hat {\mathbf j'} + z'.\vec \Omega \times \hat {\mathbf k'})$
$\left [\frac {dr(t)} {dt} \right ] = \left [\frac {dr(t)} {dt} \right ]' + \vec \Omega \times (x'.\hat {\mathbf i'} + y'.\hat {\mathbf j'} + z'.\hat {\mathbf k'})$
$\left [\frac {dr(t)} {dt} \right ] = \left [\frac {dr(t)} {dt} \right ]' + \vec \Omega \times r'(t)$

the inertial velocity $\left [\frac {dr(t)} {dt} \right]$ is the sum of the non-inertial velocity $\left [\frac {dr(t)} {dt} \right]'$ and the vector product of the angular velocity vector and the non-inertial position vector $\vec \Omega \times r'(t)$. So far, so good. Now, in my deranged idea, I though of calculating the equations of motion of a body under any accelaration vector, not only under a constant angular velocity. I'm not sure my way of thinking is correct here, I'll publish my calculations, here they go:

$\left (\frac {d^{2}r(t)} {dt^{2}} \right ) = \left (\frac {d^{2}x} {dt^{2}} \right ).\hat {\mathbf i} + \left (\frac {d^{2}y} {dt^{2}} \right ).\hat {\mathbf j} + \left (\frac {d^{2}z} {dt^{2}} \right ).\hat {\mathbf k} =$ (take a deep breath...)

$\begin{align} &= \left (\frac {d^{2}x'} {dt^{2}}. \hat {\mathbf i'} + \frac {dx'} {dt}.\frac {d\hat {\mathbf i'}} {dt} \right ) + \left (\frac {d^{2}y'} {dt^{2}}. \hat{\mathbf j'} + \frac {dy'} {dt}. \frac {d\hat {\mathbf j'}} {dt} \right ) + \left (\frac {d^{2}z'} {dt^{2}}.\hat {\mathbf k'} + \frac {dz'} {dt}. \frac {d\hat {\mathbf k'}} {dt} \right ) \nonumber \\ &+ \left (\frac {dx'} {dt}. \frac {d\hat {\mathbf i'}} {dt} + x'. \frac {d^{2}\hat {\mathbf i'}} {dt^{2}} \right ) + \left (\frac {dy'} {dt}. \frac {d\hat {\mathbf j'}} {dt} + y'. \frac {d^{2}\hat {\mathbf j'}} {dt2} \right ) + \left (\frac {dz'} {dt}. \frac {d\hat {\mathbf k'}} {dt} + z'. \frac {d^{2}\hat {\mathbf k'}} {dt^{2}} \right ) \nonumber \end{align}$
$\begin{align} = \left (\frac {d^{2}x'} {dt^{2}}.\hat {\mathbf i'} + \frac {d^{2}y'} {dt^{2}}.\hat {\mathbf j'} + \frac {d^{2}z'} {dt^{2}}.\hat {\mathbf k'} \right ) +2.\left (\frac {dx'} {dt}.\frac {d\hat {\mathbf i'}} {dt} + \frac {dy'} {dt}.\frac {d\hat {\mathbf j'}} {dt} + \frac {dz'} {dt}.\frac {d\hat {\mathbf k'}} {dt}\right ) + \nonumber \\ \left (x'.\frac {d^{2}\hat {\mathbf i'}} {dt^{2}} + y'.\frac {d^{2}\hat {\mathbf j'}} {dt^{2}} + z'.\frac {d^{2}\hat {\mathbf k'}} {dt^{2}}\right ) \nonumber \end{align}$

now, expanding on the equations described in points (1), (2) and (3):

$1) \left [\frac {d\hat {\mathbf i'}} {dt} \right ] = \vec \Omega \times \hat {\mathbf i'} \to \left [\frac {d\hat {\mathbf i'}} {dt} \right ] = [\Omega_x; \Omega_y; \Omega_z] \times \hat {\mathbf i'} \to \left [\frac {d\hat {\mathbf i'}} {dt} \right ] = \Omega_x.\hat {\mathbf i'}$​

​

so,​

​

$\begin{align} \left [\frac {d^2\hat {\mathbf i'}}{dt^2} \right ] &= \frac {d\Omega_x} {dt}.\hat {\mathbf i'} + \Omega_x.\frac {d\hat {\mathbf i'}} {dt} = \frac {d\Omega_x} {dt}.\hat {\mathbf i'} + \Omega_x.(\vec \Omega \times \hat {\mathbf i'}) \nonumber \\ &= \frac {d\Omega_x} {dt}.\hat {\mathbf i'} + \Omega_x.(\Omega_x.\hat {\mathbf i'}) = \frac {d\Omega_x}{dt}.\hat {\mathbf i'} + \Omega_x^{2}.\hat {\mathbf i'} = \left (\frac {d\Omega_x}{dt} + \Omega_x^2 \right ).\hat {\mathbf i'} \nonumber \end{align}$​

