# HOMEWORK H32.A

## 50 thoughts on “HOMEWORK H32.A”

1. Jason says:

when finding the inertia do you have to subtract the outer inertia with inner inertia?

1. Chalisa Joan Kulprathipanja says:

I am pretty sure yes because there is no area in the center since it is cut out

2. Kaushal Kamal Jain says:

That is correct, Jason and Chalisa

2. Mohnish Y Shah says:

I think so too. I am not sure if I am right but you could just find the y centroid of the shaded area and then use that directly to find the inertia.

1. Kaushal Kamal Jain says:

@Mohinish, you will need to find the inertia of the outer and the inner semi-circle about the overall centroid and then subtract inner from outer as Jason and Chalisa said

3. Russell Keith Moffat says:

I've used 1/8 pi R^4 as the formula for second moment of inertia of a circle, is this alright?

I had the same question however I think that using this formula would give us the right value for the second moment of inertia of a circle

I tried looking through the textbook and the examples and I thought that this might not be correct.

The formula 1/8 pi R^4 would give the inertia about the origin but we need to find the inertia about the centroid

1. Ruth Ivania Guerra says:

Yes, I did this as well. It 1/8pi^4 is the one that would give the second moment of inertia about the centroid.

1. Jennifer Leigh Smith says:

I used 1/8piR^4 for the second moment! Which will help you find the first half of the equation.

2. Kushal Shardul Doshi says:

So if 1/8 pi R^4 is the inertia around the origin, could i take the distance in the parallel axis theorem to be the distance from the origin to the centroid of the whole shape we find in the first part of the question? (In which case the distance in the parallel axis theorem for both the bigger and smaller semi-circle will be the same)

1. Zongxin Yu says:

It's correct, Kushal! Whenever you are using second moment of inertia from table, you will always keep in mind, which axis that's around, and around which axis I am looking for? Then apply the parallel axis theorem.

2. Ashley Molnar says:

I used 1/8pi R^4 for Ix and used the equation Ic = Ix - A * d^2 where d is the distance to the centroid!

1. Zongxin Yu says:

It's correct Ashley, but be careful, the d is the distance to the centroid of the combined shape.

1. Zach Mullen says:

So then you would have two d's for each second moment of area and they would be the difference between the overall centroid and the respective section's centroid right? And then you subtract the cutout from the bigger semicircle's second moment of area? My answer came out pretty nasty but that's the approach I used so I just wanted to make sure.

4. ringea says:

How are we supposed to find the y centroid of each circle individually?

1. Lo says:

The y centroid of the semicircles is 4R/3pi. You can find the centroid of the shape by part with the two centroids of the semicircles.

5. Deepa Jayasankar says:

For a semicircle, the moment about its centroid is 1/8 pi r^4, correct? I found this equation online and just want to check that it is the correct one before I do all of the work.

1. tc says:

Yeah I used 1/8 pi R^4 to find the moment about the centroid

1. Zongxin Yu says:

Kusha's and Ashley's discussion may give a hint on how to use this value.

6. huttonl says:

Do we find inertia for the dimension by analyzing the shape or using a specific formula?

1. Abhirakshak Raja says:

You can use the specific formula for inertia for a semicircle and with that and the parallel axis theorem you can get the inertia for the entire shaded region.

1. Ammar Al Nas says:

so basically its that I find the centroid then use PAT for the shaded region around the centroid and that's it ?

7. Jason says:

Just wanted to let everyone know that all the inertia of common shapes can be found in the textbook for corresponding chapter.

8. Ryleigh Elizabeth Norton says:

I found the centroids of both semicircles but how to we find the overall centroid of the shaded region? Is it just outer centroid - inner centroid?

1. Priyanka Jeevaretanam says:

Yes, I believe so since the smaller semicircle is a cutout we have to subtract it from the larger semicircle.

2. Kushal Shardul Doshi says:

You take a weighted average with the area of the larger semi-circle being positive and the area of the smaller semi-circle being negative as it is a cutout from the larger semi-circle.

9. Elliott Huy-Anh Eve says:

How are we supposed to find the y centroid?

1. Jessika Kathleen Wahlbin says:

You can find the y centroid of a semicircle by using the formula 4R/3pi

10. Zaid Basil Qubain says:

Is there an "I" for semicricle(or circle) or do I have to integrate that myself

1. Jessika Kathleen Wahlbin says:

Semicircle: I = 1/8 pi R^4

2. jr says:

Pg. 474 of the textbook has a helpful graphic with the geometric properties for a semicircle.

11. Jennifer Leigh Smith says:

I keep receiving an answer with R^4. I reworked it and received a value with R. I am not sure which is correct. Did anyone else receive a final answer with R^4?

1. Jennifer Leigh Smith says:

I am referring to the y-centroid not the inertia.

1. tc says:

The y-centroid I got had an R and the inertia had R^4

12. Carolyn Ann Lo coco says:

When I use the parallel axis theorem, I'm confused on what to use for d. I saw a comment saying d is the distance to the centroid of the combined shape, but isn't it the distance from the neutral axis to the top or bottom of the area?

13. Hardy says:

I believe distance is the distance from the origin to the centroid of the entire shape, referring to Kushal's comment

14. James Sun says:

I was trying to find the centroid of two circles, but I did forgot how to find out the centroid of the circle,

1. Ruth Ivania Guerra says:

centroid for half circle ybar is 4r/3pi.

15. Ruth Ivania Guerra says:

In this type of questions are we gonna have the formulas on the exam? When solving with the formula given in class I get an answer with R^3 for the centroid. I did the y_bar(At) = y1_bar(area1) + y2_bar(area2). area2 since is the not shaded portion is negative. Then I solved for y_bar. I don't know if I am moving in the right direction and if finding and answer with R^3 is correct.

1. Kaushal Kamal Jain says:

You will be provided the relevant formulas in the exam. For the centroid, you should get an answer with R.

16. Yubo Song says:

Hi, I used Yc = (A1Y1 - A2Y2) / (A1 - A2) to find Yc, with Y = 4r / (3pi). I'm not entirely sure if that's the correct way of solving this. Could someone offer some insight? thanks a lot.