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I apologize if it seems foolish, I am a beginner.

Suppose I am given an object, whose moment of inertia along the x,z axes are known. Suppose it rotates around a tilted axis, say at angle $\theta$ with the x axis. Can I find out the moment of inertia by the following method?

I resolve the rotatory motion of the object into its x and z components- $\omega \cos \theta$ and $\omega \sin \theta$. I know its MOI along the x and z axis, so I find out the work done by the rotation along the x and z axis- $2W_x= I_x \omega^2 \cos^2 \theta$ and $2W_z= I_z \omega^2 \sin^2 \theta$. If the axis passes through the CM, I know the values of $I_x$ and $I_y$ readily. If not, I use the parallel axis theorem to find out $I_x$ and $I_z$. I think the total work done will be $W= W_x + W_y$, so from here I can find the total work done (I am not sure about this, but in analogy to translational motion, it makes sense. But the total work done is also $\frac12 I \omega^2$. So making them equal, I can find I (note- the $\omega^2$ term will cancel out. Will my method give the correct answer here, or do I just have to use the classic irritating method of integration?

To be precise, I claim $$I^2= I_x \cos^2 \theta + I_z \sin^2 \theta$$

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    $\begingroup$ this formula will be wrong in the case of an ellipse whose long axis is along the line $x=z$. Consider the extreme case of a line segment along $x=z$ whose moment of inertia is zero about that axis, yet non-zero about the $x$ or $z$ axes. Read about the moment of inertia tensor $\endgroup$ – Brian Moths Oct 8 '13 at 6:10
  • $\begingroup$ @NowIGetToLearnWhatAHeadIs I couldn't understand 1 word of it. Is there any easier method (because I have just started rotational mechanics? Or do I just have to perform integration? I hate integration >_< Still, thanks for the link D $\endgroup$ – User1111 Oct 8 '13 at 6:18
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    $\begingroup$ Unless there is something special about the problem, you have to do integration. $\endgroup$ – Brian Moths Oct 8 '13 at 12:45
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So you have an object whose principal mass moment of inertia values $I_1$, $I_2$ and $I_3$ you know for the body-centered coordinates

fig1

Now you place this object on a tilted axis and rotate it about z

fig2

and you want to find out the mass moment of inertia $I_{zz}$ about z.

You correctly identified the rotational speed on the body coordinates as $$ \begin{aligned} \omega_2 &= \omega \sin \theta \\ \omega_3 &= \omega \cos \theta \end{aligned}$$

and that these cause the following components of angular momentum

$$ \begin{aligned} L_2 &= ( I_2 \sin \theta) \omega \\ L_3 &= (I_3 \cos \theta ) \omega \end{aligned}$$

except you went with energy instead of momentum, which might have obscured things because energy is a single scalar, and momentum is a vector and thus easier to understand what the terms represent (what orientation things are defined in).

The last step is to rotate the momentum components back into the world coordinate system

$$ \begin{aligned} L_{yz} &= L_2 \cos \theta - L_3 \sin \theta \\ L_{zz} &= L_2 \sin \theta + L_3 \cos \theta \end{aligned}$$

which leads to the value of MMOI once the rotational speed is factored out

$$ \begin{aligned} L_{y} &= \underbrace{\left( (I_2-I_3)\sin \theta \cos\theta \right)}_{I_{yz} } \omega \\ L_{z} &= \underbrace{\left( I_2 - (I_2-I_3)\cos^2\theta \right)}_{I_{zz} } \omega \end{aligned}$$

So your answer is

$$ \boxed{ I_{zz} = I_2 - (I_2-I_3)\cos^2\theta } $$


But it is a lot easier for me to do the transformation using the following matrix product, because it will produce all the components of the mass moment of inertia tensor.

Given the body inertia values as the following diagonal 3×3 matrix $$I_{\rm body} = \pmatrix{I_1 & & \\ &I_2 & \\ & & I_3} $$

and the rotation about the x-axis by the angle $\theta$ as a 3×3 tranformation matrix $$ R = \pmatrix{1 & 0 & 0 \\0 & \cos \theta & -\sin \theta \\ 0 & \sin \theta & \cos \theta} $$

then the mass moment of inertia tensor in the world coordinates is given by the congruent transformation $I_{\rm world} = R I_{\rm body} R^\top$

$$ I_{\rm world} = \pmatrix{1 & 0 & 0 \\0 & \cos \theta & -\sin \theta \\ 0 & \sin \theta & \cos \theta}\pmatrix{I_1 & & \\ &I_2 & \\ & & I_3} \pmatrix{1 & 0 & 0 \\0 & \cos \theta & -\sin \theta \\ 0 & \sin \theta & \cos \theta}^\top $$

$$ \boxed{ I_{\rm world} = \pmatrix{ I_1 & 0 & 0 \\ 0 & I_3 + (I_2-I_3) \cos^2 \theta & (I_2-I_3)\sin \theta \cos \theta \\ 0 & (I_2-I_3) \sin \theta \cos \theta & I_2+(I_3-I_2)\cos^2 \theta } }$$

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  • $\begingroup$ Just a suggestion, I think this answer would be much easier to understand if you have an image where you super impose the body's attached coordinated and the other coordinate axes $\endgroup$ – Buraian Feb 28 at 8:02
  • $\begingroup$ That is the reason I have the cutout on the disk because it makes it easy to orient the body coordinates in space from the view. Also to illustrate a general case where the three principal axes have different values. $\endgroup$ – John Alexiou Feb 28 at 16:03
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I have learnt that in 3D space if axis of rotation makes angle $\alpha$, $\beta$ and $\gamma$ from $x$, $y$ and $z$ axis respectively then moment of inertia can be calculated by $$I^2=(I_x\cos\alpha)^2+(I_y\cos\beta)^2 + (I_z\cos\gamma)^2$$ I hope that helps!

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  • $\begingroup$ Can you elaborate or provide a reference because I think the above is wrong. Mass moment of inertia undergoes a congruent transformation when rotated and the above is just a simple transformation. The full 3D formula is $$ I = \mathrm{R} I_{\rm body} \mathrm{R}^\top$$ with $\mathrm{R}$ the rotation matrix. $\endgroup$ – John Alexiou Oct 31 '20 at 4:16
  • $\begingroup$ Another reason I think this is wrong is because a 180° rotation in any axis should yield the same result, and your formula would introduce a negative MMOI component which is non-sensical. $\endgroup$ – John Alexiou Oct 31 '20 at 4:40
  • $\begingroup$ Feels like a dot product, the cosines look like the direction cosines $\endgroup$ – Buraian Oct 31 '20 at 7:34
  • $\begingroup$ I have learnt that this derivation comes from the resolution of angular momentum into its respective components. $\endgroup$ – Arnav Mishra Oct 31 '20 at 8:18
  • $\begingroup$ @ArnavMishra - I still doubt the corrected equation. $\endgroup$ – John Alexiou Oct 31 '20 at 22:12

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