Finding the product moment of inertia of this cylinder

[anchonee]

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 I_YZ is equal to 0, and they reason this by saying that the geometry is symmetrical.

However when looking at the YZ plane, the geometry is not symmetrical about either axis, so therefore ∫yz.dm ≠ 0?

 

[haruspex]

So the following question is attached (There is another thread with the same question but no solution to what I am asking on there)
View attachment 84362
View attachment 84363

Now according to several solutions, apparently I_YZ is equal to 0, and they reason this by saying that the geometry is symmetrical.

However when looking at the YZ plane, the geometry is not symmetrical about either axis, so therefore ∫yz.dm ≠ 0?

As I posted on the other thread, it may be that I_yz is being used to refer to the MoI about the cylinder's mass centre. The parallel axis theorem can then be used in conjunction.

 

[anchonee]

As I posted on the other thread, it may be that I_yz is being used to refer to the MoI about the cylinder's mass centre. The parallel axis theorem can then be used in conjunction.

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 H_o = H_g + (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?

 

[anchonee]

Okay let's say for example I take it from point O.

ω_x = ωi
ω_y = pj
ω_z = 0k

Therefore you would have the following for H_o:
H_o = (-I_xx - I_yx - I_zx)ω_x i + (-I_xy + I_yy - I_zy)ω_y j

From point O,
I_xoxo = I_G + md² (parallel axis theorem) = (1/4)mr² + (1/3)mb² + mh²
Due to symmetry,
I_yozo = I_zoxo = 0
I_yoyo = (1/2)mr²

Now technically from point O, the mass is not symmetrical about either axis in the YZ plane.
I would have taken it as I_zoyo = (-b/2)×h×m
This appears to also be valid when I use I_zy = ∫zy.dm

Therefore that would result in my answer being:

H_o = mω(r²/4 + b²/3 + h²) i + ((1/2)mr² - (b/2)×h×m)p j

The correct answer however is
H_o = mω(r²/4 + b²/3 + h²) i + ((1/2)mr²)p j

..any ideas?

 

[haruspex]

The p rotation is not about an axis through O.

 

[anchonee]

The p rotation is not about an axis through O.

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

 

[haruspex]

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

In the j direction, the axis of rotation is through the mass centre, so there is only an $Iyy$ term.

 

[anchonee]

In the j direction, the axis of rotation is through the mass centre, so there is only an $Iyy$ term.

perfect. thank you!