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In the first picture, there is a homogeneous metal rod of length $2L$ and mass $M$. If it rotates around a normal axis passing by $O$ (which is the center of gravity), then its moment of inertia is:

$$I_{1}=\frac{M(2L)^2}{12} = \frac{ML^2}{3}.$$

After bending the bar on $O$, will the moment of inertia for rotations around a normal axis passing by $O$ be the same, or will it dependent on the angle $\theta$ that the two segments of length $L$ make?

enter image description here

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  • $\begingroup$ integrate $r^2dm$ and you will find out $\endgroup$
    – aaaaa says reinstate Monica
    Commented Mar 14, 2015 at 0:53
  • $\begingroup$ It wouldn't change $\endgroup$
    – Sigma
    Commented Mar 14, 2015 at 1:50

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The answer depends on the direction of the axis of rotation.

If the axis is normal to the plane, then you have the same amount of material the same distance from the axis of rotation as before - and thus the moment of inertia about that axis would be unchanged.

However, if the axis of rotation you consider is in the plane of the paper, the answer will change. For the vertical axis in the plane, the projected mass per unit length will increase while the apparent length of the rod is shortened: in other words, looking at the setup from the top of the V, it looks like you have a shorter rod with more mass per unit length and the moment of inertia about that axis will decrease; similarly, the moment of inertia about the horizontal axis in the plane (originally along the axis of the unbent rod) will increase (since by the perpendicular axis theorem, their sum must equal the moment of inertia about the third, normal-to-the-paper axis).

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    $\begingroup$ I'm not convinced that the $I_{zz}$ term will change, if we definte the $z$-axis to be the one that is normal to the screen, and passing through point $O$. However, I do think the bend would cause changes in the off-diagonal terms of the moment of interia tensor... $\endgroup$
    – Sean
    Commented Mar 14, 2015 at 2:21
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    $\begingroup$ @Sean I think Floris is using the in-screen vertical axis. Of course, the question is rather unclear on this point. $\endgroup$
    – user10851
    Commented Mar 14, 2015 at 2:36
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    $\begingroup$ @ChrisWhite you are right my answer is for a vertical axis in the plane of the paper. Indeed for an axis normal to the paper the moment of inertia is independent of $\theta$ $\endgroup$
    – Floris
    Commented Mar 14, 2015 at 2:39
  • $\begingroup$ @Chris thanks for pointing out the imprecision, I edited the question. $\endgroup$
    – El Cid
    Commented Mar 14, 2015 at 4:37

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