Why Does Extracting Density Affect My Moment of Inertia Calculation?

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SUMMARY

The discussion focuses on the calculation of the moment of inertia for a cylindrical object, specifically addressing the impact of density extraction on the integral formulation. The integral I=\int r^2 dm is correctly transformed to I=\int r^2 p dV, where p is the density. The confusion arises when the user pulls p out of the integral, leading to different results: I=Mr^2 when p is not extracted and I=M/2r^2 when it is. The correct moment of inertia for a solid cylinder is I=M/2r^2, confirming that the user’s method of integrating with constant density is valid.

PREREQUISITES
  • Understanding of integral calculus, specifically integration techniques.
  • Familiarity with the concept of moment of inertia in physics.
  • Knowledge of cylindrical coordinates and volume elements.
  • Basic principles of density and mass distribution in solid objects.
NEXT STEPS
  • Study the derivation of moment of inertia for various geometric shapes, focusing on solid cylinders.
  • Learn about the application of cylindrical coordinates in calculus.
  • Explore the relationship between density, mass, and volume in physics.
  • Investigate common mistakes in integral calculus involving variable and constant parameters.
USEFUL FOR

Physics students, mechanical engineers, and anyone involved in solid mechanics or material science who seeks to deepen their understanding of moment of inertia calculations.

kuahji
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I'm attempting to derive the moment of inertia for a cylindrical object.

I know that I=[tex]\int r^2 dm[/tex]

which equals =[tex]\int r^2 p dV[/tex]

My question begins here, the derivations I seen pull p out of the integral, which makes sense to do, because in this case it's a constant. p=M/([tex]\pi[/tex]r^2L). So if I don't pull p out before integrating I get I=Mr^2, if I do pull it out, I get I=M/2r^2. I know the answer should be I=M/2r^2 because I have a solid cylindrical object. So why am I getting a different result when I leave p in, & a different result when I pull p out or am I just making a silly math error?

Below is my work when I leave p inside the integral

I=[tex]\int r^2*p*(2\pi*r)dr[/tex]
=2M[tex]\int r dr[/tex] (replacing p with M/([tex]\pi[/tex]r^2L) before integrating)
=Mr^2
 
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I believe that density is constant for each material.

p/s: you replace m = DV, then you replace D = M/V... I don't get it :(
 
I replaced dm with pdV & then p with M/V.
 

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