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Jun 9, 2023 at 0:09 comment added Amit I find the differential approach unnecessary here -- unless I'm missing something, here is the derivation I prefer: the radius of the rotation circle $R$ is clearly given by: $R=|\vec{r}|\sin{\theta}$. Where $\theta$ is the angle $\vec{r}$ makes with the $z$ axis. Now, writing the velocity's magnitude as $|\vec{v}| = wR = w|\vec{r}|\sin{\theta}$, we observe that if we define $\vec{w} = (0,0,w)$ this is the unique vector such that $|\vec{w} \times \vec{r}| = wR = |\vec{v}|$. Finally, we also observe that as a vector $\vec{v}$ is perpendicular to both $\vec{w}$ and $\vec{r}$ and we're done.
Mar 29, 2022 at 10:48 comment added VIVID @Rahul If you treat that as a normal triangle (i.e. nothing about infinitesimal rotation) $dr = r\sin\phi$. But $\sin\phi \approx \phi$ for sufficiently small angles
Mar 29, 2022 at 9:47 comment added RAHUL @VIVID, Sir, would you please elaborate why $dr=d\phi × r$
S Nov 21, 2020 at 23:42 history suggested VIVID CC BY-SA 4.0
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S Nov 21, 2020 at 23:42
Apr 8, 2015 at 18:30 vote accept CommunityBot
Mar 19, 2015 at 10:27 comment added danimal The right hand screw rule is just convention; we could define the cross product the opposite direction and it would be fine as long as the same new version was applied universally
Mar 19, 2015 at 7:56 comment added user74370 Ok then basically i've been questioning the right hand screw rule itself..and do you know a proof for the right hand screw rule? Btw this is a good approach :-)
Mar 19, 2015 at 7:46 comment added sugatasen And just so u know, if u tried v = rXω, that would be fundamentally wrong, bcuz of how the direction of ω us assumed by the right hand rule convention!, then of course you would obtain an illogical result, anyway , that's if u go by convention!!
Mar 19, 2015 at 7:42 history answered sugatasen CC BY-SA 3.0