# How much further does a ring roll up a hill

1. Nov 19, 2013

1. The problem statement, all variables and given/known data

http://puu.sh/5nLnI.png [Broken]

2. Relevant equations

I = mr2

3. The attempt at a solution

Moment of inertia:

Mass density = mass / area
= 2.59 / (π*9.52 - π*7.52)
= 0.02424772368 kg / cm2

= 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

Last edited by a moderator: May 6, 2017
2. Nov 20, 2013

### haruspex

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.

3. Nov 20, 2013

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.

4. Nov 21, 2013

### haruspex

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.