Question

Loop the Loop

http://www.youtube.com/watch?v=dA_UO86MjLY

I have a small doubt.. The calculation considering the rotational energy of the ball yields the height of release as 2.7 R.. I get that.. but what i don't get it is.. this result must be independent of mass or size of the ball... and yet.. this rotational effect is negligible if we considered like a very tinnie tiny ball bearing or something.. why is that? :O

Answer 1

@Vincent-Lyon.Fr

Answer 2

Just saying ,
A interesting experiment ^_^
I recently studied work power energy and circular motion , so i found it interesting

Answer 3

more mass more energy > less height > more centrifusal force to maintain circular motion(mv2/r) so more height so mass got cancelled

Answer 4

However small the sphere, the ratio of rotational KE to total KE will always be the same.
So it's not a matter of actual size, only of mass distribution about the axis.

Answer 5

@Vincent-Lyon.Fr but if you keep making that radius smaller and smaller and make it tend to zero .. then the particle becomes a point particle which has no rotation and hence no rotational energy..

So what i was thinking is that just like in all cases if i had a veryyyyyyyyyyyyyyy tinniiiiee tiny sphere, then i could neglect its rotation.. and hence its rotational energy..

so that means we can't ?

Answer 6

No, you can't! Sorry ;-) The ratio always remains the same.

Answer 7

if i take the limit of r--> zero.. its rotational energy should disappear right? ..

Answer 8

This argument also holds for translational KE.

All energies (TKE, RKE, PE) tend to zero, because they are all related to mass, but they always keep the same ratio relative to one another.

There is no continuity from a very small solid (ring, cylinder, sphere, etc.) to a mass-point.