0
$\begingroup$

If one can use the right hand rule to determine the angular velocity direction, and if using change in theta, and radius, would it be fair to say that it would be the cross product of r cross change in theta, or $\vec r\times\vec{\Delta\theta}$? I am currently learning physics, and I was just wondering. So if using right hand rule, does that not mean cross product?

Equations I have found online

$\endgroup$
  • 1
    $\begingroup$ en.wikipedia.org/wiki/Angular_velocity Scroll down to "Particle in three dimensions", it's explained pretty well there. $\endgroup$ – Joshua Oct 26 '17 at 19:59
  • $\begingroup$ @Joshua So it's not r cross change in theta? Nevermind it makes sense thanks, but that still leaves the question why isn't it that. $\endgroup$ – EnlightenedFunky Oct 26 '17 at 20:46
  • $\begingroup$ Actually I don't understand what are you asking... I think you are confusing angular velocity and the total angular momentum... Please explain bettere where do you want to use the cross product $\times$ and what do you want to evaluate with $r\times \Delta \theta$... $\endgroup$ – Erik Pillon Oct 26 '17 at 22:08
2
$\begingroup$

Yes the use of the right hand rule is generally related to some cross product expression. However your use of the expression $\vec r\times\vec{\Delta\theta}$ is ambiguous. I assume that you imply the definition $\vec{r} \times \frac{d\vec{x}}{2\pi r}$ (as $d\theta = \frac{dx}{2\pi r}$ where $dx$ is the infinitesimal tangential displacement). But I would argue that $\vec{{\Delta}\theta}$ has the same direction as the angular velocity (vector), as $d\vec{\omega}=\frac{d\vec{\theta}}{dt}$. And then finally the direction of the angular velocity $d\vec{\omega}$ is perpendicular to the direction of the instantaneous velocity $\frac{d\vec{x}}{dt}$. This also answers the question in your comment @Joshua.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.