Recent content by anchonee

perfect. thank you!

So how would it change what I did in the above methodology?

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...

Why is Iyz = 0? When looking at the YZ plane the geometry is not symmetrical about either axis?