# Dimensional Analysis on Rotational Motion Laws

So I was solving some rotational motion problems when I decided to include the units in solving (just realized the necessity of doing this). Then I ran into some problem regarding the units in the work-energy theorem of rotational motion ( torque $$\cdot$$ angular displacement = ∆rot.KE ). When you try doing dimensional analysis, you will see that the rotational kinetic energy side will have an extra unit of angular displacement e.g. rad.

My question is why is that the case here?

• I've removed a number of comments that were attempting to answer the question and/or responses to them. Please keep in mind that comments should be used for suggesting improvements and requesting clarification on the question, not for answering. – David Z Apr 7 at 6:48
• @MindKicks rad is not unit, rad has the unit 1., thus Work has the unit Nm and Kinetic Energy has the unit Nm, so your equation is correct. – Eli Apr 7 at 8:20

Radians are a funny unit. They are not a "real" unit like meters or kg. For example, you would never put any other unit inside of a trigonometric function like $$cos(5{\rm m})$$ or an exponential $$e^{i10{\rm kg}}$$. But radians go there with no problem. Part of the reason is that radians are more of a ratio than a proper unit. Radians are the "natural" unit of a lot of mathematical expressions. You can put your calculator in radians or degrees, but $$sin(x) = x + x^3/3! + ...$$ only really makes sense in radians.

Basically, you don't have to worry about it. If you have kept track of your other units carefully you can ignore an extra unit of radians at the end.

$$\dfrac{1}{2}I\,\left( \dfrac{d\varphi }{dt}\right) ^{2}=\tau \varphi$$
The Inertia has the unit $$[N\,m\,s^2]$$, torque has the unit $$[N\,m]$$ $$\,,\varphi$$ has no unit and time unit is $$[s]$$ thus you get
$$[N\,m]=[N\,m]$$ as it should be.