Question Number 29498 by ajfour last updated on 09/Feb/18
Commented by ajfour last updated on 09/Feb/18
$${Relate}\:{linear}\:{acceleration},\:{with} \\ $$$${position},\:{velocity},\:{angular}\:{velocity}, \\ $$$${and}\:{angular}\:{acceleration}. \\ $$
Answered by ajfour last updated on 09/Feb/18
![Acceleration:_(__________________) ds^ = r_(xy) (−sin ψi^ +cos ψj^ )dψ +r_(yz) (−sin θj^ +cos θk^ )dθ +r_(zx) (−sin φk^ +cos φi^ )dφ +dr(((xi^ +yj^ +k^ )/r)) =(zdφ−ydψ)i^ +(xdψ−zdθ)j^ +(ydθ−xdφ)k^ +(dr/r)(xi^ +yj^ +k^ ) ⇒ ds^ = [(i^ ,j^ ,k^ ),(dθ,dφ,dψ),(x,y,z) ]+ (dr/r)(xi^ +yj^ +zk^ ) (ds^ /dt)= v^ =ω^ ×r^ +(1/r)((dr/dt))r^ ⇒ a^ = α^ ×r^ +ω^ ×v^ −(1/r^2 )((dr/dt))^2 r^ +(1/r)((d^2 r/dt^2 ))r^ +(1/r)((dr/dt))v^ a^ =α^ ×r^ +ω^ ×(ω^ ×r^ )+(1/r)((dr/dt))(ω^ ×r^ ) −(1/r^2 )((dr/dt))^2 r^ +(1/r)((d^2 r/dt^2 ))r^ + +(1/r)((dr/dt))(ω^ ×r^ )+(1/r^2 )((dr/dt))^2 r^ a^ = 𝛂^ ×r^ + 𝛚^ ×(𝛚^ ×r^ ) + (1/r)((d^2 r/dt^2 ))r^ +(2/r)((dr/dt))(𝛚^ ×r^ ) −−−−−−−−−−−−−−−](Q29502.png)
$$\:\:\:\underset{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_} {\boldsymbol{{Acceleration}}:}\: \\ $$$${d}\bar {{s}}\:=\:{r}_{{xy}} \left(−\mathrm{sin}\:\psi\hat {{i}}+\mathrm{cos}\:\psi\hat {{j}}\right){d}\psi \\ $$$$\:\:\:\:\:\:\:\:+{r}_{{yz}} \left(−\mathrm{sin}\:\theta\hat {{j}}+\mathrm{cos}\:\theta\hat {{k}}\right){d}\theta \\ $$$$\:\:\:\:\:\:\:\:+{r}_{{zx}} \left(−\mathrm{sin}\:\phi\hat {{k}}+\mathrm{cos}\:\phi\hat {{i}}\right){d}\phi \\ $$$$\:\:\:\:\:\:\:\:\:+{dr}\left(\frac{{x}\hat {{i}}+{y}\hat {{j}}+\hat {{k}}}{{r}}\right) \\ $$$$\:\:\:\:\:\:=\left({zd}\phi−{yd}\psi\right)\hat {{i}}+\left({xd}\psi−{zd}\theta\right)\hat {{j}} \\ $$$$\:\:\:\:\:\:+\left({yd}\theta−{xd}\phi\right)\hat {{k}}+\frac{{dr}}{{r}}\left({x}\hat {{i}}+{y}\hat {{j}}+\hat {{k}}\right) \\ $$$$\Rightarrow\:{d}\bar {{s}}\:=\begin{bmatrix}{\hat {{i}}}&{\hat {{j}}}&{\hat {{k}}}\\{{d}\theta}&{{d}\phi}&{{d}\psi}\\{{x}}&{{y}}&{{z}}\end{bmatrix}+\:\frac{{dr}}{{r}}\left({x}\hat {{i}}+{y}\hat {{j}}+{z}\hat {{k}}\right) \\ $$$$\frac{{d}\bar {{s}}}{{dt}}=\:\bar {{v}}=\bar {\omega}×\bar {{r}}\:+\frac{\mathrm{1}}{{r}}\left(\frac{{dr}}{{dt}}\right)\bar {{r}} \\ $$$$\Rightarrow\:\bar {\boldsymbol{{a}}}\:=\:\bar {\alpha}×\bar {{r}}+\bar {\omega}×\bar {{v}}−\frac{\mathrm{1}}{{r}^{\mathrm{2}} }\left(\frac{{dr}}{{dt}}\right)^{\mathrm{2}} \bar {{r}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:+\frac{\mathrm{1}}{{r}}\left(\frac{{d}^{\mathrm{2}} {r}}{{dt}^{\mathrm{2}} }\right)\bar {{r}}+\frac{\mathrm{1}}{{r}}\left(\frac{{dr}}{{dt}}\right)\bar {{v}} \\ $$$$\bar {{a}}=\bar {\alpha}×\bar {{r}}+\bar {\omega}×\left(\bar {\omega}×\bar {{r}}\right)+\frac{\mathrm{1}}{{r}}\left(\frac{{dr}}{{dt}}\right)\left(\bar {\omega}×\bar {{r}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\frac{\mathrm{1}}{{r}^{\mathrm{2}} }\left(\frac{{dr}}{{dt}}\right)^{\mathrm{2}} \bar {{r}}+\frac{\mathrm{1}}{{r}}\left(\frac{{d}^{\mathrm{2}} {r}}{{dt}^{\mathrm{2}} }\right)\bar {{r}}+ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\frac{\mathrm{1}}{{r}}\left(\frac{{dr}}{{dt}}\right)\left(\bar {\omega}×\bar {{r}}\right)+\frac{\mathrm{1}}{{r}^{\mathrm{2}} }\left(\frac{{dr}}{{dt}}\right)^{\mathrm{2}} \bar {{r}} \\ $$$$\:\:\:\:\bar {\boldsymbol{{a}}}=\:\:\bar {\boldsymbol{\alpha}}×\bar {\boldsymbol{{r}}}\:\:+\:\:\bar {\boldsymbol{\omega}}×\left(\bar {\boldsymbol{\omega}}×\bar {\boldsymbol{{r}}}\right) \\ $$$$\:\:\:\:\:\:\:\:+\:\frac{\mathrm{1}}{\boldsymbol{{r}}}\left(\frac{\boldsymbol{{d}}^{\mathrm{2}} \boldsymbol{{r}}}{\boldsymbol{{dt}}^{\mathrm{2}} }\right)\bar {\boldsymbol{{r}}}\:+\frac{\mathrm{2}}{\boldsymbol{{r}}}\left(\frac{\boldsymbol{{dr}}}{\boldsymbol{{dt}}}\right)\left(\bar {\boldsymbol{\omega}}×\bar {\boldsymbol{{r}}}\right) \\ $$$$−−−−−−−−−−−−−−− \\ $$