Originally Posted by
Road Fan
You are starting to confuse inertia (m or I), momentum (m*v) and inertial resistance (m*a or I * alpha). It makes it harder to follow your reasoning and confirm that it's not fuzzy. You're also not commenting on what I wrote, which I had hoped to see.
Yes, I was a bit loose with the terminology. But inertia is just resistance to changes in momentum, and rotational inertia is just resistance to rotational momentum. And since we assume that the wheel isn't slipping and all the mass is at the rim, the two are very closely related.
And this "double" relationship applies if you're looking at momentum or kinetic energy (to further fuzzy things, though of course they're closely related, so much so that we often use the terms interchangeably in casual speech.)
I've also not been looking at the forces and accelerations at all -- you can, but it's easier just to look at the end result (momentum and kinetic energy), since they are trivial to calculate when you break it down into translational and rotational components, especially since the two velocities and masses are the same (in this simplified model of a wheel where all the mass is at the rim.) I could do it the harder way, but I'd have to whip out some paper.
Here's an analysis of the kinetic energy of a rotating bicycle wheel that assumes that all weight is at the rim. (Really, this is what I should have started with, but I didn't find it before.)
You can change this to a calculation of momentum just by replacing 1/2 m*v^2 with m*v and 1/2 I*w^2 with I*w, and end up with a total momentum of 2*m*v