Ron Ruff said:dvt said:Kinetic energy for linear motion is (mv^2)/2. For rotational energy, as
in the wheels spinning, the energy is (Iw^2)/2, where I is the
rotational inertia and w is the angular velocity.Do you realize that this is exactly the same? Iw^2/2 is just the
kinetic energy equation written in polar coordinates.
I hadn't thought of it that way. Now that you point it out, the mass of
rim/tyre/tubes/tape are all traveling at 40 mph or slightly less in the
scenario presented. Yes, that would have been easier to calculate.
Quoted message said:But we already
know that the only velocity that matters is at the rim/tire... and we
know that speed. You've gone to a lot of extra trouble to calculate the
same thing.Alfred Rider had the right idea, and the equation reduces to:
V= v*(m/M)^.5
He used the mass of the entire wheel (including hubs) traveling at 40
mph. That's why he came up with a much higher estimate than I.
Quoted message said:Quoted message said:Summary: I disagree with your physics, but my conclusion is the same.
Quoted message said:Guess again.
If Alfred and I used the same physics, how do you figure that we came up
with different answers? We can't both be right.
--
Dave
dvt at psu dot edu