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I understand that there are proofs (e.g. proof, another proof) of why the angular momentum about two points for an object is the same. However, could someone give an intuitive explanation of why this is? I work far better with something I can visualize than with variables and formulae.

A related question - Why doesn't the moment of inertia change when measuring momentum about different axes? Shouldn't the parallel axis theorem be used?

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We generalize the result from the second answer you linked:

$$\vec{L}=\sum_i \left(\vec{r}'+\vec{{r}_i}' \right)\times \vec{p_i}=\vec{r}' \times \sum_i \vec{p_i}+\sum_i \vec{{r}_i}'\times\vec{p_i}=\vec{r}' \times \sum_i \vec{p_i}+\vec L' $$

Now $\vec{r}'$ is some arbitrary vector, not just the vector to the center of mass. Thus when the system is moving and the sum over the momenta is not zero, the angular momentum depends on the choice of the origin of the coordinate system. When $\sum \vec p_i$ is zero then we have on the other hand $$\vec L = \vec L'. $$

The intuition behind this is the following: suppose you have one of those toys where you can balance a thing with a magnet such that it moves freely (google "Levitron"). Then you spin the freely moving object. Since it's only rotating and not moving the sum of the momenta is zero. From your point of view you then calculate $\vec L$.

Now you move your chair one meter to the left ($\vec r_i = \vec r_i' + \vec r'$) and calculate $\vec L'$. If $$\vec L \neq \vec L'$$ then this would mean that a torque had been applied on the object. But you didn't touch the object, you just moved the reference system. This is why $\vec L=\vec L'$ for systems which are at rest: otherwise you would apply a torque just by changing the origin of the reference system.

Edit (inertia part):

I think it would help to note that the moment of inertia is per definition always defined in the frame where the center of mass is at rest, so that the origin of the body-fixed coordinate system is the center of mass. Therefore it is a "property of the body", so to speak.

The parallel axis theorem is needed when the origin of the body-fixed frame is for whatever reason not at the center of mass, but at some other point. With the parallel axis theorem you can then calculate the moment of inertia about the center of mass: $$I′=I+md^2$$, from this you can calculate $I$.

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  • $\begingroup$ But can you explain why the inertia doesn't change? $\endgroup$ – ws04 Jan 26 '16 at 15:09
  • $\begingroup$ When I wrote my answer I wasn't aware of the inertia part. ^^ Wait a second. $\endgroup$ – user42076 Jan 26 '16 at 15:19
  • $\begingroup$ (it's been quite a long second) $\endgroup$ – ws04 Jan 26 '16 at 16:36
  • $\begingroup$ Alright, well that helps. Maybe add that to the answer? $\endgroup$ – ws04 Jan 27 '16 at 16:04
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Intuitively the motion of a rigid body is split between, translation of the center of mass, and rotation about the center of mass. The intrinsic inertia about those two motions are the mass and the mass moment of inertia.

To rotate a body about any other point away from the center of mass means that the center of mass has to translate in addition to the rotation. This means the effective mass moment of inertia about any other point has to include the effect of the linear motion of the center of mass. This increases the value of the MMOI as we are moving away from the center of mass by using the parallel axis theorem.

In the world of momentum the intrinsic momentum of a rigid body is the pair of vectors $$\vec{P} = m \vec{v}_{cm} \\ \vec{L}_{cm} = I_{cm} \vec{\omega}$$

If the center of mass is rotating about another point, the mass motion is $$\vec{v}_{cm} = \vec{\omega} \times \vec{r}$$ Here $\vec{r}$ represents the position of the center of mass relative to the rotation center.

The linear momentum is defined then as $$\vec{P} = -m \vec{r} \times \vec{\omega}$$

The angular momentum about the point of rotation has to include the linear momentum of the center of mass by $$\vec{L} = \vec{L}_{cm} + \vec{r} \times \vec{P} $$ $$ \vec{L} = I_{cm} \vec{\omega} - m \vec{r}\times \vec{r}\times \vec{\omega} $$ $$ \vec{L} = \left(I_{C}+m\|\vec{r}\|^{2}\mathrm{1}_{3\text{×}3}\right)\vec{\omega}$$

The above considers $\vec{r}\cdot\vec{\omega}=0$ since rotation about points parallel to the rotation axis do not change the problem. Hence the parallel axis name

This a form of the parallel axis theorem giving us the angular momentum about the center of rotation

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