# Moment of Inertia of a Spinning Record with a Hole in Center

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1. Apr 5, 2017

### Jewelz

1. The problem statement, all variables and given/known data
A 45 RPM record is a disk with a wide hole in the center. The mass of such record is 45g. It is 7 inches in diameter, and the hole in the center has a 1.5 inch diameter. (Sorry for the odd units, this was the way it was given).

Find the moment of inertia of the 45 RPM record

2. Relevant equations
$I=1/2MR^2$
ITotal=IDISC - IHOLE
Area of a circle = $πR^2$

3. The attempt at a solution
Given the fact that the moment of inertia of a disc is ½MR^2, using the information given, I calculated IDISC to be 275.63 g [cm]^2.
IDISC=0.5(45g)([3.5in])^2 = 275.63 g in^2

To find the moment of inertia of the hole, I had to find what the mass would have been if it were included.
Using the MHOLE equation, I got an answer of 2.17g.
MHOLE= (45g*0.5625π[in]^2) / (12.25π in^2 - 0.5625π in^2) = 2.17g

The mass was then used to find the moment of inertia of the hole
IHOLE=0.5(2.17g)([0.75in])^2 = 0.610g in^2

Subtracting these two values
275.63 g in^2 - 0.610 g in^2= 275.02 g in^2

Is this correct? My only concern was with finding the mass of the hole, and whether it should have been added into the mass of the disc when finding IDISC or not. Any feedback is appreciated.

Thank you

Last edited: Apr 5, 2017
2. Apr 5, 2017

### kuruman

Hi Jewelz and welcome to PF. Do you know how to derive the moment of inertia of a disk using integration? If so you can easily adapt the derivation to a disk with a hole.

Last edited: Apr 5, 2017
3. Apr 5, 2017

### Jewelz

Yes, actually, that was a previous part of the problem, getting to the 0.5MR^2. I just have never had a problem where there was a hole that needed to be accounted for

4. Apr 5, 2017

### kuruman

That makes life much, much easier. In the previous part you integrated from 0 to R. Here you need to integrate from R1 (inner radius) to R2 (outer radius), i.e. over the space occupied by the disk. You take the hole into account by omitting to add any contributions from the space it occupies.

5. Apr 5, 2017

### Jewelz

Ahhhh, that makes a lot more sense. So other than changing the radius in the equation for IDISC, which would just be the radius of the whole disc (3.5 in) minus the radius of the hole (0.75 in), which would give a new effective radius for IDISC or 2.75 in, everything else looks to be okay?

6. Apr 5, 2017

### kuruman

I don't know, what equation did you get for the moment of inertia? BTW, please see if you can learn to use LaTeX. Your equations look very confusing.

7. Apr 5, 2017

### Jewelz

I do apologize for that. I am new to the site and have never used LaTeX. I tried it for a little, but for unknown reasons, I kept getting errors with it. I cleaned up the relevant equations portion. Everything in my attempt is basically just plugging numbers straight into those equations. The moment of inertia for a disc is 0.5MR^2

8. Apr 5, 2017

### kuruman

Yes indeed it is. What about the moment of inertia of a disk with a hole in its center? Did you follow my suggestion in post #4? Your answer should have R1 and R2 in it.

9. Apr 5, 2017

### Jewelz

I just tired it out, and I got an answer of I = 0.5M((R2)^2 - (R1)^2), but I am not sure if that is completely right.

10. Apr 5, 2017

### TomHart

Doesn't that give you the moment of inertia of the portion of the disk from R1 to R2, where the total disk has a mass of M (not the portion)?

Edit: Just to clarify a little more. If you have a disk of mass M, your equation will give the moment of inertia of a portion of that disk from R1 to R2. But what you want to find is the moment of inertia of a ring (I'm not sure if ring is the correct term) of mass M that has a large radius of R2 and a small radius of R1.

So let's look at your equation where R1 is very close to R2. What happens is that your equation will approach 0. But if you have a ring of mass M (not a disk of mass M), where R1 is very close to R2, then the result should approach the same result as that of a point mass - namely, I = MR2

Last edited: Apr 5, 2017
11. Apr 5, 2017

### kuruman

Almost there. Be careful with the density expression you use when you do the integral.

12. Apr 5, 2017

### Jewelz

Looking back over at what I did, I don't really know what I should be changing about the density.

Other than simply rearranging the equation to m=ρV, then knowing volume of a disk would be πR1^2 - πR2^2, and plugging that in with the equation remaining from the integral, it still cancels out to be 0.5M(R2^2-R1^2)

13. Apr 5, 2017

### kuruman

The volume is $V=\pi (R_2^2-R_1^2)d$, d =thickness.
Can you show what you did in some detail? What does your integrand look like? Specifically what did you use for $dm$ in $dm~r^2$?

14. Apr 5, 2017

### TomHart

Rather than express density in terms of volume, sometimes it is simpler to express density in terms of area, or in terms of length. For example, sometimes the density of a wire is expressed in terms of kg/meter or g/inch.

For your problem here, since no thickness is specified for the record, and since the thickness (and therefore density) is assumed to be constant at every point in the record, I think it is simpler to express the density of the record in terms of mass/area, instead of mass/volume. So for a disk, for example, how would you calculate density (in terms of mass/area)? You would take the total mass of the disk (mass) and divide it by the total area of the disk, (πr2).

Now what if you had the situation as in this problem where the object is a not a disk, but the shape of a record. Same general method - to find the density you would take the total mass and divide it by the total area. What is the area of the record? Well it's simply (the area of a disk of the same diameter as the record) minus (the area of the hole in the record).

So now that you know the density (mass per unit area) of the material that the record is made of, you can calculate the mass of any piece of this material by simply multiplying the area of that piece times the density.

So now if we want to integrate, how can we do it? Well, for me, the most straightforward method would be to break up the record into very thin rings. So for a ring that is at a distance r from the center of the record, that ring would have a length of (2)(π)(r). And multiplying the length of that ring by its width, we can find the area of that ring. So using an infinitesimally small width of dr, then the area of that thin ring will simply be its length times its with, or A = 2πrdr. Once we know the area of the thin ring, we can calculate its mass by multiplying area times density. And if you know the formula for inertia of a thin ring, you can simply integrate that from R1 to R2 and, voila, there you have it.

I recommend that when you work this out, you do not plug any values in until the very end. So when you calculate density of the record, find the expression for density in terms of mass, R1 and R2. It will help simply your overall general expression at the end.