ω is just v/r (v = rω), and translational KE is ½mv² or ½mr²ω², so if rotational KE is ⅕mr²ω², then rotational KE is 10⁄35 or 29% of total KE.
I guess if we assume the ball is rolling without slipping as it exits the track, then the ratio of translational KE to rotational KE is fixed regardless of what happened earlier in the drop, so maybe it doesn’t matter after all.
Sorry that I didn’t notice your comment before. You took it the one extra step of getting kinetic and rotational energy in the same units. (I had been trying to compare potential and rotational energy and gave up when there were quantities that would have to be numerically evaluated.)
Yeah, I follow your algebra. The radius of the ball cancels and we only have to compare 12 and 15. Indeed, a uniformly solid sphere (an assumption I made) rolling without sliding without change in potential energy (at the end of the ramp) has 29% rotational energy and 71% linear kinetic energy, independently of its radius and mass. That’s a cute theorem.
It also means that my “physics intuition trained on similar examples in the past” was wrong, because I was imagining a “negligible” that is much smaller than 29%. I was imagining something less than about 5% or so. So the neural network in my head is apparently not very well trained. (It’s been about 30 years since I did these sorts of problems as a physics major in college, if that can be an excuse.)
As for your second paragraph, it would matter for solving the article’s problem because if you used the ball’s initial height and assumed that all of the gravitational potential energy was converted into kinetic energy to do the second part of the problem, “how far, horizontally, will the ball fly (neglecting air resistance and such)?” you would overestimate that kinetic energy by almost a third, and how much you overestimate would depend on how much it slipped. Still, though, the floppy track would eat up a big chunk, too.
ω is just v/r (v = rω), and translational KE is ½mv² or ½mr²ω², so if rotational KE is ⅕mr²ω², then rotational KE is 10⁄35 or 29% of total KE.
I guess if we assume the ball is rolling without slipping as it exits the track, then the ratio of translational KE to rotational KE is fixed regardless of what happened earlier in the drop, so maybe it doesn’t matter after all.
Sorry that I didn’t notice your comment before. You took it the one extra step of getting kinetic and rotational energy in the same units. (I had been trying to compare potential and rotational energy and gave up when there were quantities that would have to be numerically evaluated.)
Yeah, I follow your algebra. The radius of the ball cancels and we only have to compare 12 and 15. Indeed, a uniformly solid sphere (an assumption I made) rolling without sliding without change in potential energy (at the end of the ramp) has 29% rotational energy and 71% linear kinetic energy, independently of its radius and mass. That’s a cute theorem.
It also means that my “physics intuition trained on similar examples in the past” was wrong, because I was imagining a “negligible” that is much smaller than 29%. I was imagining something less than about 5% or so. So the neural network in my head is apparently not very well trained. (It’s been about 30 years since I did these sorts of problems as a physics major in college, if that can be an excuse.)
As for your second paragraph, it would matter for solving the article’s problem because if you used the ball’s initial height and assumed that all of the gravitational potential energy was converted into kinetic energy to do the second part of the problem, “how far, horizontally, will the ball fly (neglecting air resistance and such)?” you would overestimate that kinetic energy by almost a third, and how much you overestimate would depend on how much it slipped. Still, though, the floppy track would eat up a big chunk, too.