______________________________________________________________________________________________________________________​

​

$2) \left [\frac {d\hat {\mathbf j'}} {dt} \right] = \vec \Omega \times \hat {\mathbf j'} \to \left [\frac {d\hat {\mathbf j'}} {dt} \right ] = [\Omega_x; \Omega_y; \Omega_z] \times \hat {\mathbf j'} \to \left [\frac {d\hat {\mathbf j'}} {dt} \right ]= \Omega_y.\hat {\mathbf j'}$​

​

so,​

​

$\begin{align} \left [\frac {d^2\hat {\mathbf j'}} {dt^2} \right] &= \left [\frac {d\Omega_y}{dt}\right ].\hat {\mathbf j'} + \Omega_y.\left[\frac {d\hat {\mathbf j'}}{dt}\right ] = \left [\frac {d\Omega_y}{dt}\right ].\hat {\mathbf j'} + \Omega_y.(\vec \Omega \times \hat {\mathbf j'}) \nonumber \\ &= \left [\frac {d\Omega_y}{dt}\right ].\hat {\mathbf j'} + \Omega_y.(\Omega_y.\hat {\mathbf j'})= \left [\frac {d\Omega_y}{dt}\right].\hat {\mathbf j'} + \Omega_y^2.\hat {\mathbf j'} = \left (\frac {d\Omega_y}{dt} + \Omega_y^2 \right ).\hat {\mathbf j'} \nonumber \end{align}$​

______________________________________________________________________________________________________________________​

​

$3) \left [\frac {d\hat {\mathbf k'}}{dt}\right ] = \vec \Omega \times \hat {\mathbf k'} \to \left [\frac {d\hat {\mathbf k'}}{dt}\right ] = [\Omega_x; \Omega_y; \Omega_z] \times \hat {\mathbf k'} \to \left [\frac {d\hat {\mathbf k'}}{dt}\right ] = \Omega_z.\hat {\mathbf k'}$​

​

so,​

​

$\begin{align} \left [\frac {d^2z'}{dt^2}\right ] &= \left [\frac {d\Omega_z}{dt}\right ].\hat {\mathbf k'} + \Omega_z.\left [\frac {d\hat {\mathbf k'}}{dt}\right ] = \left [\frac {d\Omega_z}{dt}\right ].\hat {\mathbf k'} + \Omega_z.(\vec \Omega \times \hat {\mathbf k'}) = \left [\frac {d\Omega_z}{dt}\right ].\hat {\mathbf k'} + \Omega_z.(\Omega_z.\hat {\mathbf k'}) \nonumber \\ &= \left [\frac {d\Omega_z}{dt}\right ].\hat {\mathbf k'} + \Omega_z^2.\hat {\mathbf k'} = \left (\frac {d\Omega_z}{dt} + \Omega_z^2\right ).k' \nonumber \end{align}$​

therefore,

$\begin{align} \left [\frac {d^2r(t)}{dt^2}\right ] &= \left (\frac {d^2x'}{dt^2}.\hat {\mathbf i'} + \frac {d^2y'} {dt^2}.\hat {\mathbf j'} + \frac {d^2z'}{dt^2}.\hat {\mathbf k'}\right ) + 2.\left (\frac {dx'} {dt}.\Omega_x.\hat {\mathbf i'} + \frac {dy'}{dt}.\Omega_y.\hat {\mathbf j'} + \frac {dz'}{dt}.\Omega_z.\hat {\mathbf k'}\right ) + \nonumber \\ &\left [x'. \left (\frac {d\Omega_x}{dt} + \Omega_x^2\right ).\hat {\mathbf i'} + y'.\left (\frac {d\Omega_y}{dt} + \Omega_y^2\right ).\hat {\mathbf j'} + z'.\left (\frac {d\Omega_z}{dt} + \Omega_z^2\right ).\hat {\mathbf k'}\right ] \nonumber \end{align}$

$$\begin{align}
\left [\frac {d^2r(t)}{dt^2}\right ] &= \hat {\mathbf i'}.\left [\frac {d^2x'}{dt^2} + 2.\frac {dx'}{dt}.\Omega_x + x'.\left (\frac {d\Omega_x}{dt} + \Omega_x^2 \right )\right ] \nonumber \\ &+ \hat {\mathbf j'}.\left [\frac {d^2y'}{dt^2} + 2.\frac{dy'}{dt}.\Omega_y + y'.\left (\frac {d\Omega_y}{dt} + \Omega_y^2\right )\right] \nonumber \\ &+ \hat {\mathbf k'}.\left [\frac {d^2z'}{dt^2}.\hat {\mathbf k'} + 2.\frac {dz'}{dt}.\Omega_z + z'.\left (\frac {d\Omega_z}{dt} + \Omega_z^2 \right )\right ] \nonumber
\end{align}$$

[...]

