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In classical physics we often cast an analogy between translational and rotational systems

Force < > Torque

Energy < > Rotational Energy

Momentum < > Angular Momentum

and considering SI units we have [Force] = N, [Torque] = N-m, [Energy] = [Rotational Energy] = N-m (Joules), [Momentum] = N-sec and [Angular Momentum] = N-m-sec.

Physically this analogy seems to make sense, but if you ponder the units in a simplistic way, questions come up like:

Why does torque, which is an analogy of force have the same units as energy, but force does not?

and

If there are differences in units between the analogy for force and torque, why not also a difference between energy and rotational energy?

Is there a simple way to reconcile these questions, or do you have to step outside classical physics?

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    $\begingroup$ physics.stackexchange.com/q/37881 Short-short version the cross-product develops a different kind of entity than the dot-product so they are not the same units at all. $\endgroup$ Commented Oct 19, 2014 at 16:47
  • $\begingroup$ Thanks, but looking for more of a physical 'why' rather than mathematical. $\endgroup$
    – docscience
    Commented Oct 19, 2014 at 16:57
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    $\begingroup$ In that case you want the answer to 37881 that talks about $dW = \tau \cdot d\theta$ implying that torque is in Joules per radian. $\endgroup$ Commented Oct 19, 2014 at 17:05
  • $\begingroup$ @dmckee Still, the units are the same even if the beast is an entirely different thing. The answer cited above makes that case very well. I would say that the coincidence in units has no deep meaning, and points to no connection. $\endgroup$
    – garyp
    Commented Oct 19, 2014 at 17:08
  • $\begingroup$ @garyp I suppose that's a matter of interpretation. As far as I'm concern you can't add a torque to an energy so they are not the same no matter that you write the units as mass*distance^2/time^2 in each case. In other words, I'm including the identification of the mathematical class in the units because both are about identifying what sort of critter you're talking about. $\endgroup$ Commented Oct 19, 2014 at 17:12

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This is a side-effect of treating angles as dimension-less.

For translational systems, we have

\begin{align*} [\text{linear momentum}] &= [\text{action}][\text{length}]^{-1} \\ [\text{force}] &= [\text{linear momentum}][\text{time}]^{-1} \\&= [\text{energy}][\text{length}]^{-1} \end{align*}

Correspondingly, for rotational systems, we have

\begin{align*} [\text{angular momentum}] &= [\text{action}][\text{angle}]^{-1} \\ [\text{torque}] &= [\text{angular momentum}][\text{time}]^{-1} \\&= [\text{energy}][\text{angle}]^{-1} \end{align*}

If $[\text{angle}] = 1$, obviously $[\text{torque}] = [\text{energy}]$, even though these quantities are rather different, both from a physical as well as geometrical point of view.

In contrast, translational and rotational energies both contribute to total energy and it doesn't really make sense to introduce a distinct unit for each type of energy.

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  • $\begingroup$ Thanks @Christoph ! Very clear and to the heart of my question. In terms of action the analogy is complete, less confusing and the lesson learned that radians are not 'unitless'. $\endgroup$
    – docscience
    Commented Oct 19, 2014 at 21:49
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Torque is a cross product, and work is a dot product. So one big difference is that torque is a vector and work is a scalar. Another way to think about it is that work is a force being applied over a length interval, where only the force applied in the direction parallel to the displacement counts toward the work performed.

On the other hand, torque is best thought of as a force applied _at_a distance away from an axis of rotation. Only the part of the force applied perpendicular to the lever arm distance counts toward torque. These really are two very different concepts, and despite the apparent match-up in units, are not analogous at all. Since in torque, the distance that you are using only states how far away the force is from the axis of rotation, and not how much the rigid body actually rotates, you can see the mismatch. Think of a static situation where a rigid body is experiencing balanced torques. Since the object is not moving, obviously no work is being done, but there are torques on the object (although admittedly not a net torque).

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  • $\begingroup$ Why the down vote? There's nothing in this answer that is incorrect $\endgroup$
    – Sean
    Commented Oct 19, 2014 at 23:22
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Why does torque, which is an analogy of force have the same units as energy, but force does not?

The magnitude of (net) force is, in a sense, work done per unit displacement thus we can think of it as, e.g., Joules per meter.

In a similar sense, the magnitude of (net) torque is work done per unit angular displacement, e.g., Joules per radian.

But the unit of work is simply Joules so, conceptually, it isn't really correct to say that torque has the same units as work (or energy).

However, they do have the same dimensions since angular displacement is dimensionless quantity.

From the Wikipedia article "Torque":

For example, the quantity torque may be thought of as the cross product of force and distance, suggesting the unit newton metre, or it may be thought of as energy per angle, suggesting the unit joule per radian."

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We could standardise Torque at 1 meter and it would then only be a force. It is in fact only a force, but one that can be varied with r (radius). As a vector it has direction, but we do not use this by convention. As engineers, we want to divide this force, or multiply it. Gears, pulleys, chains. The units are the units most suitable for our calculations.

Do not assume that because there is torque, there is rotation.

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Torque and work are different physical quantities,so it makes sense to use different units. Since torque is a vector and work is a scalar, one idea would be to use "$\mathrm{N\times m}$" for torque, instead of "$\mathrm{N \cdot m}$". $\mathrm{N\times m}$ would be consistent with torque, or cross product, while work is dot product! Moreover, $\mathrm{N\cdot m = J}$. But it is never easy to change units, as history teaches us.

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  • $\begingroup$ Angles are dimensionless, and $W = \tau \Delta \theta$ for rotational motion (and constant torque). So you'd need some additional rule along the lines of "you have to convert $\mathrm{N \times m}$ to $\mathrm{N \cdot m}$ when you multiply a torque by an angle." This could get complicated pretty quickly. $\endgroup$ Commented May 25, 2022 at 14:59
  • $\begingroup$ To add to these complications, Torque units translate to "m x N", since the cross product is not commutative: r x F =-F x r... $\endgroup$
    – stefanee
    Commented May 26, 2022 at 11:12

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