I have a question regarding angular momentum and torque in the following example. This is not a homework problem, I just believe understanding some parts of this particular problem would help me understand the two concepts better.

I have a rod of length $l$ oriented at an angle $\phi$ about an axis that passes through its center of mass. The rod is rotating about this axis, and I'm supposed to find the magnitude of the angular momentum and the torque.

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In this example, the given angle is $\pi/4$, but I'm looking for a general solution.

Using the inertia tensor, I can easily find the moment of inertia of this rod, in the direction of the axis. let us have the $x$ axis along the rod, and the $y$ axis perpendicular to it. Let us have the axis of rotation along some direction $\vec{a}$. We can see, that $\vec{a}=a\cos\phi\hat{i}+a\sin\phi\hat{j}$.

If $I$ denotes the inertia tensor, and $I_a$ represents the moment of inertia about this axis $\vec{a}$, then I can easily write :

$$I_a=( cos\phi, \sin\phi) I ( cos\phi, \sin\phi)^T $$

In the case of our rod, the moment of inertia along the length ($x$ axis) is $0$. So, we can just say :

$I_a=I_y \sin^2\phi=\frac{1}{12}Ml^2\sin^2\phi$

So, I think I've managed to find the moment of inertia about the $a$ axis.

Now I'm confused regarding how should I calculate the angular momentum in this case.

On one hand, I have $$\vec{L}=I\vec{\omega}$$

So, I can simply multiply $\omega\hat{a}$ to the moment of inertia along $\hat{a}$.

Using this, I get $$\vec{L}=\frac{1}{12}Ml^2\omega \sin^2\phi\hat{a}$$

However, I could also have done the following :

$$L=I_x\omega_x \hat{i}+I_y\omega_y\hat{j}$$ Here $I_x=0$ and $\omega_y=\omega\hat{a}.\hat{j}=\omega\sin\phi$

In this case, I get: $$\vec{L}=I_y\omega_y=\frac{1}{12}Ml^2\omega\sin\phi\hat{j}$$

I'm inclined to believe that the second one is correct, but I have no idea why the first method seems to be wrong here. Any explanation would be highly appreciated.


1 Answer 1


Your projection of the inertial moment into the rotational axis, $\hat a$, is not correct.

Starting with your inertial moment in your $\hat i$ (along the rod) and $\hat j$ (perpendicular to the rod) coordinate. $$ \bf{I} = \begin{bmatrix} 0 & 0 \\ 0 & \frac{1}{12}Ml^2 \end{bmatrix} $$

The transform to the $\hat a$ and $\hat b$ (perpendicular to $\hat a$) by a rotation matrix $\bf R$: \begin{align} \bf I_a & = \bf R^T \bf I \bf R \\ &= \begin{bmatrix} \cos \phi & \sin\phi \\ -\sin\phi & \cos\phi \end{bmatrix} \begin{bmatrix} 0 & 0 \\ 0 & \frac{1}{12}Ml^2 \end{bmatrix} \begin{bmatrix} \cos \phi & -\sin\phi \\ \sin\phi & \cos\phi \end{bmatrix} \\ &= \frac{1}{12}M l^2 \begin{bmatrix} \sin^2 \phi & \sin\phi\cos\phi \\ \sin\phi\cos\phi & \cos^2\phi \end{bmatrix} \end{align}

Then angular momentum in $\hat a$ and $\hat b$ coordinate become \begin{align} \vec L_a & = \bf I_a \, \vec \omega_a \\ &= \frac{1}{12}M l^2 \begin{bmatrix} \sin^2 \phi & \sin\phi\cos\phi \\ \sin\phi\cos\phi & \cos^2\phi \end{bmatrix} \begin{bmatrix} \omega \\ 0 \end{bmatrix}\\ &=\frac{1}{12}M l^2 \, \omega \sin\phi \left\{ \sin\phi \,\hat a + \cos\phi \,\hat b \right\} \end{align}

You may also check that $\vec L_a = {\bf R^T} \vec L $.

  • $\begingroup$ Suppose I take a line element $dx$ along the rod, at some distance $x$ from the center along the rod. I know, moment of inertia about any axis is $I=\int dm r^2$, where $r$ is the perpendicular distance from the axis. In our case, $dm=\frac{M}{l}dx$ and $r=x\sin\phi$. Integrating over this, I'd get $I_a=\frac{1}{12}Ml^2\sin^2\phi$ $\endgroup$ Feb 2, 2022 at 0:44
  • $\begingroup$ This seems to be different from what you have done, can you point out where I'm going wrong here? I've seen many professors derive the moment of inertia of rods, discs, etc about tilted axes, using integration as I've done here, or by defining the tilted axis as $|n\rangle$, and the moment of inertia about this new axis as $\langle n|I|n\rangle$ $\endgroup$ Feb 2, 2022 at 0:48
  • $\begingroup$ (sorry to use the bra-ket notation to denote the transpose of the unit vector along the new axis ) $\endgroup$ Feb 2, 2022 at 0:48
  • $\begingroup$ You can do that only for the principle axis of rotation. For this problem. the angular momentum is not a constant of time. It precesses around the $\hat a$ axis. Your projection is only a component of the total angular momentum. $\endgroup$
    – ytlu
    Feb 2, 2022 at 0:50
  • $\begingroup$ Because all the inertial moment calculated in the text book are based on a principle axis, a symmetric axis of rotation. $\endgroup$
    – ytlu
    Feb 2, 2022 at 0:52

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