Is the radius equal to length/radians? [closed]

Question

Is the radius equal to the length/radians. Since the circumference is 2 pi times r and radians of an entire circle is 2 pi r should be equal to lentgh/radians I needed this proof to understand why we use radians/s and not degrees/second when calculating velocity using angular velocity This way the units cancel out and I get the answer in metres per second

Answer 1

It is often the case that when something new is introduced the question "Why bother?" is asked.
That is true of a lifetime of using degrees then to find that the radian suddenly appears.
Use of the radian makes life easier.

Consider $v=r\omega$ where $v$ is the speed in metre per second, $r$ is the radius in metres and $\omega$ is the angular speed in radians per second.
If the angular speed $\omega'$ was measured in degrees per second the equivalent formula would be $v = r \omega'\,\dfrac{\pi }{180}$.

In calculus $\dfrac{d \sin(x)}{d x} = \cos (x)$ with $x$ in radians would become $\dfrac{d \sin (x')}{d x'} = \cos(x')\dfrac{\pi }{180}$ where $x'$ is in degrees.

So the "proof" you seek is really a matter of convenience/simplicity.

Answer 2

What we call "radian" is a dimensionless unit for the measure of an angle. An angle is defined by the ratio of two lenght $s$ (the arc) and $R$ (the radius) and so it is dimensionless: \begin{equation} \theta=\frac{s}{R} \end{equation} To distinguish one dimensionless scale to another (e.g. radians from degrees) we say that $\theta$ in this definition is expressed in radians, but it is only our nomenclature to remind the choice of the scale used for this numbers.

When you study a circular motion (e.g. uniform motion) you start from this definition of an angle. The equation of motion is: \begin{equation} s(t)=s_0+\dot{s}(t-t_0) \end{equation} dividing everything by R: \begin{equation} \theta(t)=\theta_0+\omega(t-t_0) \end{equation} where the angular velocity $\omega=\frac{d\theta}{dt}$ is typically expressed in unit of $\frac{rad}{s}$ and it has dimension of an inverse of a time (again: angle is dimensionless, so in the ratio $\frac{d\theta}{dt}$ only the denominator gives dimensions).

In conclusion, you don't need to cancel out radians because they are "fictious" dimensions in the sense that they are dimensionless and in this way you can find everything you already know!

Answer 3

If you picture a circle and consider an arc of length $s$, this defines an angle $\theta$ at the centre of the circle; its value in radians is just the arc length $s$ divided by the radius of the circle $r$. Now consider that $s(t)$ represents the displacement of a point on the circumference, then the "angular displacement" is just $\theta(t)$ in radians, and so you'd have: $$ \theta(t)=\frac{s(t)}{r} $$

Note that by this definition $\theta$ is dimensionless, as both $s,t$ have units of length; indeed the radian is a "dimensionless" unit, which means you just use it when you need it.

Also note that similarly we can define the angular velocity $\omega(t)$, measured in $\mathrm{rad}\ \mathrm{s}^{-1}$, as the velocity $v(t)$ (in $\mathrm{m}\ \mathrm{s}^{-1}$) of the point on the circumference divided by the radius: $$ \omega(t)=\frac{v(t)}{r} $$

In general any "angular quantity" can be defined as the corresponding "linear quantity" just by dividing for the radius of the motion.