Okay let's say for example I take it from point O.
ωx = ωi
ωy = pj
ωz = 0k
Therefore you would have the following for Ho:
Ho = (-Ixx - Iyx - Izx)ωx i + (-Ixy + Iyy - Izy)ωy j
From point O,
Ixoxo = IG + md2 (parallel axis theorem) = (1/4)mr2 + (1/3)mb2 + mh2
Due to symmetry,
Iyozo = Izoxo = 0...
Thanks for your reply.
I have tried the question by taking the angular momentum from the centre of mass, then using the parallel axis theorem.
e.g. I used Ho = Hg + (r x G)
That works perfectly fine, the answer was correct. But why is it not the case when I try to take it from point O?
So the following question is attached (There is another thread with the same question but no solution to what I am asking on there)
Now according to several solutions, apparently IYZ is equal to 0, and they reason this by saying that the geometry is symmetrical.
However when looking at the...