i'd split the gravity study and centrifugal study and figure out a way to superimpose the stresses later. the reason i say this is because you really have 2 gravity cases that are "worst case" one when the part is in the orientation you have now, and one when it is in the other direction.
i'm not sure i totally understand where you're getting hung up on the restraint but it sounds like you want to see how the bolts behave and since you fix the cylinder they are held together with that instead of the bolts. try adding the next part in the assembly (the shaft). but i still think you might not get exactly what you want. start there and let's see what you get.
Thanks for the response.
I have one other worst case which is when the system is rotated 90 degrees, which creates tension in one of them and compression in the other.
Your are right, what I'm supposed to test are the bolts... but also the parts themselves.
How can I make a study with the bolts with centripetal force provided from the parts?
If do calculations by hand is true to assume the centripetal aceleration at the center of the mass for the bolts?
i'm not totally following what you are suggesting but think about it this way, centripetal force is a body force that is applied to the nodes in a direction perpendicular to the axis that you choose. if you really wanted to do a test that was similar, you could pull on the part with a force close to the base in an equal and opposite way of the centriptal force of the component to understand what is happening at the bolts.
So for the bolts I think we have the same idea, for example:
if both parts weight 100 kg I can calculate the centripetal force and the center of mass of which one of those parts?
for example assuming a velocity of 1 m/s and a distance from the shaft equal on both parts to the center of mass of 1 m the force on the bolts would be
2*100*1^2/1=200 N (total bolts).
But if I want to calculate the tension in the parts the centripetal force applied will vary along the "radius" so the far end will have to support less than the closer nodes to the shaft...
It´s "like" a triangle supporting is own weight, we can calculate him so the tension is always the same along the heigh. This is just an example if I confused you with this don't care about this!
Jared thank you very much again for the help!