Finding Moment of Inertia of Infinitely Thin Hoop using a Double Integral

In summary, the problem asks how to determine the moment of inertia of a hoop with an infinitely thin width. The user is trying to find a way to modify the region R to take into account the "uniform thin hoop" that has no width. If they use a line integral, they get an answer for one of the perpendicular axes but need to use symmetry arguments to check their answer.
  • #1
opticaltempest
135
0
Here is the problem:

http://img141.imageshack.us/img141/3830/problemsm5.jpg

Is it possible to determine this moment of inertia in this problem using double integrals of the form:

http://img172.imageshack.us/img172/1219/momented0.jpg

I could do this problem using double integrals if the region R was a solid circle using [tex]x^2+y^2=R^2[/tex] and then simplifying the integration by switching to polar coordinates. However, the region is a hoop whose width is infinitely thin.

How do I modify my region R in order to take into account the "uniform thin hoop" that has no width? Is there a way I could evaluate a hoop of width [tex]\Delta R[/tex] and find moment of inertia in terms of [tex]\Delta R[/tex] then let find the limiting value as [tex]\Delta R[/tex] approaches zero?

If I used that approach, what would I use for the density of the infinitely thin hoop?
 
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  • #2
I think you are talking about comparing two disks of radius R and R+dr... and look as the limit of dr goes to 0.

When I first read this tho, I went about it a slightly different way...
Note how their example has you integrating over little pieces of mass (along the whole loop). In this case since it's infinitely thin, in some sense you are really doing a line integrals, not area integrals. If you want to use the pendendicular axis theorem the line intergrals are just more difficult because they have constraints that are more difficult, but you still use density as mass per unit length... not mass per unit volume... and what would that value be? (Hint -- what is the circumference of the loop?)

Actually, if you are familiar with "delta functions", the constraints and integrals are pretty easy.

Then - once you get an answer for one of the perdendicular axes, can you use symmetry arguments to check your answer? (Hint: How should the magnitudes of Ix, Iy compare?)
 
  • #3
I think I'm almost there if I use a line integral. I am off by a factor of [tex]R[/tex]. Why?

The density will mass per unit length

[tex]\rho = \frac{M}{L}[/tex]

The length of the curve is [tex]2\pi R[/tex] which gives a density of

[tex]\rho = \frac{M}{{2\pi R}}[/tex]The equation of the circle is

[tex]x^2 + y^2 = R^2[/tex]

Parameterizing the curve gives

[tex]\vec r(t) = x\left( t \right)\hat i + y\left( t \right)\hat j[/tex][tex]\vec r(t) = R\cos \left( t \right)\hat i + R\sin \left( t \right)\hat j[/tex]Setting up the line integral we have

[tex]\int_C^{} {\left( {\frac{M}{{2\pi R}}} \right)} ds[/tex]

[tex]ds = \sqrt {x'(t)^2 + y'(t)^2 }dt [/tex]

[tex]ds = \sqrt {\left( { - R\sin t} \right)^2 + \left( {R\cos t} \right)^2 }dt[/tex]

[tex]ds = \sqrt {R^2 \sin ^2 t + R^2 \cos ^2 t}dt[/tex]

[tex]ds = \sqrt {R^2 \left( {\sin ^2 t + \cos ^2 t} \right)}dt[/tex]

[tex]ds = \sqrt {R^2 }dt [/tex]

[tex]ds = Rdt[/tex]

The integral for the moment of inertia of the line

[tex]
\frac{M}{{2\pi R}}\int_0^{2\pi R} {Rdt} = \frac{M}{{2\pi }}\int_0^{2\pi R} {dt = } \frac{M}{{2\pi }}\left[ t \right]_{t = 0}^{t = 2\pi R} = MR
[/tex]
Where does the other factor of R need to come from?

What are delta functions?
 
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  • #4
opticaltempest said:
Where does the other factor of R need to come from?
It comes from the definitiuon of moment of inertia, and using the correct limits of integration. t is not a length. It is a parameter that ranges from zero to 2*pi

[tex]\frac{M}{{2\pi R}}\int_0^{2\pi} {R^2 Rdt} [/tex]
 
  • #5
opticaltempest said:
Here is the problem:

I could do this problem using double integrals if the region R was a solid circle using [tex]x^2+y^2=R^2[/tex] and then simplifying the integration by switching to polar coordinates.

