# conceptual doubt in method to find moment of inertia about an axis

I asked this question before about whether I can take a component of angular velocity along another axis and say that the body spins about that axis with that component.

Now I have another doubt:

Consider a rigid body having an inertia $I_0$ and angular velocity $\omega_0$ about some axis. So according to the answer to my question above, I can say that the object has an angular velocity $$\omega_0\cos\theta$$ about an axis inclined at $\theta$. And I can also say that the angular momentum about that axis will be $$I_0\omega_0\cos\theta$$ by taking the component of the angular momentum about the original axis, $I_0\omega_0$ along the axis at $\theta$.

So why can't I say that the inertia about that axis will be $$I = \frac L{\omega}=\frac{I_0\omega_0\cos\theta}{\omega_0\cos\theta} = I_0$$

Where is the problem in this?

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You need the entrire tensor of inertia's as well as the vector of angular velocities to do what you are trying to do. –  ja72 Jul 29 '13 at 16:37
Also note that angular momentum is a vector and not a scalar so direction is important as well as magnitude. Best to consider all the components at the same time. –  ja72 Jul 29 '13 at 17:06

well if you look through your argument you will see that the second equation you wrote is wrong.angular momentum is moment of inertia times angular velocity and if you see this carefully you will realise that you have taken moment of inertia to be the same as before and then you prove that moment of inertia is the same. it is very important to realize that physics is not just a bunch of equations and you punch in the values of different quantities and grind out the answers.you must realise that every equation represents a real physical situation .Take for instance your above question ,you have proved that moment of inertia does not depend upon the orientation of the axis which is of course false.

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the second equation comes from taking the component of the angular momentum $I_0\omega_0$ along the axis inclined at $\theta$ which gives me $I_0\omega_0\cos\theta$. –  udiboy1209 Jun 29 '13 at 7:49
Well i guess you must be knowing that angular momentum is a vector and angular velocity is also a vector so you can always take components as you wish but in general angular momentum is not equal to moment of inertia times angular velocity . –  Sahil Chadha Jun 29 '13 at 7:59
so when is $L=I\omega$ valid? –  udiboy1209 Jun 29 '13 at 8:00
It is only when you have fixed axis rotation.The general relationship involves tensors. –  Sahil Chadha Jun 29 '13 at 8:01

Here are the fundumentals you are missing. Consider a rigid body with (fixed) inertia tensor about the body axes as

$$I_{body} = \begin{bmatrix} I_{11} & I_{12} & I_{13} \\ I_{12} & I_{22} & I_{23} \\ I_{13} & I_{23} & I_{33} \end{bmatrix}$$

If the 3D orientation of the rigid body is given by 3 unit vectors of the inertial axes $\hat{u}_x$, $\hat{u}_y$ and $\hat{u}_z$ then the 3x3 rotation matrix is $$E = \left[ \begin{matrix} \hat{u}_x & \hat{u}_y & \hat{u}_z \end{matrix} \right]$$ and the inertia tensor in the world coordinates is (congruent transformation)

$$I = E \, I_{body} \, E^\top$$

If the body is rotating about an axis $\hat{k}$ by $\Omega$ then $\vec{\omega} = \Omega\,\hat{k}$ and the angular momentum vector is

$$\vec{L} = I\, \vec{\omega}$$

Now you want to change coordinates such that $\hat{k}^\star = R\, \hat{k}$ where $R$ is a 3x3 rotation matrix. To maintain the above relation you need

$$\vec{\omega}^\star = R\,\vec{\omega}$$ $$\vec{L}^\star = R\,\vec{L} = R\, I \vec{\omega} = \left(R\, I R^\top\right) \vec{\omega}^\star = I^\star \vec{\omega}^\star$$ $$I^\star = R\, I R^\top$$

Which is again the congruent transformation. So to define the inertia at a different axis you need to transform the components as shown above in order to maintain Newton's Laws intact.

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I'm not so familiar with tensor math. I got this doubt because of a solution my professor gave about calculating the inertia of a cube about the body diagonal. What he did was assumed that the cube had an $\vec \omega$ about the body diagonal. So the angular momentum vector is $I_d\vec\omega$, which can be broken into three components about the three edges of the cube as $I_d\vec\omega=I_e\frac{\omega}{\sqrt 3}\hat i+I_e\frac{\omega}{\sqrt 3}\hat j+I_e\frac{\omega}{\sqrt 3}\hat k$. Then he equated the moduli and said $I_d=I_e$. Is this a correct method? –  udiboy1209 Jul 30 '13 at 12:22
I'm getting the same answer using integration so I thought this might be correct. –  udiboy1209 Jul 30 '13 at 12:23