Moment of Inertia for a solid circular disc

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Discussion Overview

The discussion revolves around calculating the moment of inertia (MOI) for a solid circular disc rotating about an axis that passes through its edge and is parallel to its diameter. Participants explore the application of relevant equations and the parallel axis theorem in this context.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant expresses confusion about the problem and suggests there might be a typo in the question paper regarding the moment of inertia to be shown as I=1.25mr^2.
  • Another participant asks if the original poster is familiar with using integration to calculate the moment of inertia.
  • A different participant suggests looking into the parallel axis theorem as a potential method to solve the problem.
  • One participant realizes they had forgotten to include a previous question about proving that a disc rotating about an axis through its diameter has I=0.25mr^2, and wonders if this value should be used in their calculations.
  • A later reply indicates that the participant successfully applied the parallel axis theorem, arriving at the correct moment of inertia of I=1.25mr^2.
  • Another participant reflects on their earlier confusion and acknowledges that the problem was easier than initially thought.

Areas of Agreement / Disagreement

Participants generally agree on the use of the parallel axis theorem and the calculations involved, but there is no explicit consensus on the initial interpretation of the problem or the presence of a typo.

Contextual Notes

Some participants express uncertainty about the syntax of the question and the assumptions underlying the calculations. There are references to prior problems that may influence the current discussion.

Elmowgli
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hey kinda new to this and I know the rules say I am not allowed to be told how to do this but I am totally stumped and its to be handed in tomorrow. I've looked through everything and cannot find out how to do it anywhere I am starting to think there is a typo in the question paper :S

show that a disc rotating about an axis that passes through the edge of the disc and parallel to its diamter is I=1.25mr^2


Homework Equations



I=0.5mr^2


The Attempt at a Solution


the only thing I have found that's in any way similar is the MOI of a sphere which equals 2/5MR^2, I think if there's some way of multiplying the two fifths by the regular equ then it would work, but i can't find anyway of doing that.

Any help is appreciated :)

 
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Elmowgli said:
hey kinda new to this and I know the rules say I am not allowed to be told how to do this but I am totally stumped and its to be handed in tomorrow. I've looked through everything and cannot find out how to do it anywhere I am starting to think there is a typo in the question paper :S

show that a disc rotating about an axis that passes through the edge of the disc and parallel to its diamter is I=1.25mr^2


Homework Equations



I=0.5mr^2


The Attempt at a Solution


the only thing I have found that's in any way similar is the MOI of a sphere which equals 2/5MR^2, I think if there's some way of multiplying the two fifths by the regular equ then it would work, but i can't find anyway of doing that.

Any help is appreciated :)

Welcome to the PF. Are you familiar with how to use integration to calculate the MOI?
 
thanks :) I am vaguely familiar with it but I am not exactly great at it.
 
Well, I don't really understand the question, i.e., the syntax seems a bit awkward, but never mind. I think it's OK to give you this suggestion: Look at the parallel axis theorem.
 
TVP45 said:
Well, I don't really understand the question, i.e., the syntax seems a bit awkward, but never mind. I think it's OK to give you this suggestion: Look at the parallel axis theorem.

Yes, good hint :wink:
 
hi that's really helping, I just realized that it in my confused state I forgot to include that the question before this was to prove that,
a disc rotating about an axis that coincides with a diamter is I=.25mr^2.

and I was wondering would this give me the value that I would sub in for D in the equation?
 
I got it eventually, I was just looking at it the wrong way round,

ended up with

Io=Ic + md^2
= 0.25mr^2 + mr^2
= 1.25mr^2

feel very stupid now after seeing how easy it was!
 
All my problems have been easy once I saw how to do them.:blushing:

The only dumb questions are the ones unasked.
 
haha, too true! :)
 

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