Recent content by lsie

Ok, I've worked out equation number 4. My solution was to square equations one and two, add them together, and reduce them down with the help of equation three. This worked well, because squaring Equation One produces a few squared cos and sin values that can be converted to '1,' some x and y...

Got it. I saw some google-images of a hoop with some thick-ish edges and it threw me off. Cheers!

That worked! Thanks. I was certain the solution would be more elegant. Out of curiosity, are there good reasons why we can neglect the difference between r and R? Surely there are hoops where r and R are (how's that for a tongue twister) significantly different?

I've worked out how to derive the formulas for a solid cylinder and a solid sphere rolling down a hill. E.g., for a cylinder: Emech = KE + PE mgh = 1/2 mv^2 + 1/2 Iw^2 gh = 1/2 v^2 + 1/2 (1/2r^2) v^2/r^2 gh = 3/4 v^2 v^2 = 4/3 gh I then performed a derivative with respect to time and found a...

I've worked on this and still haven't figured it out. I'm wondering if it's basically a matter of substitution tetris or if there's something I'm missing. Here's what I've tried: Substituting equation one (V1 = V1'cos theta + V2'cos phi) and equation two (0 = V1' sin theta - V2' sin phi)...

You're absolutely right that the OP wrote equation 2 down wrong. I'll be back to look over what you've written tomorrow, when I have more energy. Thank you for the help!

I think, for equation 6, I need to end up with something like sin (theta) - cos (theta) = 0. But I can't seem to get there. Ditto for equations 4 and 5. I've tried all kinds of substitutions and the equations seem to get longer and longer without resolving.

Did anyone manage to figure this out? I'm hitting a brick wall.