Recent content by Panda1175

Thank you! I think that is what I did above?

Thank you so much! Oh! So the ##\vec\omega## and ##\vec R## are cross-multiplied while the ##\vec v_{A}## is dot produced with itself? If that is the case, then ##\vec\omega## x ##\vec R = <r, 0, 0>## x ## <0, 0, \omega> = <0-0, -wr+0, 0-0> = <0, -wr, 0>## Letting ##v_{cm}## = <a, b, 0> ##\vec...

I don't understand the extra - in the second part of the equation, before the 2. From my understanding, the squaring of the first part of the first equation is so that the equations consist of magnitudes. I'm just not quite sure where the negative came from. If I had to guess, it would be, w...