Rings Vs Disks In Moment of Moment of Inertia Integral

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SUMMARY

The discussion centers on the differences in volume element calculations for center of mass and moment of inertia problems involving a cone. The volume element for the center of mass is expressed as dv = (πr²)dz, while for moment of inertia, it is dv = (2πr)drdz. The teacher explains that rings are used in moment of inertia calculations due to their mass being concentrated at a single radius, which contrasts with the disk approach used for center of mass calculations. The participant seeks clarity on why differentials are treated differently and confirms that both methods yield correct results when triple integrals are properly set up.

PREREQUISITES
  • Understanding of volume elements in calculus
  • Familiarity with moment of inertia concepts
  • Knowledge of center of mass calculations
  • Basic integration techniques, particularly triple integrals
NEXT STEPS
  • Study the derivation of volume elements for various geometric shapes
  • Learn about the application of triple integrals in physics problems
  • Explore the differences between mass distribution in disks and rings
  • Investigate the implications of using different volume elements in physics calculations
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Students and educators in physics and engineering, particularly those focusing on mechanics and calculus applications in real-world scenarios.

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Homework Statement


Not really homework, but putting it here to be safe. While doing a center of mass calculation of a cone centered about the origin (tip touching origin, z axis through it's center) my teacher put on the board that the volume element was

dv= (pi*r^2)*dz using the area of a disk. The variable r was then eliminated by substituting r as a function of z.

However, when doing a moment of inertia problem, he used

dv= (2*pi*r)*dr*dz using rings of width dr

I asked him why the differentials were different and he said, because when doing moment of inertia problems, we use rings because they have their mass concentrated at one radius.

Wouldn't the same argument require us to use rings for the center of mass calculation, which is also dependent upon r ?

He is changing the differentials to avoid making people do triple integrals. Which I prefer here because I can get the correct answer in both cases by setting up the proper triple integral and working it out. I just want to understand it his way in case a test is more easily worked that way.

Thank you
 
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In the center of mass calculation, you're finding ##\langle z \rangle## whereas in the moment of inertia calculation, you're finding ##\langle r^2 \rangle##. If you set up the triple integrals and then integrate so that z or r is the last variable of integration, you'll see you end up with the two different volume elements.
 
Ok thank you.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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