Moment of Inertia of a partial disk

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SUMMARY

The discussion focuses on calculating the moment of inertia for a partial disk missing one sixth of its area. The correct limits of integration for the angle are established as 0 to 5π/3 radians, eliminating the need for the parallel axis theorem in this scenario. The integration approach involves using cylindrical coordinates, specifically the volume element dV = r dr dφ dz. The conversation highlights the importance of persistence in mastering complex calculus concepts related to physics.

PREREQUISITES
  • Understanding of moment of inertia concepts
  • Familiarity with calculus integration techniques
  • Knowledge of cylindrical coordinates
  • Basic physics principles related to rotational motion
NEXT STEPS
  • Study the derivation of moment of inertia for various shapes
  • Learn advanced integration techniques in calculus
  • Explore applications of the parallel axis theorem in physics
  • Practice problems involving cylindrical coordinates in physics
USEFUL FOR

Students in physics and engineering, particularly those studying mechanics and calculus, will benefit from this discussion as it addresses the complexities of calculating moment of inertia for partial shapes.

abertram28
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Let me first start by saying that this problem is a challenge problem and the due date has passed (friday). I've still been working on it because it stumped me. in my physics class we did some integration based inertia problems. my calculus isn't all that great so its a bit tough for me. what i did understand was finding the moment of inertia of the rod through calculus. and i followed along to see the moment of inertia of a disk through the axis of revolution. it was a bit confusing on using dTheta instead of a dx and dy.

the teacher asked us to show the moment of inertia of a disk missing one sixth. so i guess the new limits of integration of the dTheta part are going to go from 0 to 5pi/3? is that it? no parallel axis theorem?

if you reply, please try to have a little understanding that my math lags behind my physics a bit. sorry bout that, I am taking a strong math semester next to make up for it.
 
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the teacher asked us to show the moment of inertia of a disk missing one sixth. so i guess the new limits of integration of the dTheta part are going to go from 0 to 5pi/3? is that it? no parallel axis theorem?
Correct !

[tex]I = \int {\rho (\vec{r})r^2dV[/tex]
and in cylindrical co-ordinates you have :

[tex]dV = r dr d\phi dz[/tex]
 
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First of all, it's great that you're still working on this problem even though the due date has passed. It shows determination and a desire to fully understand the concept. Calculus can be challenging, but with practice and persistence, you will surely improve.

Now, to address your question, yes, the new limits of integration for the dTheta part would be from 0 to 5pi/3. This is because the missing one sixth of the disk would result in a sector of 5pi/3 radians. And you are correct, there is no need to use the parallel axis theorem in this case.

It is understandable that your math skills may not be at the same level as your physics skills, but don't worry. As you mentioned, you are taking a strong math semester next to catch up, and that shows commitment to improving. Keep up the hard work and don't be afraid to ask for help or clarification when needed. Good luck!
 

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