Recent content by satisf

Allright, the error was of course that for a disk I=\frac{MR^2}{2}. I relied to blindly on the formulaium i got. I=mr² is for point masses, so for a disk you get following equation for I: [ŧex]I=int_0^\pi int_0^R \frac{M}{\pi r^2} r^2 r \mathrm{d}r\mathrm{d}\theta = \ldots =...

Homework Statement Given is the setup in the attachment, I think the values speak for itself. Additionally, there is a friction coefficient of 0.36 that works in on the 2 blocks. So in short: 2 masses (1 one a slope), connected through a pulley, that does have friction and to complete...