How much further does a ring roll up a hill

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The discussion focuses on calculating how far a ring rolls up a hill by determining its moment of inertia and energy conservation. The moment of inertia is calculated using the difference between two discs, emphasizing that the mass density approach is unnecessarily complex. Participants highlight the importance of keeping calculations symbolic until the final step to minimize errors and improve clarity. A correction was made regarding the inclusion of a factor of 1/2 in the calculations. The conversation stresses the need for a solid understanding of the physics involved, particularly in treating the annulus correctly.
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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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