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The title is a weird question, i know. But hear me out: Take the equation that relates the tangential velocity $\vec v$ and the angular velocity $\vec \omega$ of an object moving on a circle with radius $r$.

$\vec v = \vec \omega × \vec r$

If we want to know the magnitude of the axial quantity $\omega$, we need to divide the tangential quantity $v$ by the radius $r$.

$v = \omega r \Leftrightarrow \omega = \frac{v}{r}$

Now let's take a look at the case of a torque.

$\vec \tau = \vec r × \vec F$

Here, we can just multiply the radius $r$ by the tangential quantity $F$ to get the axial quantity $\tau$.

$\tau = r F$

Why is it that we multiply in one case and divide in the other, if we want to know the magnitude of the axial quantity, even though the overall structure of these three vectors is the same in both cases. I mean, you can more or less make an ignorant one to one correspondence beteween $\vec \tau$ and $\vec \omega$, $\vec F$ and $\vec v$ and well, $\vec r$ and $ \vec r$, if you comapre both scenarios. This ignorant comparison is depicted below. I mean you can literaly draw these two scenarios in the exact same diagram. But we still end up with two different equations. I find that a bit odd.

enter image description here

I reckon it has something to do with the fact that in the case of the velocity we have a differntial relationship, which is not present in the case of the torque. I. e. the fact that $\vec v = \frac{d\vec r}{dt}$ but $ \vec \tau ≠ \frac{d \vec F}{dt}$. But i can't really pin down a complete explanation.

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    $\begingroup$ You could just recast torque as $\tau =I \alpha= I \frac{a}{r}$. $\endgroup$
    – Triatticus
    Nov 2, 2021 at 8:40
  • $\begingroup$ Read this answer and maybe you get some insight into what these quantities represent. $\endgroup$ Nov 2, 2021 at 13:27
  • $\begingroup$ Two angular quantities don't have to be defined in the same way. You may see a pattern across all angular motion parameters. But not necessarily with other parameters $\endgroup$
    – Steeven
    Nov 2, 2021 at 16:12
  • $\begingroup$ Your ponit with $\tau = I \alpha = I \frac{a}{r}$ is quite nice! But it is still a bit odd to me, that you need to use an entierely different equation to get to an analogous $\tau = I \frac{a}{r}$. Not to mention, that in one case, we have a cross product, in the other we have a product between a tensor/scalar and an axial vector, to get another axial vector. Still weirdly unsemetric, considering i can draw $\vec \tau = \vec r \times \vec F$ and $\vec v = \vec \omega \times \vec r$ in the exact same diagram. $\endgroup$
    – Lukas G.
    Nov 2, 2021 at 19:22

5 Answers 5

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Here's one way to think about why they transform seemingly oppositely. Force and velocity form a pair in the sense that when multiplied you get power: $$W = \int F \cdot v\,dt.$$

The same is true for torque and angular velocity: $$W = \int \tau \cdot \omega\,dt.$$

So if to go from force to torque we multiply by $r$, we need to compensate by dividing velocity by $r$.

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  • $\begingroup$ This is a weirdly practical explanation. Nice one! $\endgroup$
    – Lukas G.
    Nov 2, 2021 at 18:55
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The magnitude of $ \,\vec c= \vec a\times \vec b $ is

$$|c|=|a||b|\sin(\phi)$$

Where $\phi$ is the angle between vector a and vector b

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  • $\begingroup$ I know that much too. Probably phrased my question bad. Suppose for the sake of simplicity, that $\phi = 90°$ in both cases. Than we get just $v = \omega r$ and $\tau = r F$ (i just did not bother writing the two lines indicating the magnitude of the vectors). So the question boils down - for the lack of a better term - to the "order of operations" in the two original cross products. Phrased differently: why is the axial quantity in one case on the same side of the equation as the radius and in the other case, the radius is on the same side as the tangential quantity. $\endgroup$
    – Lukas G.
    Nov 2, 2021 at 8:21
  • $\begingroup$ you can only compare the magnitude of the vector r, because r is the common vector in both cases $\endgroup$
    – Eli
    Nov 2, 2021 at 8:51
  • $\begingroup$ It's not really a direct comparison that i am after. I just find it a bit odd, that in these two seemingly symmetric situations, we get two completely different equations for the axial quantity. $\endgroup$
    – Lukas G.
    Nov 2, 2021 at 9:03
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The overall structure of the two equations aren't actually the same despite both using the cross product. The first equation $v = \omega \times r$ is multiplying an axial vector with a vector, and the second equation $\tau = r \times F$ is multiplying a vector with a vector. In three dimensions both of them can be expressed with the cross product, but this isn't true in other dimensions.

