Deriving velocity in rolling without slipping in another approach

I want to derive the velocity of a point P on a surface of a cylinder rolling on a flat plane, by considering the rolling as instantaneous rotation with repect to the contact edge line E of the cylinder. The cylinder with radius R rolls with a linear velocity u. When the centre of mass, contacting edge E, and point P make a angle θ, the velocity(given by the solution) is $$v = 2R \cosθ × ω$$ where $$ω = \frac{u}{R}$$ But if we take the edge as rotating axis, isn’t $$\frac{dθ}{dt}$$ the angular velocity of the system, and the velocity is instead $$v = 2R \cosθ × \frac{u}{2R}$$ ? I have used the reason “angle at centre twice the angle of circumference” in geometry for $$\frac{dθ}{dt}$$ here. The answer given by the solution is consistent with the expression we get from the conventional “tanslational + rotational” approach, and my version lacks a factor of 2. Can anyone tell me why ω is used instead of $$\frac{dθ}{dt}$$ ? Thanks a lot.

• No, ${\rm d}\theta / {\rm d}t$ is not equal to $\omega$. At any instant $\theta$ is fixed and it is not a function of time. Commented Jan 8, 2019 at 15:55

The comment is clear: the $$\theta$$ parameter you are using is a geometric parameter to locate the position of the point at a given moment. It does not depend on time.
To evaluate the rotational angular velocity, one must look at how much turns a segment for a short time $$dt$$ . For example, if you assume that the speed at the top is $$2u$$ , the higher part of the vertical diameter advances by $$2udt$$ and its distance to the axis is $$2R$$. So the angle of rotation is $$d\alpha =\frac{2udt}{2R}$$and angular velocity is $$\omega =\frac{d\alpha }{dt}=\frac{u}{R}$$