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# Moments of Inertia watch

1. If i want to find the inertia of the beam above as it rotates about the z-axis, what sides do i have to use? Would it be (mass/12)(a^2 + c^2)? or (mass/12)(b^2 + c^2)? ... or neither!?
2. You need to find the perpendicular distance of the centre of mass away from the z-axis, square it and multiply it by the mass. That is, .

You haven't given us enough information to solve the problem though so at the moment this is the best I can do.
3. (Original post by nuodai)
You need to find the perpendicular distance of the centre of mass away from the z-axis, square it and multiply it by the mass. That is, .

You haven't given us enough information to solve the problem though so at the moment this is the best I can do.
I know i have to use parallel axis theorem like you said, but to do that i need the inertia of beam first, and then add that to mr^2 so i can find the inertia about z-axis.
I am trying to find the inertia of the beam.
Lets say the centre of mass of the beam was located on the z-axis (i.e. mr^2 not needed), and it was rotating about the z-axis at the angle it is in the image i drew. What would the inertia be then?
4. This is what i mean:

If the box spins around the z-axis, then which one of those equations do i use to find its inertia?
5. If the side of length c is parallel to the z axis, then 1/12 m (a^2+b^2), but if it's not parallel, then none of them; you are into the realms of tensors which is outside my areas of expertise.
6. (Original post by ghostwalker)
If the side of length c is parallel to the z axis, then 1/12 m (a^2+b^2), but if it's not parallel, then none of them; you are into the realms of tensors which is outside my areas of expertise.
what if length c is perpendicular to the z-axis and length a was parallel to it, then would it be (b^2 + c^2)?
7. (Original post by Desert Eagle)
what if length c is perpendicular to the z-axis and length a was parallel to it, then would it be (b^2 + c^2)?
Yep, with the 1/12m.

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