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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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