- #36

erobz

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Disregard. I think I have some wires crossed in my cross product...erobz said:For the angular velocity at any instant in time if the wheel is not slipping, I can't see how the relationship is invalid?

$$ \hat{v} = \hat{r} \times \hat{\omega} = | \hat{r} | ~ |\hat{\omega}| \sin \theta = r \omega \sin \theta \hat s = v(s) \hat s $$View attachment 329251I think the angular kinetic energy is simply as it was:

$$KE_{rot.} = \frac{1}{4} m(s) v(s)^2 $$

and it was the translational energy that needed to be adjusted?

I'm not sure how to fix it formally, but its only the component of ##\hat v ## perpendicular to ##\hat r ## and ##\hat \omega## which contributes to the angular kinetic energy, correct?

I still think the angular energy term is

$$ \frac{1}{4} m(s) v(s)^2 $$

But I can't seem to show what I want formally.

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