Using Calculus To Derive Forumlas

In summary, the proof of the moment of inertia of a sphere uses integration to divide the sphere into a lot of small masses and then calculates the contribution to the moment of inertia for each mass. The approach is general, but easier to understand if you can exploit symmetry in the system. There are many online examples of this sort of calculation.
  • #1
Periapsis
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Can someone explain this proof of the moment of inertia of a sphere? And how you can use the same principals to solve for other formulas?

https://www.physicsforums.com/attachment.php?attachmentid=62987&stc=1&d=1381895479http://

I understand each individual variable. But I don't understand how they were able to get the moment of inertia by the orginal integrand, and where did that integrand come from?
 
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  • #2
The image didn't come out.

The general principle is to divide the object into a lot of small masses dm.
Work out the contribution to the moment of inertia for each dm and add them up.
Since it is a continuous sum, you use an integration sign instead of a sigma.

The approach is general - however, the math is easier if you can exploit some aspect of the symmetry of the system. i.e. for the sphere, they probably exploited spherical symmetry by using spherical-polar coordinates.

There are many examples of this sort of calculation online - have you had a look?
Mainly you just have to understand how integration works as a sum of small bits rather than as a collection of fancy tricks and formulas.
 
  • #3
I followed along with a proof of the moment of inertia of a solid cylinder, but I suppose the only thing i don't understand, is what formulas they start with. I get that you always start with I=∫r2dm is it all just a matter of logic, and how you can substitute dm with ρdV?
 
  • #4
"how you can substitute dm with ρdV? "

Mass equals density times volume, does it not?
 
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  • #5
It does if density is a constant. More generally, "mass equals the integral of density over the volume" which immediately leads to the differential formula "[itex]dm= \rho dV[/itex]".
 
  • #6
It does, i think you just answered my ultimate question. How does substitution work in any integral/derivative? so you can substitute mass with ρdV, simply because mass = ρv? why do you say its with respect to volume and not density? Same with velocity as a derivative of position. d/dx s(t) = v. why is that?
 
  • #7
@Hallsoflvy why are you saying mass equals the integral of density over volume? ρ = m/V. rearranged, m = ρV.
 
  • #8
Remember that we always have M/V=density (when we regard density as the average density over V)

When we in maths shrink, in the limiting process, the volume element, the right hand side (i.e, the density) goes towards a non-zero, constant value we call the LOCAL density. The LHS has then become dm/dV if you like.
 
  • #9
ohhh okay, so we take the differential of Volume? because we are splitting the sphere up into tiny spheres with near-zero volume? And since density stays constant, its simply with respect to volume?
 
  • #10
Periapsis said:
ohhh okay, so we take the differential of Volume? because we are splitting the sphere up into tiny spheres with near-zero volume? And since density stays constant, its simply with respect to volume?

It "becomes" a constant, if we make the volume small enough, within our mathematical modelling.
--------------------------------------
If we look at the product:
[tex]\rho(x)(\bigtriangleup{x}|_{x=x_{0}}) (*)[/tex]
where the volume [itex]\bigtriangleup{x}|_{x=x_{0}}[/itex] is CENTERED about [itex]x=x_{0}[/itex], while we let the density be allowed to vary across the volume, we have, approximately:
[tex]\rho(x)\approx\rho(x_{0})+\frac{d\rho}{dx}|_{x=x_{0}}(x-x_{0})[/tex]
and (*) can then be written as:
[tex]\rho(x)(\bigtriangleup{x}|_{x=x_{0}})=\rho(x_{0})(\bigtriangleup{x}|_{x=x_{0}})+\frac{d\rho}{dx}|_{x=x_{0}}(x-x_{0})(\bigtriangleup{x}|_{x=x_{0}})[/tex]

Note that the second term is, basically, of magnitude volume SQUARED, while the first is of magnitude "volume".

Thus, as you let "volume" shrink in magnitude, the second term gets negligible, relative to the first, sincy tiny*tiny=even tinier.

For EACH number x_0, then, each with an infinitesemal volumedV attached to it, dM=rho*dV, where the density function is evaluated at x_0 (and we drop the subscript _0 in the following integration)
 
  • #11
Periapsis said:
ohhh okay, so we take the differential of Volume? because we are splitting the sphere up into tiny spheres with near-zero volume? And since density stays constant, its simply with respect to volume?
Almost - splitting the sphere into tiny spheres is unlikely to help because spheres don't tesselate well.

