Conceptual Problem (Integration)

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SUMMARY

The discussion centers on the derivation of the moment of inertia of a cylinder, specifically addressing the relationship between the surface area of cylindrical shells and the change in volume. The equation for the change in mass is established as (change in mass) = 2(π)(radius)(height)(change in radius)(density). The key insight is that the term "dr" represents an infinitesimal thickness of each cylindrical shell, allowing for the calculation of volume as dV = 2(π)(radius)(height)(dr). This integration leads to the total volume of the cylinder.

PREREQUISITES
  • Understanding of calculus, specifically integration techniques.
  • Familiarity with the concept of cylindrical coordinates.
  • Knowledge of the physical properties of density and mass.
  • Basic understanding of geometric shapes, particularly cylinders.
NEXT STEPS
  • Study the process of integration in calculus, focusing on volume calculations.
  • Learn about cylindrical coordinates and their applications in physics.
  • Explore the derivation of the moment of inertia for various geometric shapes.
  • Investigate the relationship between density, mass, and volume in physical systems.
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Students of physics and engineering, particularly those studying mechanics and material properties, as well as educators looking to explain the concept of moment of inertia in a clear and structured manner.

anonymousphys
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1. Why is 2(pie)rh multiplied by the dr (change in r) equal to the change in volume?
This is in the case of deriving the moment of inertia of a cylinder; the equation is (change in mass)=2(pie)(radius)(height)(change in radius)(density).

Homework Equations


The Attempt at a Solution


I don't quite understand how 2(pie)(radius)(volume)(change in radius) equals change in volume? Is there a proof for this? I can imagine this working mathematically but not conceptually.2(pie)(radius)(height) is just the surface area for each layer, but how does the change in radius come in? I know it has something to do with accounting for the thickness of each hollow cylinder?

All replies are much appreciated. thanks.
 
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anonymousphys said:
1. Why is 2(pie)rh multiplied by the dr (change in r) equal to the change in volume?
This is in the case of deriving the moment of inertia of a cylinder; the equation is (change in mass)=2(pie)(radius)(height)(change in radius)(density).

Homework Equations


The Attempt at a Solution


I don't quite understand how 2(pie)(radius)(volume)(change in radius) equals change in volume? Is there a proof for this? I can imagine this working mathematically but not conceptually.2(pie)(radius)(height) is just the surface area for each layer, but how does the change in radius come in? I know it has something to do with accounting for the thickness of each hollow cylinder?

All replies are much appreciated. thanks.

Well let's start off simple. We know that 2*pi*r is the circumference of a circle. So the circumference of a circle multiplied by a height h means that we now the have the surface area of a cylinder (excluding tops). Do you see that at least? Once we have the surface area, it's just one more dimension to get to volume. So if you multiply that surface area by an infinitesimal distance element dr, you get an infinitesimal volume element = dV = dA*dr. Integrate over all that and you get the total volume.
 
Hi anonymousphys! :smile:

(have a pi: π :wink:)
anonymousphys said:
I don't quite understand how 2(pie)(radius)(volume)(change in radius) equals change in volume? … how does the change in radius come in? I know it has something to do with accounting for the thickness of each hollow cylinder?

I suspect you're confused because you're calling dr the "change in" radius.

When you're integrating, "d" just means "a small amount of" …

change doesn't come into it.

You're dividing the big cylinder into a lot of tiny cylindrical shells, each of thickness dr.

Each shell has surface area 2πrh, so multiply that by the thickness (dr) to get the volume dv = 2πrh(dr), and by density to get dm = 2πρrh(dr).

(Then integrate to get ∫ 2πρrh dr)

In all these cases, "d" simply means a small amount. :wink:
 
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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