Curvature (math) and angular velocity (physics)

Question

I was reading a bit about curvature these days. One of its definitions says that curvature is the limit of the angle (between two tangent lines) when the distance travelled (along the curve) tends to zero.

This made me recall about Angular velocity from high-school physics.
I have only some vague memory about it.

https://en.wikipedia.org/wiki/Angular_velocity

So... can we say that if $f(t)$ is some function of time so that $(t, f(t))$ describes the $(x,y)$ position of a moving point $M$ at time $t$, then the curvature of the graph of $f(t)$ at the point $t = a$ is equal to the angular velocity of the moving point $M$ when it's at the point $(a, f(a))$?

I feel like the two concepts are related but I am not quite sure how.
I mean, I cannot quite express formally how they are related.
I am not good at physics at all.
Is the statement which I made above correct?

Answer 1

Angular velocity is the rate of change of the angle w.r.t time whereas the curvature is the rate of change of angle w.r.t arc length. The expressions may look similar but they very clearly are not related closely.

Answer 2

Note that for the following parametrization $(x,y)=\left(t,f(t)\right)$ the curvature is given by

$$\kappa =\frac{|f''(t)|}{\left(1+f'(t)^2\right)^\frac32}$$

which is a number while the angular velocity $\vec \omega$ is a (pseudo)vector.

We can consider the modulus of the angular velocity and therefore

$$|\vec \omega|=\left|\frac{d\theta}{dt}\right|=\kappa\left|\frac{ ds}{dt}\right|=\kappa |\dot s| \neq \kappa$$

the two quantities are equal when the speed $|\dot s| =1$, that is

the curvature ($m^{-1}$) is equal to (the modulus of) the angular velocity (rad$/s$) for a point travelling the trajectory with speed equal to one ($m/s$)