Hoop rolling down a hill

I've worked out how to derive the formulas for a solid cylinder and a solid sphere rolling down a hill.

E.g., for a cylinder:
Emech = KE + PE
mgh = 1/2 mv^2 + 1/2 Iw^2
gh = 1/2 v^2 + 1/2 (1/2r^2) v^2/r^2
gh = 3/4 v^2
v^2 = 4/3 gh
I then performed a derivative with respect to time and found a = 2/3 g sin theta

But I'm having trouble proving the formula for spherical shells and hoops. The rub is that extra radius nested in the inertial moment (1/2m (r^2 + R^2). I get to gh = 3/4 v^2 + 1/2 R^2 v^2 and I don't know how to deal with the R^2.

I've tried using various equations (v = wr, a = αr, and α = w/t) but can never get rid of that radius.

Any pointers would be appreciated.

Cheers!

Neglect the difference between r and R.

That worked! Thanks.

I was certain the solution would be more elegant. Out of curiosity, are there good reasons why we can neglect the difference between r and R? Surely there are hoops where r and R are (how's that for a tongue twister) significantly different?

lsie said:

That worked! Thanks.

I was certain the solution would be more elegant. Out of curiosity, are there good reasons why we can neglect the difference between r and R? Surely there are hoops where r and R are (how's that for a tongue twister) significantly different?

(It sounds like) You're looking for a "thin hoop approximation". Thats why it can be neglected. If moment of inertia never mattered, then they wouldn't bother with it. Perhaps I'm not seeing what you are asking.

i.e. I can't think of a reason to say ##r \approx R## unless it's true for the model you are analyzing?

Got it. I saw some google-images of a hoop with some thick-ish edges and it threw me off. Cheers!