Deriving velocity in rolling without slipping in another approach

Question

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.

Answer 1

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 $\omega =\frac{d\alpha }{dt}=\frac{u}{R}$