How much further does a ring roll up a hill

  • Thread starter Thread starter Esoremada
  • Start date Start date
  • Tags Tags
    Hill Ring Roll
Click For Summary

Homework Help Overview

The discussion revolves around a physics problem involving the rolling motion of a ring up a hill, specifically focusing on the calculation of its moment of inertia and energy considerations. The subject area includes concepts from rotational dynamics and energy conservation.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of the moment of inertia for an annulus and question the method of using mass density. There are suggestions to treat the annulus as the difference of two discs, and some participants express the importance of maintaining symbolic representations in calculations to avoid errors.

Discussion Status

The discussion is ongoing, with participants providing guidance on the calculation methods and emphasizing the importance of symbolic manipulation. There is recognition of mistakes in initial calculations, and some participants are revising their approaches based on feedback.

Contextual Notes

Participants are navigating through the complexities of calculating the moment of inertia and energy for a non-standard shape, with some constraints related to the assumptions made about mass density and the geometry of the problem.

Esoremada
Messages
50
Reaction score
0

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
 
Last edited by a moderator:
Physics news on Phys.org
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.
 
  • Like
Likes   Reactions: 1 person