Why Is Deriving the Motion Formula for Hoops More Complex?

AI Thread Summary
Deriving the motion formula for hoops is more complex due to the additional radius in the moment of inertia, which complicates the equations of motion. The discussion highlights the successful derivation for solid cylinders and spheres but struggles with spherical shells and hoops. The key issue arises from the relationship between the inner and outer radii, leading to confusion in the calculations. Participants suggest using a "thin hoop approximation" to simplify the problem, allowing for the neglect of the difference between the inner radius (r) and outer radius (R). Ultimately, this approximation is valid for certain models, clarifying the complexity involved in deriving the motion formula for hoops.
lsie
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Homework Statement
Conservation of mechanical energy proof
Relevant Equations
a = 1/2 g sin theta
a = 3/5 g sin theta
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!
 
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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?
 
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Got it. I saw some google-images of a hoop with some thick-ish edges and it threw me off. Cheers!
 
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