The "correct" equations, in the language of geometric algebra, and the correct generalization beyond three dimensions, are actually $v = r \cdot \omega$, where $\cdot$ is the inner product between a vector and a bivector, and $\tau = r \wedge F$ where $r \wedge F$ is the exterior product between a vector and a vector. (Note that geometric algebra generalizes the inner product between a vector and a bivector in a way that doesn't mean you just dot product the vector $r$ and the axial vector $\omega$.)

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  • $\begingroup$ Yes, but why though? Why is the structure of these equations different, even though the picture of these vectors are simmilar, if i would draw them. Or to put it bluntly: Why is ist $\vec \tau = \vec r × \vec F$ and not $\vec F = \vec \tau × \vec r$ (ignore the units for now). One could certainly think this at first if someone only knew the equation $\vec v = \vec \omega × \vec r$ and then saw the picture of those vectors. Is this just a result of some things being defined in a certain way, so that is has to be this way. Or is there an actual, physical reason why it is so dissimmilar. $\endgroup$
    – Lukas G.
    Nov 2, 2021 at 9:54
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Velocity is the moment of rotation, just as torque is the moment of force, as I explained in this answer.

$$\begin{aligned} \vec{v} & = \vec{r} \times \vec{\omega} \\ \vec{\tau} & = \vec{r} \times \vec{F} \\ \end{aligned}$$

where $\vec{r}$ points from the measuring/reference point (the origin) to the axis (rotation axis, or line of action axis).

If you assume the perpendicular distance to the axis is $d$ then

$$\begin{aligned} \|\vec{v}\| & = d\, \|\vec{\omega}\| \\ \|\vec{\tau}\| & = d\, \|\vec{F}\| \\ \end{aligned}$$

so your question is odd to me. In both cases you multiply by the moment arm $d$ to get the magnitude of the moment-of quantity. Just in your post you re-wrote the second expression as $\| \vec{F} \| = \frac{ \| \vec{\tau} \|}{d}$, which means you are interested in the axial vector and not the moment-of vector. You can do the same with motion and write $\| \vec{\omega} \| = \frac{ \| \vec{v} \|}{d}$ if you are interested in the axial vector of motion.

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Generally torque is just force (angular force that's what I call it) and angular velocity is moment of rotation( as other have described)

Torque occurs in rotational body if some kind of force was occurred which we denoted as $\vec F$. If you remember your study than you may remember that we write $F=dp/dt$. So force is actually changes in momentum with respect to time. The same thing happens with torque. We denote angular momentum $\vec L$. The equation of angular momentum is $\vec L=m\vec r \times \vec v$. You can write torque something just like this : $\tau=\dfrac{dL}{dt}$

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  • $\begingroup$ I know the analogies you mentioned between $\vec F = \dot{\vec p}$ and $\vec \tau = \dot{\vec L}$.That's not really what i am after. I just find it a bit curious that i can literaly draw the exact same picture (see my edited question above) for both cases, i. e. the cross product for the velocity and the one for torque. But despite the same picture, i end up with two different equations. $\endgroup$
    – Lukas G.
    Nov 2, 2021 at 19:14
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    $\begingroup$ @LukasG. they are the same equation, you just have to consistent on your definition of $\vec{r}$ $\endgroup$ Nov 3, 2021 at 0:37
  • $\begingroup$ @LukasG you are thinking it harder. Just look at equations properly, as we know omega=v/r it would be more clear if we write omega=(1/r)(dx/dt) omega is representing a velocity which is occuring by a rotational body where radius of that object is r. As we know that sometimes object move with same velocity. If object's velocity is constant than acceleration will be 0 so newtonian force will be 0 $\endgroup$ Nov 3, 2021 at 6:14

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