so I need to use the eq.: omega_pr = ##mgr / L_s## with r = 6cm.
But from how I imagine/see it, the rod pierces the disk through the middle so how does this create precession if the weight will not cause a torque then and thus a hortizontal change in angular momentum?
Thanks in advance!
should've examined it more properly basically.. the problem is not reallly difficult if I would've thought a bit simpler about it. i.e. from the basic principles up.
well.. that is exactly what I don't understand..
I didn't understand why the solution has used two integrals in such a way.. I haven't used a volume integral or smtn. I did differently by using a mass element and summing over that with one definite integral.
It's gravity, isn't it? That the one that creates the torque. If I were to take the 'rotation axis' in f.e. disk2, the gravitational force of disk 1 would still exert an angular impulse
Well, friction will occur. But I can't see how this creates non-conservation of AM
Or.. is it the torque/moment that the force of your hand will apply to the disks when smooshed together that creates an angular impulse?
Yep, that is what I did I expressed in terms of y^2 and then went ahead with that. But I just didn't understand how this solution came about basically.
yeah apologies it should have been 2 times the symbolic equation, just forgot to type it out.
Omg... totally forgot that... thanks a lot... I needed to use the moment of inertia about the center of mass basically..,.
so calculated, the moment of inertia for a rod about an axis at the end of the rod is I = 1/3 * M * L^2
here for case 1: arms to the side
I is calculated to be ##I = 0.224##
for case 2: arms stretched
## I = 1 / 3 * M * L^2 + M * d^2 ## with L = 0.6 m (length of rod) and d = 0.2 (dinstance from...