# Rotational Dynamics: Calculating a Moment of Inertia

1. Jan 3, 2009

### Haths

I feel like a div. Quite simply I've forgotten how to use integration to calculate a moment of inertia (MOI). Ok I want to calculate the MOI of say a solid disk, about an axis perpendicular to the plane of the disk, running through its center of mass.

I remember that for any point particle in the disk its MOI is going to be:

mr2

I can show this because if we consider a point mass at some distance r from a point of rotation, it will still subscribe to F=ma in the tangential direction. Knowing that:

T= I [alpha]

Where T is the torque, I the MOI, and alpha the angular acceleration.

T=Fr

Therefore:

T=mar

As

[alpha] = a / r

mar=I a/r

I=mr2

Hence I'm certain of that. Thus I can appreciate that for a solid disk I am looking at a summation of the MOI's between the axis and the edge of the disk. Hence I need an integration factor such that;

IT= Integr0{ dm/dr r2 } dr

I think. Yes? Hence I need an expression for the rate of change of mass with respect to radial displacement. Hence I need an area integral, and this is where I'm left blank, because I have no idea how to express the circle's mass with respect to increasing radius. Unless...

I make my integral:

IT= Integ2PI0{ dm/dr r2 } d[theta]

Where dm/dr = r2PI

Therefore dm = 1/3 r3PI |r0 hence:

IT= Integ2PI0{ 1/3 r3 PI r2 } d[theta]

= 1/3 PI r5 [theta] |2PI0

= 2/3 PI2 r5

Which I'm certain is wrong. I'm not asking why it's wrong. But how should I construct my dm/dr and integrate that through with reasoning why that's a good method. Because I have a feeling I'm forgetting something fundamental in this part of the calculation.

Cheers,
Haths

2. Jan 4, 2009

### chrisk

Moment of inertia is found from summing each mass element times the square of the distance from the axis of rotation to the mass element. Typically, the mass element is expressed as the product of density and a volume (or area) element. If the density is uniform it can be moved outside the integrand and the volume element is expressed in terms of the radius and can be readily integrated. For a thin disk using polar coordinates is easiest and gives an area element of r*dr*d{theta}. The limits of double integration are from zero to R (the radius of the disk) and from zero to 2pi for theta. The result is the density times pi times radius to the fourth power all divided by two. But the density is the total mass, M, divided by the area (pi times radius squared). So I = (MR^2)/2.

3. Jan 4, 2009

### Haths

Yeah I already knew what the MOI was. Ok

So your saying I can express my Integral as:

I = [rho] Integ2PI0 [ Integr0{ r } dr ] d[theta]

+++Quick working+++
1/2 r^2
eval?
1/2 r^2 [theta]
eval?
r^2 PI
*[rho]
=m only?...
+++/ends+++

I don't get it. I can see that intergrating r*dr*d{theta} is esentailly summing up all the possition vectors of the point masses* in the circular disk (which was my initail question answered). But my quick working came out wrong. *Tries again long hand, non-mentally*

I still come out with I=m only.

Perhaps my intergration skills are bad. Ok First I intergrate r with respect to dr and get;

1/2 r2 to be evaluated between r and 0. Yes?

This is still just 1/2 r2

Then I intergrate this with respect to d[theta] and evaluate between 2PI and 0 and get;

1/2 2PI r2

Now I multiply this by [rho] which is m / PI r2

Which leaves me with m. Ok so where is my mistake here? Is it that I didn't start with r2 in my integral *penny drops*. It's Integ Integ{ r2*r*dr*d[theta]} dr d[theta]

Because r*dr*d[theta] is dm/dr. Yes ok. I'm I right with this? Because *mental calculation* r^3 => 1/4r^4 when multiplied 2PI: 2/4=1/2, PI drops out with Rho.

Cheers man, that was a bad moment when I forgot how to this.

[hr]

I = [rho] Integ2PI0 [ Integr0{ r2 dm/dr } dr ] d[theta] where dm/dr = r dr d[theta]

*That's the learning point I take away.

Haths

4. Jan 4, 2009

### Nick89

You're post(s) are really hard to read for me (maybe it's because it's late lol...) but I'm going to give it a try.

The moment of inertia can be calculated as:
$$I = \int r^2 dm$$
Here, dm is the mass of a small piece of your disk at radius r from the center of mass (origin).

The mass of the disk is unknown, but you do know that it will be equal to
$$m = \rho V$$
(density times volume)

So the mass of a small piece of disk (volume dV) is
$$dm = \rho dV$$
(Note that rho could still depend on the radius, but I assume it is a uniform disk which means rho is constant)

So
$$I = \int \rho \, r^2 \, dV = \rho \int r^2 dV$$

Now, what do you know about the volume of a small peice of your disk (dV)?
If you take a circular segment of your disk (a loop around your disk), and take its volume as dV, then it is not very hard to see that its volume is
$$dV = 2 \pi \, h \, r \, dr$$
This is just the circumference of your loop (2 pi r) times the width of the loop (dr) times the height (h) which gives you the volume.

Entering this into the equation for I we get:
$$I = \rho \int r^2 2 \pi \, h \, r \, dr = 2 \pi \, h \rho \int r^3 dr$$

Now if you look at the density \rho again, that is also equal to the mass divided by the total volume, which is just the area of the disk times its height:
$$\rho = \frac{m}{V} = \frac{m}{\pi R^2 h}$$

To sum over the entire disk, you need your loops to run with a radius from r = 0 to r = R (radius of the disk), so we integrate from 0 to R:
$$I = 2 \pi \, h \rho \int r^3 dr = \frac{2 \pi h m}{\pi R^2 h} \int_0^R r^3 dr = \frac{2m}{R^2} \frac{R^4}{4} = \frac{mR^2}{2}$$

5. Jan 5, 2009

### Haths

Sorry if I knew how to display the equations like you'd done then I'd of been able to make it easier to read.

It's alright I understand. I wasn't sure on how to express dm/dr, at first though, I was thinking of strips of [delta]r decreasing in height on the y-axis. Which left me rather confused. So then I thought that you simply intergrate the formula for the area of the shape with respect for r. But of course it's not that.

Once chrisk reminded me that it's not the area formula, you are actually using intergration to express the area as part of this calculation (the penny dropped). Then your dm/dr is an expression which when intergrated expresses the area. Then [rho] is used to link the mass of the object into this, and yes if [rho] was a function of r then you can't take that out of the integral.

Anyhow thanks people,
Haths