# Polar moment of inertia

1. Oct 9, 2013

### Woopydalan

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

2. Relevant equations

3. The attempt at a solution
I know some of you may have given up on me understanding moment of inertia/second moment of area, but here is another problem. I am using the this table to get the equations for the polar moment of area of a hemicircle around the x-axis, then I am applying the parallel axis theorem to find the polar moment of inertia. The answer in the back of the book doesn't even have the term a in it, which is what I don't get.

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• ###### 7.58 attempt 1.pdf
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Last edited: Oct 9, 2013
2. Oct 9, 2013

### SteamKing

Staff Emeritus
You had calculated the correct location for the centroid of each of the semicircular cutouts (d = 2a - 4a/(3π)), but then you didn't use this value when you subtracted the polar moments of inertia from that of the square. You used 2a instead.

3. Oct 9, 2013

### Staff: Mentor

I'm not sure, but I think the OP also used the wrong formula for the area of the semicircle.

4. Oct 9, 2013

### Woopydalan

The reason I didn't use the centroid was because I decided to use the formula as if I was considering the parallel axis at what would be point O in the "table of inertias'', so the distance from the centroid of the square to the edge of the semicircle is a distance 2a

Last edited: Oct 9, 2013
5. Oct 9, 2013

### SteamKing

Staff Emeritus
Chestermiller is right, and I checked into the polar moment of inertia for a semicircle. Your value is indeed about an axis coincident with the diameter and not the centroid of the semicircle.

Still, you need to correct your calculation of the polar moment for the figure, and then calculate the gyradius as requested by the problem statement.

6. Oct 10, 2013

### Woopydalan

here is my attempt #2.

The back of the book says the moment of inertia is 25.1 in^4 and the radius of gyration is 1.606a. I don't see how the moment of inertia could be independent of a. Perhaps the solution is wrong?

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7. Oct 10, 2013

### SteamKing

Staff Emeritus
I would think so. I will check your calcs tomorrow (or today).

8. Oct 10, 2013

### Staff: Mentor

My understanding is that the parallel axis theorem only works relative to the centroid. This is because of the distance squared times the area term. So, if as SteamKing says, the moment of inertia for the semicircle in the figure is about the origin, then you first have to use the parallel axis theorem to get it back to the centrioid, and then apply the parallel axis theorem again to get it to your specific axis of rotation. I hope this makes sense.

Chet

9. Oct 10, 2013

### SteamKing

Staff Emeritus
I did my check calculation for the polar moment, and J = 25.1*a^4 in^4 and the radius of gyration = 1.606*a. It appears the book omitted the factor of a^4 in their answer for J.

NB for Woopydalan: I looked at your calcs, and there is something off with your calculation for J.
In my check calculations, because of symmetry for this problem, Ix = Iy, and because J = Ix + Iy, J = 2*Ix = 2*Iy. Therefore, I calculated Ix and used that to obtain J. You might like to try this approach as well.

10. Oct 10, 2013

### Woopydalan

So what is wrong with my calculation for J?? Is it true what chestermiller said regarding the parallel axis theorem?

Edit: I found an equation online for the polar moment of inertia around the centroid of a semicircle
http://www.efunda.com/math/areas/circlehalf.cfm

and used that with the parallel axis theorem and got it..finally it's been like 2 days for one problem, except I don't know why I have to use centroids for the parallel axis theorem???

Last edited: Oct 10, 2013
11. Oct 10, 2013

### SteamKing

Staff Emeritus
I looked at your calculations (and there appears to be an arithmetic mistake in your final calculation of J), but there was something still I could not put my finger on. This is why I did a calculation of Ix, because it simplifies the use of the parallel axis theorem.

By calculating the moment of inertia about the x-axis, for example, the inertia of the two semicircles on the x-axis do not require the use of the parallel axis theorem: their inertia may be subtracted from the inertia of the square. The two semicircles on the y-axis will required the addition of A*d^2 to their inertias before subtraction from the inertia of the square.

It's easy in these calculations to make a mistake and not notice it. That's why it is advisable to set up a tabular form calculation so that the calculations are laid out in an orderly and regular fashion. Unfortunately, all of the textbooks I've seen do not show this approach, which is to the detriment of the student.