Recent content by Exath

Ah, that's what i was looking for, i was looking at the forces the same way i was looking at torques, that's obviously wrong lol

So I'm looking at a problem that involves a situation that looks like this the cylinder rolls without gliding. And there are these following equations that apply to it (1) mg - T = ma (for the block hanging vertically) (2) T + f = Ma (for the cylinder f = friction force, T = String force) (3)...

Thanks for the help i got the rest from there :D

Ah, the inertia should be Icm + Md^2 (d = distance between new axis and center of mass) = 1/2*MR^2 + M*(R + L)^2

Ok, so what's left is the inertia for the disk, how would i go about finding that?

It also falls like the disk, but counting from the center of mass it would be m*g*(L/2)?

Homework Statement [/B] A thin cylindrical rod with the length of L = 24.0 cm and the mass m = 1.20 kg has a cylindrical disc attached to the other end as shown by the figure. The cylindrical disc has the radius R = 8.00 cm and the mass M 2.00 kg. The arrangement is originally straight up...