Help with finding Moment of Inertia


by yang09
Tags: inertia, moment
yang09
yang09 is offline
#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
Phys.Org News Partner Science news on Phys.org
Simplicity is key to co-operative robots
Chemical vapor deposition used to grow atomic layer materials on top of each other
Earliest ancestor of land herbivores discovered
kuruman
kuruman is offline
#2
Mar30-10, 03:03 PM
HW Helper
PF Gold
kuruman's Avatar
P: 3,444
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
Doc Al is online now
#3
Mar30-10, 03:10 PM
Mentor
Doc Al's Avatar
P: 40,875
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
yang09 is offline
#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
Doc Al is online now
#5
Mar30-10, 04:20 PM
Mentor
Doc Al's Avatar
P: 40,875
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
yang09 is offline
#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
Doc Al is online now
#7
Mar30-10, 04:55 PM
Mentor
Doc Al's Avatar
P: 40,875
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
yang09 is offline
#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
Doc Al is online now
#9
Mar30-10, 05:21 PM
Mentor
Doc Al's Avatar
P: 40,875
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
yang09 is offline
#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
Doc Al is online now
#11
Mar30-10, 05:46 PM
Mentor
Doc Al's Avatar
P: 40,875
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
yang09 is offline
#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


Register to reply

Related Discussions
finding moment of inertia Introductory Physics Homework 1
moment of inertia - disk - finding dm Introductory Physics Homework 2
I need help finding my mistake...moment of inertia Advanced Physics Homework 1
Finding moment of inertia Introductory Physics Homework 1
Finding the moment of inertia Introductory Physics Homework 10