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I have a quick question about the parallel axis theorem.

Let's say I have a box-like object rotated about the following pivot point

enter image description here

I was wondering, by applying the parallel axis theorem, would the moment of inertia about the pivot point be the following,

$$M = I_{cm} + M_d^2+M_h^2$$

where $I_{cm}$ is the moment of inertia about the box's centroid and $M$ is its mass.

Or, does the parallel axis theorem not hold true for two axis shifts?

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  • $\begingroup$ Rotate the system by arctan(d/h) and what do you get? $\endgroup$ – Phil Sweet Jul 19 '19 at 17:45
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Moment of inertia is based on both a point of reference and an axis of rotation.

So the answer to your question depends on the axis you're asking about.

For the moment of inertia around the horizontal axis, you only care about the vertical shift (and visa-versa for the vertical axis, of course):

$$I_{xx} = I_{cm} + M\cdot h^2$$

For any other axis, things get much more complicated, as suggested by @kamran's answer.

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  • $\begingroup$ Thanks for your help on this! $\endgroup$ – Justin Jul 19 '19 at 15:29
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Parallel axis would apply if you first calculate the $I$ of the box about an axis perpendicular to a line connecting the pivot point to the center of mass of the box.

This I is not the same as the one fore a thin plate, the box has volume.

It gets a bit complicated, having to triple integrate in polar coordinate the $\text{d}v \cdot r^2$.

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  • $\begingroup$ Thank you for your help with this problem! $\endgroup$ – Justin Jul 19 '19 at 15:29
  • $\begingroup$ He has this, it's Icm in his formula. He just wants to know if he can decompose the radius into components along arbitrary axes the way he has. Simple chain rule or algebra is all it takes. $\endgroup$ – Phil Sweet Jul 19 '19 at 18:15
  • $\begingroup$ @PhilSweet, the Icm mus be the one about the actual axis of rotation, not the axis of symmetry. $\endgroup$ – kamran Jul 19 '19 at 18:25

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