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Help with finding Moment of Inertia

by yang09
Tags: inertia, moment
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yang09
#1
Mar30-10, 02:49 PM
P: 76
1. The problem statement, all variables and given/known data

Determine the moment of inertia of a cylinder
of radius 0.37 m, height 1.8 m and density
(1.11 − 0.555 r + 0.284 r2) kg/m3 about the
center.
Answer in units of kg m2.

2. Relevant equations

I=Integral[(r^2)dm]
I = mr^2
Density = Mass/Volume
Volume of a cylinder = (Pie)(radius^2)(Height)

3. The attempt at a solution

I used the density formula and solved for mass, Mass = (Density)(Volume). To get the volume, I used the volume for a cylinder formula and plugged in my numbers. From the volume and density, I got mass = (1.11-0.555r+0.284r^2)(Pie)(.37^2)(1.8). I then plugged my mass into the Inertia equation, getting I = (mass)(.37^2). I calculated it all out and got I = 0.099997.
This is wrong and I don't know what I'm doing wrong. Am i supposed to integrate the radius, but that wont do anything because r is a constant.

1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
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kuruman
#2
Mar30-10, 03:03 PM
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The formula Mass = Density*Volume is applicable only when the density is constant. More generally, when the density depends on position

[tex]M=\int \rho(r) dV[/tex]

So to calculate the moment of inertia, first you need to replace dm with ρ(r)dV and then you need to do the (triple) integral

[tex]I=\int \rho(r) r^2 dV[/tex]
Doc Al
#3
Mar30-10, 03:10 PM
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Quote Quote by yang09 View Post
I=Integral[(r^2)dm]
This is good. To set up the integral, treat 'dm' as the mass of a cylindrical shell. How would you express the volume and then the mass (dm) of that shell in terms of r and h?

yang09
#4
Mar30-10, 03:12 PM
P: 76
Help with finding Moment of Inertia

Ok so by doing so, I got

I = (Pie)Integral[(1.11r^4)-(0.555r^5)+(0.284r^6)h].
But how would I integrate the function if there are two variables?
Doc Al
#5
Mar30-10, 04:20 PM
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Quote Quote by yang09 View Post
Ok so by doing so, I got

I = (Pie)Integral[(1.11r^4)-(0.555r^5)+(0.284r^6)h].
Looks off a bit to me. Please describe your volume element.
But how would I integrate the function if there are two variables?
I only see one variable. (h is a constant.)
yang09
#6
Mar30-10, 04:30 PM
P: 76
From Integral[(Density)(r^2)dV], I replaced dV with the cylinder volume, (Pie)(r^2)(height), and replaced density with (1.11-0.55r+0.284r^2).
After doing so, I got Integral[(1.11-0.55r+0.284r^2)(r^2)(Pie)(r^2)(h)]. I brought Pie in front of the integral because pie is a constant and multiplied the terms out. Ultimately getting Integral[(1.11r^4)-(0.555r^5)+(0.284r^6)h].
So what you're saying is to bring "h" out in front and treat it like pie?
Doc Al
#7
Mar30-10, 04:55 PM
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Quote Quote by yang09 View Post
From Integral[(Density)(r^2)dV], I replaced dV with the cylinder volume, (Pie)(r^2)(height),
No. dV is not the volume of the entire cylinder. (As kuruman pointed out, the density is not constant.) dV is volume of a thin cylindrical shell of radius r. Hint: Its thickness will be dr.

(And yes, you can treat h like any other constant.)
yang09
#8
Mar30-10, 05:08 PM
P: 76
As you can tell, I'm new to this subject. Just please bear with me.
So I'm still confused. You're saying that dV is not the volume of the entire cylinder so I guess I'm not allowed to plug in the cylinder volume into dV. But if I'm not allowed to do that, then how I would I incorporate h into the integral?
Doc Al
#9
Mar30-10, 05:21 PM
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dV will be the volume of a cylindrical shell of height h and radius r. So your integral will still have an h. Step one is to write dV in terms of r and h.
yang09
#10
Mar30-10, 05:41 PM
P: 76
Tell me if this is correct, if I'm trying to find the change in volume, then that means that I am finding the area within the cylinder. And to do so, I would integrate the area of the cylinder, 2(pie)(radius)(height), making my dV become the area of the cylinder. Is this process right or is it not?
And then from here, I would distribute my terms out and integrate?
Doc Al
#11
Mar30-10, 05:46 PM
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Quote Quote by yang09 View Post
Tell me if this is correct, if I'm trying to find the change in volume, then that means that I am finding the area within the cylinder. And to do so, I would integrate the area of the cylinder, 2(pie)(radius)(height), making my dV become the area of the cylinder. Is this process right or is it not?
Let's tune that up a bit.

Looked at from above (along the axis of the cylinder) the shell is just a thin circular strip. Its radius is r, thus its length is 2πr. The thickness of the strip is dr, thus its area is 2πr dr. The volume of the corresponding shell is that area times the height, h. So dV = 2πr dr * h = 2πhr dr.

Make sense?
yang09
#12
Mar30-10, 05:53 PM
P: 76
Ok thanks for all the help. It seems to make sense as of now. But I have to go to class in a couple of minutes so I can't really work out the problem now. But if I don't understand, then I'll just message again tomorrow. Thanks again


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