The exact shape of the volume element is something you choose according to the coordinate system (which you choose according to symmetry).

In cartesian coordinates, you divide the sphere into lots of little cubes.

A rectangular volume would be given by ΔV=ΔxΔyΔz [with one corner at position (x,y,z)]
A smaller volume would be represented by a small delta: δV=δxδyδz ... it's a notation meaning "small change in..."

An infinitesimal volume would be dV=dx.dy.dz

The amount of mass in the infinitesimal volume dV at position (x,y,z) is dm=ρ(x,y,z)dxdydz

In your example the density is uniform so you just get dm=ρdxdydz


In cylindrical coordinates, a volume element is a kind-of curved brick: dV=dr(r.dθ)dz
Because rdθ is the arc-length at radius r inside angle dθ... dr is the width and dz is the height.


It can take a while to get used to - but it is key to understanding how to set-up integrals.

i.e. an area would be given by: ##A=\int dA## ... "doh!" right?
But let's say it's the area of a rectangle with sides b and w... then dA=dxdy and:
$$A=\int_0^w\int_0^b dxdy$$
How about the area of a circle radius R ... then dA=r.dθdr
$$A=\int_0^{2\pi}\int_0^R r.drd\theta$$
 
  • #12
I think i would understand this a lot better if I had a much better understanding of integrals. In high school AP calculus, we are just learning the product and quotient rule of derivatives, I just read ahead and try and scrounge the internet for more information
 
  • #13
Ah right ... so right now integration is a spooky magic black-box process where you apply a bunch of rules.

You know yu can find the area of an irregular polygon by dividing it up into rectangles and triangles, finding the areas of each of them then adding them all up? Integration is a way of doing that in a more systematic way.

If you want the area under f(x) inside a<x<b ... then you can divide the regeon into N smaller ones with width Δx = (b-a)/N and use those to make a bunch of rectangles.

The 1st rectangle has height f(a+Δx), the third has height f(a+2Δx) and the third f(a+3Δx) and so on ... the nth one has height f(a+nΔx) and the Nth one f(a+NΔx)=f(b) ... all these rectangles sort-of follow the curve.

So the nth rectangle has area: f(a+nΔx)Δx ... and the area of the entire curve is approximately $$\sum_{n=1}^N f(x_n)\Delta x:x_n=a+nΔx$$ If we make N bigger and Δx smaller and smaller in such a way that NΔx = (b-a) doesn't change, then the approximation gets closer and closer to the actual area you want to find.

When Δx is infinitesimally small ... the sum becomes an integral sign and the Δx becomes a dx and the xn's are just values on the x-axis and you write: $$A=\int_a^bf(x)\; dx$$... which should be familiar to you.

When you study sums like this, you discover there are lots of shortcuts that mean you don't have to go back to dividing a function into strips every single time. That's what you are learning.Dividing into strips was just a handy shortcut also. You can also divide the y-axis up into M sections and you get a lot of small boxes - then you count up the boxes and mutiply by the area.
... that will require two sums, which leads to two integrals as you make the boxes very small.See also:
http://tutorial.math.lamar.edu/Classes/CalcI/CalcI.aspx
 
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  • #14
okkayy, i totally understand that! thank you so much!
 
  • #15
It takes a while to get used to it.
Enjoy.
 

What is calculus?

Calculus is a branch of mathematics that deals with the study of change and motion. It involves the use of mathematical models to understand and solve problems related to rates of change, derivatives, and integrals.

How is calculus used to derive formulas?

Calculus is used to derive formulas by understanding the relationship between the variables involved in a problem and how they change over time. By taking the limit of small changes in these variables, calculus can help determine the precise formula that describes the relationship between them.

What are some common applications of calculus in deriving formulas?

Calculus is used in many fields, such as physics, engineering, economics, and statistics. Some common applications include finding the velocity and acceleration of a moving object, determining the maximum or minimum value of a function, and calculating the area under a curve.

What are the different techniques used in calculus to derive formulas?

The two main techniques used in calculus are differentiation and integration. Differentiation involves finding the rate of change of a function, while integration involves finding the area under a curve. These techniques can be applied to derive various formulas and solve problems.

Why is calculus important in scientific research?

Calculus is important in scientific research because it provides a powerful tool for modeling and analyzing complex systems and phenomena. It allows scientists to make accurate predictions and understand the behavior of different variables in a system. Many scientific theories and principles, such as Newton's laws of motion and the laws of thermodynamics, are based on calculus.

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