Today I came in class and in one of the problems the teacher used $\vec{a}=\vec{\omega}\times \vec{r}$ which made me very confused because I don't know where it comes from, it seems pulled out of thin air, the only one I know of is $\vec{v}=\vec{\omega}\times \vec{r}$ which then as well confuses me more since that would imply that acceleration and velocity are the same and that's not true. What's going on here?
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7$\begingroup$ The formula is wrong, but maybe he forgot (or you did not see) the little dot on top of $\omega$ to signify a time derivative such as $\vec{a} = \dot{\vec{\omega}}\times \vec{r}$ (assuming the radius does not change). $\endgroup$– sazanCommented Sep 12, 2023 at 9:29
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1$\begingroup$ The units don't balance on both sides of the equation. I suggest you ask in class in this point and allow the professor to correct his error. $\endgroup$– jalexCommented Sep 12, 2023 at 13:33
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2 Answers
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From my knowledge your professor might have just been wrong. Or maybe he used his own notation for the angular acceleration as omega, thing that I never saw in any textbook, which practically makes it wrong. There is nothing about the cancellation of the radian unit (it can be canceled for dimensional analysis purposes as of the "recent" physics congress) either, because we are talking about the powers of the second, so he just is probably wrong.
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1$\begingroup$ so $\vec{a}=\vec{\alpha}\times \vec{r}$ is a thing? $\endgroup$– UlshyCommented Sep 12, 2023 at 9:10
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1$\begingroup$ yes,it is,the alpha is called angular acceleration,which is rate of change of angular velocity with respect to time $\endgroup$ Commented Sep 12, 2023 at 9:11
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1$\begingroup$ To @Ulshy. That is the tangential acceleration. $\endgroup$– AlvCommented Sep 12, 2023 at 16:37
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Your professor is (probably) wrong. If he is talking about angular acceleration, then we can derive an expression using the chain rule.
if $\vec{v} = \vec{\omega} × \vec{r}$, we can differentiate this with respect to time to give
$$d\vec{v}/{dt} = \omega (dr/dt) + r(d\omega/dt)$$
$$ a = \omega(dr/dt) + r(d\omega/dt)$$
this is the general explanation of the acceleration vector. We could also substitute $d\omega/dt$ as 0, implying that the particle is under uniform circular motion.
It is also worth noting that the directions of these accelerations is different as well, with the $\omega(dr/dt)$ being directed tangential to the path covered and the r alpha part being along the (-ve) radius vector to the center of the circle from the particle.
