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Question: in the simple equations $\tau = r \times f$ and $L = r \times p$, what is the origin of the vector $r$? In many textbook discussions on rotation of a rigid body about a fixed axis, $r$ is taken to be the closest point on the axis of rotation. Hence, $r$ is always perpendicular to the axis of rotation and has no component in the direction of $\omega$. However, in other discussions (e.g., Halliday and Resnick 5th edition Section 12-3), $r$ originates from a constant fixed point. I've heard this latter version called "generalized torque" and "generalized angular momentum." Another way of putting this question is: do you take angular momentum about a fixed point or about an axis of rotation?

Let me make this very concrete. Consider a rigid object made of two 1kg point masses; at t=0 one point mass is at the point (2,0,0) and the other is at (2,0,4). The object rotates around the z axis with $\omega$=3 rad/s. At t=0, both thus have velocity $v=r \times \omega = (0,-6,0)$. In the "non-generalized" scheme, the angular momentum about the z axis would be $L = r_0 \times mv_0 + r_1 \times mv_1$ = $(2,0,0) \times (0,-6,0) + (2,0,0) \times (0,-6,0)$ = $(0,0,-24)$. However, in the "generalized" scheme, the angular momentum about the origin x=y=z=0 would be $L = r_0 \times mv_0 + r_1 \times mv_1$ = $(2,0,0) \times (0,-6,0) + (2,0,4) \times (0,-6,0)$ = $(24,0,-24)$. The difference between the two answers stems from using (2,0,0) for $r_1$ in the first case and (2,0,4) for $r_1$ in the second case.

So which is it? Are both correct, but in different circumstances? Is it legal to get torque and angular momentum either about a point or about an axis? Is the former more general? If so, then how come an introductory textbook like H&R mentions both, and somewhat-more-advanced textbooks (Symon's Mechanics and the Feynman Lectures) mention (as far as I've found) only computing $r$ from an axis of rotation rather than from a point?

If you really can choose the origin of the vector $r$ to be in either place, then it would seem to me that most theorems about angular momentum (of which there are many) should have two versions, which is of course not the case.

I want to tell myself that the answer is as simple as "sometimes objects are rotating around an axis and sometimes they're not, and you use generalized torque and angular momentum when objects are not just rotating around an axis." But that doesn't seem quite right, either.

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  • $\begingroup$ What do you mean by the "generalized" torque? Torque is always defined by $\mathbf{r} \times \mathbf{F}$, with both $r$ and $F$ being vectors. It doesn't make a difference whether it's in 2D or 3D, it's still defined in the same way. $\endgroup$ – Kyle Arean-Raines Jun 24 '15 at 13:21
  • $\begingroup$ @Kyle: yes, certainly torque is $r \times f$. However, the question then is where the vector $r$ originates. Is it a single fixed point (as indicated by H&R section 15-3), or does $r$ originate from the nearest point on an axis of rotation (as Symon indicates in, e.g., his Section 5-2)? $\endgroup$ – JoelG Jun 24 '15 at 14:55
  • $\begingroup$ I think I might understand what you're asking. $\mathbf{r}$ is a vector that points out radially from the rotation point or axis of the object to the point of contact where you're applying the torque. So in the case of a circular disc in the $xy$-plane that is free to rotate about its center clockwise or counterclockwise, $r$ is a vector in the $xy$-plane that points from the center of the circle, the point $(0,0,0)$ to the point at which you're applying the torque, say $(D/2,0,0)$ with $D$ being the diameter. So then $\mathbf{r}=D/2 \mathbf{\hat x}$. Does that make sense? $\endgroup$ – Kyle Arean-Raines Jun 24 '15 at 18:25
  • $\begingroup$ @Kyle: I've reworded the question again to try to make myself more clear. Your response makes sense, but doesn't quite answer the question I'm asking. So I've included an example that's a bit more complex than the one you've picked. Thanks. $\endgroup$ – JoelG Jun 24 '15 at 19:50
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Torque is the angular force (the moment) - i.e. $\tau = r \times F$, not $r \times p$, which is angular momentum.

When you are dealing with a rotating or rotational system (like a disc on an axis), then when the torque and angular momentum are always aligned along that axis, it is much easier to use the magnitude of the torque and the magnitude of the angular momentum than to keep using the vector forms. But it is merely a device to simplify the maths, there is no separate non-general form of either measure.

Perhaps the textbooks are simplifying things by using 2D instead of 3D, thus the $r$ vector is w.r.t. a point instead of an axis. As long as everything aligns, the maths is the same.

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  • $\begingroup$ @PhilHL Sorry for the typo, using "p" rather than "f" for force. I've fixed that now. But I'm not convinced that it's merely a 2D vs. 3D issue. For example, when H&R discuss generalized torque (Section 12-3 in the 5th ed), they mention nothing about 2D vs. 3D. And common identities such as that the kinetic energy of rotation = L.L/(2I) don't seem to always work unless you use the non-generalized version. $\endgroup$ – JoelG Jun 24 '15 at 12:59
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By definition, both torque and angular momentum are relative to a point. So when you say $\tau=r \times f$ and $L=r \times p$, the vector $r$ originates from some fixed point. In light of these definitions, you can also easily prove that $dL/dt = \tau$ and various other useful properties. In classical mechanics, these are always true.

One very useful special case is a rigid body rotating around a fixed axis (which for simplicity we will assume to always be the $z$ axis). In this specific case (but not the general case), we can make several additional claims. First, we can define an angular position $\theta$ and a moment of inertia $I$. These quantities may make less sense in the general case. Once we’ve defined $\theta$, the entire set of purely-kinematic properties follow just as they did with translational position (e.g., $\theta_f = \theta_i + \alpha t$ for a constant acceleration $\alpha$).

A set of dynamic properties also exists for the rigid-body-fixed-axis case; e.g., $L_z=I \omega$. This did not exist in the general case, since I is only defined for an axis of rotation. However, note the subscripted $L_z$; while $\omega$ by definition points along the axis of rotation, $L$ need not. This formula only predicts the component of $L$ along the axis of rotation.

In the rigid-body-fixed-axis case, it is easy to compute $L_z$ directly. Whereas, in general, $L = r \times p$ and $r$ is the direction vector from a fixed reference point, it is also true that $L_z = r_p \times p$, where $r_p$ is the direction vector originating from the fixed axis and perpendicular to it. For better or worse, people frequently use one formula $L = r \times p$ both for its usual use, and also as a way of saying $L_z = r_p \times p$. I.e., they say $L$ where they could more clearly say $L_z$ and say $r$ where they could more clearly say $r_p$. Likewise, the general formula $\tau = r \times f$ is often used to mean $\tau_z = r_p \times f$. Thus, one formula can mean two different things, and it’s always useful to carefully look at the context. When $r$ is used to mean the perpendicular to a fixed axis of rotation, then the associate $L$ or $\tau$ usually actually means $L_z$ or $\tau_z$.

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