I'm trying to use the mechanical impedance analogy to describe a spring mass damper system as an electrical circuit.

Mechanical Impedances for linear motion are

$$ Z_{lin}(s) = \frac{F(s)}{\dot X(s)} = Ms + B + \frac{K}{s} $$

where M has units of kg, B has units of N-s/m and K has units N/m. The units of mechanical impedance then are N-s/m.

When I translate this to the rotational domain, I start having problems.

Rotational impedance will again be the ratio of an effort (torque) to a flow (rotational velocity).

Similarly, we have

$$ Z_{rot}(s) = \frac{\tau(s)}{\omega(s)} = Js + B_{rot} + \frac{K_{rot}}{s} $$

Based on dividing torque by angular velocity, the units of rotational impedance are [N-m-s/rad], or $\frac{ML^2}{T}$. When I try to verify that the units of rotational impedance are consistent, I get

$$J\cdot s = \frac{kg\cdot m^2\cdot rad}{s} = N\cdot m\cdot s \cdot rad = \frac{ML^2}{T}$$

$$B = N\cdot m\cdot\frac{s}{rad} =\frac{ML^2}{T} $$ and $$\frac{K}{s} = \frac{N\cdot m\cdot s}{rad^2}=\frac{ML^2}{T}$$

I can't sleep with all those radians being all over the place! It seems awfully inconsistent to have them in different place and I haven't been able to find any sources on the web or in my textbooks that adequately deal with this since mechanical impedance is such a niche topic. In doing this, I've assumed that the $s$ has units of [rad/s], which some people dislike and just use $s= \frac{1}{T}$ Is there a way for me to make this a little more consistent, possibly by using radians in the units for rotational inertia and the spring constant?

I've also looked at the discussions here:

  1. https://physics.stackexchange.com/questions/36079/unit-of-torque-with-radians
  2. https://physics.stackexchange.com/questions/11500/simple-harmonic-motion-what-are-the-units-for-omega-0
  3. https://physics.stackexchange.com/questions/33542/why-are-radians-more-natural-than-any-other-angle-unit
  • 1
    $\begingroup$ You should not use rad/s for $s$. Radians should be avoided at all because it is not actually a unit. If you introduce it to all the other parameters you will mess up everything. So get rid of the radiants and you are done. $\endgroup$
    – MrYouMath
    Commented Apr 23, 2017 at 10:22

1 Answer 1


Radians and unit analysis

The comment by MrYouMath is correct. For the purposes of dimensional analysis, the radian is considered a dimensionless quantity and therefore does not need to be included. This is because the radian is defined by an arc length (L) divided by a radius (also L) to give (L/L). The units of the Laplace domain complex number, $s$, are technically rad/s which, as the OP states, are considered to be $1/T$ for dimensional analysis.

$$ J\cdot s = \left[\frac{N\cdot m \cdot s \cdot rad}{rad}\right] = \frac{ML^2}{T}\left(\frac{L}{L}\right)\left(\frac{L}{L}\right) = \frac{ML^2}{T}$$

So from a dimensional analysis perspective we are fine. You can double check the units for the other terms as well.

Mathematically, radians are often ignored/assumed because they are so convenient (you don't end up with a bunch of constant factors in your equations). In engineering, radians serve mostly as a reminder that we are using them instead of degrees! So you can ignore them until you are figuring out what the units of your final answer are.

Mechanical impedance to blame?

The reason that the units don't work out has nothing to due with the fact that this is a mechanical system. Check out the units for the transfer function of an RLC circuit:

$$ \frac{V}{I} = Ls + R + \left(\frac{1}{C}\right)\frac{1}{s} $$

$$\left[ \frac{kg\cdot m^2}{s^3 \cdot A^2}\right] = \left[\frac{kg \cdot m^2}{s^2 \cdot A^2}\right]\cdot\left[\frac{rad}{s}\right] + \left[\frac{kg\cdot m^2}{s^3\cdot A^2}\right] + \left[\frac{kg\cdot m^2}{s^3\cdot A^2}\right]\cdot\left[\frac{s}{rad}\right]$$

Notice once again those pesky radians sticking their noses where they don't belong! In all seriousness though, you will see this "problem" regardless of what domain you work in.

The fact of the matter is simply this: when doing unit analysis in the Laplace domain (transfer functions, etc.), ignore the radian units for the complex number $s$. If you do this, the units will work out as long as you use consistent units for your physical constants $J$, $B$ and $K$ (you'll have to check it for the mechanical system yourself, I've wasted far too much time at work already).

Final note: Just remember that when you do your frequency analysis the units of frequency are still rad/s (because we use $\sin{\omega t}$ instead of $\sin{2\pi ft}$).


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