As you said it yourself, you have simplified the integration by switching to polar coordinates. The additional R comes from polar coordinates. When switching the variables of integration, you have to include the Jacobian determinant in the integral. The Jacobian determinant is the determinant of the matrix of all first-order partial derivatives of a vector-valued function.

In our case, we have to find the partial derivatives of [tex]x=R\cdot \cos(\phi)[/tex] and [tex]y=R\cdot \sin(\phi)[/tex] (x and y expressed in polar coordinates) using R and [tex]\phi[/tex] as derivative parameters:
[tex]\left|\frac{dx}{dR} \ \frac{dx}{d\phi}\right|[/tex]
[tex]\left|\frac{dy}{dR} \ \frac{dy}{d\phi}\right|[/tex]

For the integration parameters:
[tex]\phi[\tex] then ranges from 0 to 2[tex]\pi[\tex] and R ranges from 0 to R.
 
  • #6
I still got some doubts about this.. I've tried many ways to find the moments of inertia [tex]I_{x} = I_{y} [/tex] but i always get some constants wrong.
If i parametrize [tex] x(t) = R Cos(t) [/tex], what's wrong with taking the integral [tex] \int _{0} ^{2 \pi} \rho x^{2} dt [/tex]? If i do that, i get the correct factor of [tex] 1/2 [/tex] but R instead of [tex] R^{2} [/tex] :
[tex] \int _{0} ^{2 \pi} \rho x^{2} dt = M/{2 \pi R} \int _{0} ^{2 \pi} {R^{2}cos^{2} (t) dt = {M R}/{4 \pi} [/tex] .. i don't even get the 1/2 not to mention that [tex] \pi [/tex]
If i try to set x as a function of y, then i get some [tex] 1/3 [/tex] around from taking the integral of [tex] y^{2} [/tex]
this calculations frustate me...
 
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  • #7
i've found this theorem helpful:
http://www.rzg.mpg.de/~rfs/comas/lectures/SportundPhysik/materialien/Aerodynamics/Schulprojekt/Magnus/ajp00181.pdf
 
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  • #8
I've found it! I'm going to post it just ftw.
I had some mistakes in my first post, for a start, the integral [tex] \int _{0} ^{2 \pi} x^2 \rho dt [/tex] should be done with ds instead (so the [tex] \rho ds [/tex] would give the mass of that bit, and all makes physical sense). Since [tex] ds = R dt [/tex] we have:
[tex] I_y = \int_{0} ^{2 \pi} R^2 cos^2 (t) \rho R dt =
{M R^2}/{2 \pi} \int_{0} ^{2 \pi} cos^2 (t) dt [/tex]
i thought that the integral was just 1/2, but i was wrong. applying integration by parts twice i get that the integral is [tex] \pi [/tex] and it all comes out nice...
got to be careful with those integrals and differentials!
 

Related to Finding Moment of Inertia of Infinitely Thin Hoop using a Double Integral

What is the concept of moment of inertia?

The moment of inertia is a physical property of an object that describes its resistance to changes in rotational motion. It is a measure of the object's distribution of mass around an axis of rotation.

How is the moment of inertia of an infinitely thin hoop calculated?

The moment of inertia of an infinitely thin hoop can be calculated using a double integral, which involves integrating the mass of the hoop over its entire surface area. The equation for the moment of inertia of an infinitely thin hoop is I = mr^2, where m is the mass of the hoop and r is the radius.

What is the significance of using a double integral to calculate the moment of inertia of an infinitely thin hoop?

A double integral allows for the calculation of the moment of inertia of an infinitely thin hoop, which is a continuous object with no specific point masses. This method takes into account the distribution of mass over the entire surface area of the hoop, resulting in a more accurate calculation.

What are the assumptions made when calculating the moment of inertia of an infinitely thin hoop?

The calculation of the moment of inertia of an infinitely thin hoop assumes that the hoop has a constant density and a uniform shape with no variations in its thickness. It also assumes that the hoop is rotating around its central axis.

Can the moment of inertia of an infinitely thin hoop be calculated using other methods?

Yes, the moment of inertia of an infinitely thin hoop can also be calculated using the parallel axis theorem or the perpendicular axis theorem. However, the double integral method is often preferred as it takes into account the entire surface area of the hoop and does not require any additional assumptions.

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