# Units don't match in the torsional spring energy!

According to Wikipedia's description of torsion springs and according to my understanding of physics the energy of a torsional spring can be written as $$U=\frac{1}{2}k \varphi^2$$ where $k$ is a constant with units of $\rm N\,m/rad$.

I am freaking here because if the energy of a torsional spring is really $k \varphi^2$ than the units are $\rm (N\,m/rad) \cdot rad^2=Joule\cdot rad$. ??

What on earth am I missing here?

• Related: physics.stackexchange.com/q/37881/2451 , physics.stackexchange.com/q/36079/2451 and links therein. Mar 21, 2016 at 12:02
• Interesting fact - you'd never be able to take sin or cos of a number in radians if radians were an actual "dimension" - consider adding up the Taylor Series terms. Lots of different powers of that dimension.
– Joel
Mar 21, 2016 at 21:23
• @Joel yes, that makes sense, but the question is as follows (is written somewhere here in the comments but wasn't answered). If my $k$ is defined as $k=EI/L$ where $E$ is Young's elasticity module and $I$is moment of inertia and $L$ is unit length in meters, than $k$ is in units of Nm. Now the question is: does one have to divide the previously defined $k$ by $2\pi$ in order to get the correct units (Nm/rad) or not? I do understand the meaning of radians, at least I think I do, but sometimes the numbers matter too! Or do i simply imagine that the radians are already there since rad =1. Mar 21, 2016 at 21:28
• What I'm saying is basically an extension of the accepted answer - radians aren't a dimension. I'm just highlighting one of the things that would really go wrong if people tried to treat them as if they were.
– Joel
Mar 21, 2016 at 21:38
• @Joel Well, if radians were considered a dimension, that formula would have a lot of negative powers of radians in it. It would be the same computation, but written in a way that would make the units work. Mar 21, 2016 at 22:33

Radians are a pure number, so they do not contribute to your dimension considerations.

The units of the torsion constant are $\mathrm{Nm}$ which are equivalent to Joules.

• I don't understand why this has two downvotes, it is the correct answer as far as I can see. Mar 21, 2016 at 11:27
• @lemon consider a much simpler case: length of arc = radius * angle in radians. You've got two lengths (in meters) and a dimensionless angle. Mar 21, 2016 at 11:39
• @PhillS Yes, I understand that the dimensions are fine, but the units are not. How would you modify $k$ if you wanted to use degrees instead of radians? Mar 21, 2016 at 11:40
• @lemon The same way you'd convert $L=R\theta$, or trig functions. Put your angles to radians, since that is what the formulae use. Sure, you can convert a formula to use degrees, but all you're doing is arbitrarily scaling your angles by $\frac{180}{\pi}$ and multiplying $k$ by the appropriate factor to cancel that out. Mar 21, 2016 at 11:45
• @PhillS The fact that your dimensional analysis is incomplete doesn't mean there isn't one. Length of arc can be in "distance along arc". radius can be in "distance radially". Then radians becomes along arc/radially. Now it might be very questionable to have such quirky dimensions, but dimensional analysis can be done with things other than time/length/etc. (and it can be done with time and length being the same dimension!) Dimensional analysis is mostly appending a free (usually abelian) group x the reals, and isomorphims for "unit" conversion.
– Yakk
Mar 21, 2016 at 15:17

You are correct that, in terms of units, $k$ should be [E]/rad$^2$.

So why is it given as [E]/rad? Sloppiness/convenience.

The unit [E]/rad is equivalent both in dimension and in value to [E]/rad$^2$, so it probably makes life easier for engineers to pretend that $k$ is the same for both the torque $\tau=k\theta$ and the energy $U=\frac{1}{2}k\theta^2$. But they are not, which is something you would have to keep in mind when converting from radians to degrees.

Another motivation for this sloppiness is that it's presumably an attempt to emulate the standard harmonic spring $f=kx$ and $U=\frac{1}{2}kx^2$, for which the $k$ values are indeed equal.

• So... if my $k$ is defined as $k=EI/L$ where $E$ is in Pascals, $I$ is in meters$^4$ and $L$ is in meters, than my $k$ is in units of Nm. But in order to do my calculations correctly (torsional energy) I should probably divide it by additional $2\pi$ meaning $k=EI/(L2\pi)$ to produce units Nm/rad. Right? Mar 21, 2016 at 12:19
• @skrat I don't recognise that equation (what does it relate to?), but $k$ would appear to be in units of Pascals... Mar 21, 2016 at 12:25
• $E$ is young's elasticity module, $I$ is moment of inertia and therefore k=Pam$^4$/(m)=Nm$^4$/m$^3$ =Nm. Mar 21, 2016 at 12:29

An angle is just the ratio of the length of a circular arc to its radius, so the radian has units of length/length, which means it's a dimensionless quantity.

In the SI system of units, the radian is a special name for 1 (see SI brochure), that is, $$\mathrm{rad}=1.$$

Therefore, $$[k] = \mathrm{N\,m/rad} = \mathrm{N\,m}$$ and $$\mathrm{J\,rad} = \mathrm{J}.$$

Since the last revision of the SI, the radian is no longer a supplementary unit: an angle is now defined as the ratio of two lengths, and the unit radian is now maintained for convenience. However, it's just a synonym of 1, and can be used (but it's not necessary) to convey, or to strengthen, the information that the quantity of interest is an angle.

Let's see how the units work out if we convert a linear spring (where we know everything) to a torsion spring, by attaching our linear spring to a stiff rod some distance $R$ from a pivot:

The (linear) force due to the spring is $\vec F = -k\Delta \vec x$, for spring constant $k$ having units of newtons per meter. The torque is $$\tau = RF = -R \cdot k (R \Delta\theta) \equiv -\kappa \Delta\theta$$ So apparently the torsion spring constant $\kappa = kR^2$ has units of newton-meters, which is equivalent to newton-meters per radian, because the radian is a dimensionless ratio. If you mistrust the apparatus of calculus and would like to do a lot more work you could use the appropriate trig function $\sin\Delta\theta = \Delta x /R$; in that case only follow my argument in the small-angle approximation $\lim_{|x|\ll1} \sin x = x$.

The energy stored in the linear spring is \begin{align} U = -\int_0^{\Delta x} \vec F \cdot d\vec x = \frac12 k(\Delta x)^2 = \frac12 k(R\Delta\theta)^2 = \frac12 \kappa (\Delta\theta)^2 \end{align} which is exactly the same thing you get if you integrate the torque $$U = -\int_0^{\Delta\theta} \vec\tau \cdot d\vec\theta = \frac12 \kappa (\Delta\theta)^2$$ As another answer says much more succinctly: it all works because the radian, a ratio between two lengths, is dimensionless.

lemon asks in a comment elsewhere how you would convert $\kappa$ in each case if you were tied to a log in a sawmill and commanded either to use degrees or die a hideous bloody death. (My paraphrase; I would probably choose death, myself.) In that case you might grudgingly admit that the relevant constant in the torque equation has units of foot-pounds per degree, while in the energy equation you have picked up another angular factor so that the unit is b.t.u. per degree squared. I don't think there's anything profound about the coincidence that torque has units of energy; I do think there's something profound about the fact that we have invented the SI units to make these pointless problems go away.