Recent content by sparkle123

Thank you to both! :D

Homework Statement How did this approximation come about? It doesn't seem like it's by L'Hopital's rule. Thanks! Homework Equations The Attempt at a Solution

Thanks a lot!

Wait dimension10, how do you solve it your way, after getting u = x/sqrt(m^2+x^2/c^2)?

Figured it out thanks! :))

I simply don't understand this conversion. I've only worked with du, dx, dy, etc. before. How do you change a d(γmu) into some kind of du? (This is in the contest of relativity, although this info is not essential i think) Thanks! :)

^bump

Isn't a the double derivative of r (so they are parallel)? Thanks again!

Then there is no angular momentum and only linear momentum? EDIT: actually i made an error with my question. the new image is attached. If ri` and ai` are parallel, shouldn't the cross-product be 0? So in the last line, the Ʃ miri` × A would be left instead?

Background: we're trying to show that the rate of change of angular momentum of an object about its center of mass (position given by R) is equal to the total torque about R. Why are the terms in red equal to 0? If anything, shouldn't the terms circled in in blue be equal to zero since the...

Oh thank you! I should brush up on matrices hehe :)

Okay, I got $$\left( \begin{array}{ccc} x^2 & xy & xz \\ xy & y^2 & yz \\ xz & yz & z^2 \end{array} \right)\mathbf{α}$$ How does $$r^2 - matrix$$ work? Thanks! :)

Hi I like Serena! :smile: Do we get: $$ (\mathbf{r} \cdot \mathbf{α})\mathbf{r} = ([x\ y\ z] \cdot \begin{bmatrix}x \\ y \\ z \end{bmatrix})\mathbf{α} = (x^2 + y^2 + z^2)\mathbf{α}$$ $$∴ mr^2 \mathbf{α} - m(\mathbf{r} \cdot \mathbf{α})\mathbf{r} = m(r^2 - x^2 - y^2 - z^2)\mathbf{α}$$ I still...

We have the representation of torque attached. The components of r are (x,y,z). Where did the matrix come from and how did we get the stuff in the matrix? (Basically I understand all the steps except the step from the 3rd line to the 4th line.) Thank you very much!

Thanks dikmikkel! :)