If I have a particle of a rigid object rotating about a fixed axis, how do I show that the tangential and radial vector components of the linear acceleration are:
$$\bar{a}_{tan}=\bar a\times \bar r \;\;\; \bar{a}_{R}=\bar w \times \bar v$$
I know this is more for the physics part of stack exchange but I did not get any answers over there. The x's mean cross-product by the way.
I'm assuming anyone who answers this has probably taken a physics course before and these are basics about angular momentum.
Sorry if this is more physics than math, I just can't figure out how to do this, I looked online for anything similar, my textbook does not do a good job explaining this, and the text does not have any answers for me to check against.
Thanks for any help, I really do appreciate it.
I'm not a mathematician, so this is probably not the "proof" one would use in an article, but at least this should be logical and easy to follow:
Without loss of generality, we can assume the rotation axis corresponds to the $z$ axis, and that the particle is at $(r, 0, 0)$ at time $t = 0$. (In other words, we can rotate and translate any system OP described to this orientation, without adding any new constraints; so, for the purposes of this "proof", we can simply assume such a coordinate system.)
If the angular velocity $\omega$ is constant, the location of the rigidly rotating particle as a function of time $t$ is $$\vec{r}(t) = ( r \cos \omega t ,\, r \sin \omega t ,\, 0 )$$ The velocity vector $\vec{v}(t)$ of the particle is $$\vec{v}(t) = \frac{d\,\vec{r}(t)}{d\,t} = ( -r \omega \sin \omega t ,\, r \omega \cos \omega t ,\, 0 )$$ and the acceleration vector $\vec{a}(t)$ is $$\vec{a}(t) = \frac{d^2\,\vec{r}(t)}{d\,t^2} = \frac{d\,\vec{v}(t)}{d\,t} = ( - r \omega^2 \cos \omega t ,\, -r \omega^2 \sin \omega t ,\, 0 )$$
The radial component $\vec{a}_r(t)$ of the acceleration vector $\vec{a}(t)$is the part that is in the direction of the radial vector $\vec{r}(t)$: $$\vec{a}_r(t) = \vec{a}(t) \left\lvert \frac{\vec{a}(t)}{\left\lVert\vec{a}(t)\right\rVert} \cdot \frac{\vec{r}(t)}{\left\lVert\vec{r}(t)\right\rVert} \right\rvert = \vec{a}(t) \lvert 1 \rvert = ( -r \omega^2 \cos \omega t ,\, -r \omega^2 \sin \omega t ,\, 0 )$$ The tangential component $\vec{a}_t(t)$ is zero, because $\vec{a}(t) = \vec{a}_r(t)$.
This is as expected, because a particle rotating at a constant angular velocity $\omega$ at a radius of $r$ around a fixed point has acceleration $r\omega^2$ towards the fixed point.
This also means that $$\vec{a}(t) \times \vec{r}(t) = ( 0 , 0 , 0 )$$
Note that angular velocity $\omega$ has units angle/time. Particle speed $v = \left\lVert\vec{v}(t)\right\rVert$ and angular velocity $\omega$ are related by $$v = \left\lVert\vec{v}(t)\right\rVert = \sqrt{\vec{v}(t) \cdot \vec{v}(t)} = r \omega$$ i.e. $$v = r \omega \quad \iff \quad \omega = \frac{v}{r}$$
If we use $\hat{n} = (0, 0, 1)$ for the rotation axis, then $$\hat{n} \times \vec{r}(t) = ( -R \sin \omega t ,\, R \cos \omega t ,\, 0 ) = \frac{1}{\omega} \vec{v}(t)$$ $$\hat{n} \times \vec{v}(t) = ( -R \omega \cos \omega t ,\, -R \omega \sin \omega t ,\, 0 ) = \frac{1}{\omega} \vec{a}(t)$$
If we use $\vec{w} = (0, 0, \omega)$ to describe the rotation axis, with magnitude reflecting the angular velocity ($\omega = \lVert\vec{w}\rVert$), then $$\hat{w} \times \vec{r}(t) = \vec{v}(t)$$ $$\hat{w} \times \vec{v}(t) = \vec{a}(t)$$
In summary, if we consider the cross product between the acceleration vector $\vec{a}(t)$ and the location vector $\vec{r}(t)$ (wrt. to the point the particle rotates around), it does indeed match the tangential component of the acceleration vector $\vec{a}_t$ (because they are both zero): $$\vec{a}(t) \times \vec{r}(t) = ( 0 , 0 , 0 ) = \vec{a}_t$$
If we use $\vec{w}$ to describe the axis of rotation, with its magnitude being the angular velocity, $\lVert\vec{w}\rVert = \omega$, then the cross product between $\vec{w}$ and the velocity vector $\vec{v}(t)$ does indeed match the radial component $\vec{a}_r(t)$ of the acceleration vector $\vec{a}(t)$ (because the acceleration vector only has the radial component, $\vec{a}_r(t) = \vec{a}(t)$): $$\vec{w} \times \vec{v}(t) = ( -r \omega^2 \cos \omega t ,\, -r \omega^2 \sin \omega t ,\, 0 ) = \vec{a}(t)$$