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Suppose an angular oscillatory motion. In the function below, $\alpha$ and $\alpha_o$ are angles measured in radian, $\omega$ is circular frequency ($2\pi/T$) measured in [radian/s] and $t$ is time.

$$ \alpha = \alpha_o\sin(\omega t) $$

Angular velocity is obtained by taking derivative with respect to time:

$$ \frac{d\alpha}{dt} = \alpha_o\omega\cos(\omega t) $$

I get confused with the unit of angular velocity. To me it looks like:

$$ \frac{d\alpha}{dt} = \alpha_o[rad]\omega[rad/s]\cos(\omega t)[1] =[rad^2/s] $$

Why I cannot get $rad/s$ for angular velocity?

Qmechanic
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Shibli
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  • If $sin$ and $cos$ are unitless, then the argument must also be unitless. Else, as you have seen, you will introduce arbitrary units simply by differentiating more and more. Remember that radian measure is dimensionless because it is arc length divided by radius, so length divided by length. – Marius Ladegård Meyer Oct 09 '21 at 13:21
  • Radians are dimensionless. – Vincent Thacker Oct 09 '21 at 13:21
  • @MariusLadegårdMeyer trigonometric functions takes radian as input and produce non-dimensional output as far as I know. So I don't see why argument must be unitless. – Shibli Oct 09 '21 at 13:30
  • @VincentThacker So do you mean rad^2/s = rad/s? What is the physical interpretation of this? – Shibli Oct 09 '21 at 13:32
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    @Shibli See this post and this post. Radian is not a unit. – Vincent Thacker Oct 09 '21 at 13:38
  • Again, one radian is defined as the angle inside a circle sector with arc length equal to radius, more specifically the ratio of the two. A ratio of two lengths is dimensionless. Hence, 1^2/s = 1/s. Makes sense, no? – Marius Ladegård Meyer Oct 09 '21 at 13:38

2 Answers2

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It is radians squared. The problem is you are using two different radians. One is the physical angle $\alpha$, which oscillates with amplitude $\alpha_0$. Fine...you could measure that in degrees, say if it were a latitude or longitude. That might make the point clearer, or at least less abstract.

The other angle is the temporal phase angle of the oscillatory function:

$$\phi(t) = \omega t $$

So, these two angles live in different spaces, and their radians aren't the same thing. Of course, you can also just not use radians, and they are dimensionless (with good reason: space radians and time radians don't mix well, as you discovered).

JEB
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Radian is defined in such a way that it has no unit. So we cannot mathematically differentiate between a numerical constant and a unit. And this leads to many confusions. Using degree as unit of angle will make this more clear.

Let $\alpha,\alpha_o,\omega,t$ be numerical value

$$ \alpha^\circ = \alpha_o^\circ\sin\left(\left(\frac{\omega^\circ}{s}\right)\left(\frac{\pi}{180^\circ}\right)\left(t\space s\right)\right) $$

Angular velocity is obtained by taking derivative with respect to time:

$$ \left[\frac{d\alpha}{dt}\left(\frac{^\circ}{s}\right)\right] = \left[\alpha_o^\circ\left(\frac{\omega^\circ}{s}\right)\left(\frac{\pi}{180^\circ}\right)s\cos\left(\left(\frac{\omega^\circ}{s}\right)\left(\frac{\pi}{180^\circ}\right)\left(t\space s\right)\right)\right]$$

$$\implies \left(\frac{^\circ}{s}\right)=\left(\frac{^\circ}{s}\right)$$

SHARDUL VIKRAM SINGH
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