# Finding the product moment of inertia of this cylinder

1. Jun 2, 2015

### 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 IYZ 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?

2. Jun 2, 2015

### haruspex

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

3. Jun 2, 2015

### anchonee

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?

4. Jun 2, 2015

### anchonee

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 - Izxx i + (-Ixy + Iyy - Izyy j

From point O,
Ixoxo = IG + md2 (parallel axis theorem) = (1/4)mr2 + (1/3)mb2 + mh2
Due to symmetry,
Iyozo = Izoxo = 0
Iyoyo = (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 Izoyo = (-b/2)×h×m
This appears to also be valid when I use Izy = ∫zy.dm

Therefore that would result in my answer being:

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

Ho = mω(r2/4 + b2/3 + h2) i + ((1/2)mr2)p j

..any ideas?

Last edited: Jun 2, 2015
5. Jun 2, 2015

### haruspex

The p rotation is not about an axis through O.

6. Jun 2, 2015

### anchonee

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

7. Jun 2, 2015

### haruspex

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

8. Jun 2, 2015

### anchonee

perfect. thank you!