How much further does a ring roll up a hill

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SUMMARY

The discussion focuses on calculating how far a ring rolls up a hill using principles of physics, specifically the moment of inertia (MoI) and energy conservation. The moment of inertia for the annulus is derived from the difference of two discs, with the formula I = (areaA)(density)(radiusA) - (areaB)(density)(radiusB). The final height calculation is performed using energy equations, resulting in a height of approximately 0.975695 meters. Participants emphasize the importance of maintaining symbolic representations until the final calculation to minimize errors.

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  • Understanding of moment of inertia (MoI) calculations
  • Familiarity with energy conservation principles in physics
  • Knowledge of annulus geometry and density calculations
  • Proficiency in symbolic manipulation of equations
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Homework Statement



http://puu.sh/5nLnI.png

Homework Equations



I = mr2

The Attempt at a Solution



Moment of inertia:

Mass density = mass / area
= 2.59 / (π*9.52 - π*7.52)
= 0.02424772368 kg / cm2I = (areaA)(density)(radiusA) - (areaB)(density)(radiusB)
= π*9.52*0.02424772368*0.0952 - π*7.52*0.02424772368*0.0752
= 0.0379435

Ek = 0.5*m*v2
= 0.5*2.59*2.72
= 9.44055

Er = 0.5*I*ω2
= 0.5*0.0379435*(v/r)2
= 0.5*0.0379435*(2.7/0.095)2
= 15.32455

Eg = Ek + Er
mgh = 9.44055 + 15.32455
(2.59*9.8)h = 9.44055 + 15.32455
h = (9.44055 + 15.32455) / (2.59*9.8)
= 0.975695

sin(37.8) = 0.975695 / d
d = 0.975695 / sin(37.8)
= 1.59m
 
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Esoremada said:
Mass density = mass / area
= 2.59 / (π*9.52 - π*7.52)
= 0.02424772368 kg / cm2


I = (areaA)(density)(radiusA) - (areaB)(density)(radiusB)
You don't get the MoI of a disc by multiplying its mass by its radius. (That would be dimensionally wrong.) Besides, computing the density is rather a long way round.
Treat the annulus as the difference of two discs. What is the MoI of a disc about its centre?
Also, please don't plug in numbers until the final step. It's much easier to follow the logic and spot mistakes if you keep everything symbolic as long as possible. It can also avoid some calculation steps, so reduce numerical error.
 
haruspex said:
You don't get the MoI of a disc by multiplying its mass by its radius. (That would be dimensionally wrong.) Besides, computing the density is rather a long way round.
Treat the annulus as the difference of two discs. What is the MoI of a disc about its centre?
Also, please don't plug in numbers until the final step. It's much easier to follow the logic and spot mistakes if you keep everything symbolic as long as possible. It can also avoid some calculation steps, so reduce numerical error.

Oh wow I left out the 1/2 on the disks that was dumb. I fixed that and got the right answer. My initial post was treating the annulus as a difference of two disks. How could I do this without finding the mass density? I need the mass of the two disks.
 
Esoremada said:
How could I do this without finding the mass density? I need the mass of the two disks.
Yes, I see. I suppose what I would have done effectively does that, but not so explicitly:
M(Router2 - Rinner2*(Rinner2/Router2))/2.
More importantly, I urge you to get into the habit of working entirely symbolically, as mentioned.
 
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