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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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udiboy1209
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2 Answers2

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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.

Sahil Chadha
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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.

John Alexiou
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