Please god let this bunch of stuff make any sort of sense!

Thank you for reading all of it =)

 

[VicenteMMOS]

I’m going crazy with this problem. I believe I spotted a mistake in my previous calculations, also had an idea building upon the previous one, and started from scratch. But I don’t really know if anything of what I’m doing makes any sense at all. Here’s what I’m trying to do, to create a mathematical description in inertial coordinate system of a variable motion of an object (which has its own non-inertial coord system), with variable linear and angular accelerations. Now, I restarted from two 3x3 matrices, describing position, velocity and acceleration of angular and linear movements in 3 directions. But then I keep confusing myself, trying to answer “what is the correlation between linear and angular displacements, velocities and accelerations?” I’m not sure if my math is right, but I’m eve less sure about my reasoning of the physics involved… :P I don’t know where to find help with this.

 

[vanhees71]

$$\newcommand{\uvec}[1]{\underline{#1}}$$
It's much simpler to use some matrix algebra and Ricci calculus. Let $\vec{e}_j$ be the Cartesian basis of a fixed inertial reference frame and
$$\vec{e}_k'(t)=D_{jk}(t) \vec{e}_j$$
the basis of the frame rotating wrt. the inertial one. Here I use Einstein's summation convention, i.e., over indices appearing twice in a formula you have to sum from 1 to 3. $(D_{jk})=\hat{D}$ is a rotation matrix, i.e., fulfilling $\hat{D} \hat{D}^{\text{T}}=\hat{1}$ and $\mathrm{det} \hat{D}=1$. From this you get
$$\vec{e}_j=D_{jk} \vec{e}_k'.$$

The time derivative of the rotating basis vectors is
$$\dot{\vec{e}}_k'(t) = \dot{D}_{jk} \vec{e}_j =\dot{D}_{jk} D_{jl} \vec{e}_l'=D^{\text{T}}_{lj} \dot{D}_{jk} \vec{e}_l'. \qquad (*)$$
Now from $\hat{D}^{\text{T}} \hat{D}=\hat{1}$ you get by taking the time derivative
$$\hat{D}^{\text{T}} \dot{\hat{D}}=-\dot{\hat{D}}^T \hat{D} = -[\hat{D}^{\text{T}} \dot{\hat{D}}]^{\text{T}},$$
which means that $\hat{D}^{\text{T}} \dot{D}$ is an antisymmetric matrix, and you can write
$$(\hat{D}^{\text{T}} \dot{D})_{lk}=-\epsilon_{lkj} \omega_j'.$$
This means that (*)
$$\dot{\vec{e}}_k'=+\epsilon_{jkl} \omega_j' \vec{e}_l'.$$
So for an arbitrary vector, decomposed wrt. the rotating basis you get
$$\dot{\vec{V}}=\dot{V}_k' \vec{e}_k' + V_k' \dot{\vec{e}}_k' = \dot{V}_k' \vec{e}_k' - \omega_j' V_k' \epsilon_{jkl} \vec{e}_l'=[\dot{V}_l' + (\uvec{\omega}' \times \uvec{V}')_l]\vec{e}_l'.$$
Here $\uvec{V}'$ means the column vector of the components of $\vec{V}$ wrt. the rotating basis $\vec{e}_k'$.

Defining a "covariant time derivative" for vector components in the rotating frame by
$$\mathrm{D}_t \uvec{V}'=\dot{\uvec{V}}' + \uvec{\omega}' \times \uvec{V}'$$
is a nice shortcut summarizing the above somewhat involved calculation to get the time derivative of a vector in terms of its components wrt. the rotating frame.

This rule you can now apply twice to the position vector $\vec{r}$. You get
$$\uvec{v}'=\mathrm{D}_t \uvec{r}' = \dot{\uvec{r}}'+\uvec{\omega}' \times \uvec{r}'$$
and then
$$\uvec{a}'=\mathrm{D}_t \uvec{v}' = \ddot{\uvec{r}}' + \uvec{\omega}' \times \dot{\uvec{r}}' + \dot{\uvec{\omega}}' \times \uvec{r}' + \uvec{\omega}' \times \dot{\uvec{r}'} + \uvec{\omega}' \times (\uvec{\omega}' \times \uvec{r}').$$
Simplifying this finally leads to the formula you want to derive
$$\uvec{a}'=\ddot{\uvec{r}}' + 2 \uvec{\omega}' \times \dot{\uvec{r}}' + \uvec{\omega}' \times (\uvec{\omega}' \times \uvec{r}') + \dot{\uvec{\omega}}' \times \uvec{r}'.$$