Center of gravity - conflicting result

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Homework Help Overview

The discussion revolves around calculating the center of mass for a wire frame shaped like a half-circle and two straight lines. Participants are comparing their results with a solution manual, noting discrepancies in their calculations of the center of mass coordinates.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants are exploring different methods for calculating the center of mass, questioning the definitions and assumptions regarding the shapes involved, particularly the treatment of areas versus lengths in the context of a wire frame.

Discussion Status

Some participants have identified potential misunderstandings regarding the nature of the shapes involved and the choice of axes for calculations. There is ongoing exploration of how these choices impact the results, with various interpretations being discussed.

Contextual Notes

Participants are working under the constraints of a specific problem statement and are attempting to reconcile their calculations with those presented in a solution manual, which may have different assumptions regarding the coordinate system used.

Femme_physics
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From some reason Yc's result is different than mine. But, the solution manual and me are using the same X axis to calculate Yc (we're using a different Y axis to calculate Xc, though). Can someone explain to me what's wrong here?The question:


Calculate the center of mass of the material line in the drawing. The line is made out of uniform thin wire bent in the shape of a half-area and two straight lines. Measurements are in the drawing
http://img9.imageshack.us/img9/6158/tarshim4question.jpg

My answer

http://img827.imageshack.us/img827/4899/ycxcfix.jpg

Solution manual's answer

http://img804.imageshack.us/img804/8776/comnotmine.jpg
 
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Hi Femme_physics! :smile:

erm … πr2 for a length? :redface:
 
tiny-tim said:
Hi Femme_physics! :smile:

erm … πr2 for a length? :redface:

No no, it's my mass (or "area" in this case since we're in 2D without defined masses). πr2 for a length? (area) x 4R/3π (length to the axis). That should give me the correct distance.

Do note that I divided it to two shapes (half a circle and a rectangle), whereas in the solution manual from some reason they did 3 shapes (quarter of a circle, quarter of a circle, and rectangle)
 
The problem states that the shape consists of a wire frame. No areas, just lines.
 
Not following you. :confused:

The solution has L1 = πr = 20π = 62.83
 
Well, what is L1? I'm not even sure I understand what they did in the solution.

But, I just used my formulas. I solved many COM problems like that. It was always "area" multiplied by the "distance of the center of gravity of the object to the axis". I thought I did exactly that. I'm not sure what's wrong with this picture. :-/
 
Femme_physics said:
It was always "area" multiplied by …

There's no area here.

It's a "material line … made out of uniform thin wire". :wink:
 
tiny-tim said:
There's no area here.

It's a "material line … made out of uniform thin wire". :wink:
Ohhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh.

I get it.

Let me work on that then.

Thanks, hero :)
 
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  • #10
Femme_physics said:
Now I got the same Yc :)

But, for Xc, I'm not sure whether I should've added the half circle shape or minus the half-circle shape. What's the verdict here?

It depends on the choice of coordinates. Where is the origin?

You have chosen the center of the half circle as your origin, so the verdict with this choice is minus.

In the solution manual they have apparently chosen the origin as the leftmost point of the object.

I can't read the Hebrew, but does it say anything about the choice of the origin?
 
  • #11
I can't read the Hebrew, but does it say anything about the choice of the origin?

The solution manual has taken it upon itself to change the axes. It should be the one in my solution. :) And thanks, I'll change it to a minus then!
 
  • #12
Femme_physics said:
The solution manual has taken it upon itself to change the axes. It should be the one in my solution. :) And thanks, I'll change it to a minus then!

So you've found yet another mistake in the solution manual then? :smile:
 
  • #13
Most likely ;) But, I wouldn't sent it to the author since he'd just claim that it's perfectly legit to change the axes, he'd just get a different numeral result. I just think it's more proper to use the defined axes :)
 
  • #14
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  • #15
Femme_physics said:
Okay, I'm trying to find Xc with my Y-axis the same as in the solution manual so I'd get the same answer. From some reason I was a bit off with the answer :(

http://img840.imageshack.us/img840/1765/xcxcxcxcxcxcxcxcxcxcx.jpg

Good morning Fp! :smile:

You have forgotten to adjust the x coordinates of the other segments.

Btw, if you had finished your previous calculation with the minus sign you already mentioned, you should have the same result as the solution manual, except for a difference of 20...
 
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  • #16
Good morning Fp!

Haha, you've turned to tiny-tim ILS ;D

Ahem,

Good morning, I Like Serena! :wink:

You have forgotten to adjust the x coordinates of the other segments.

Actually, the x coordinate is the same as the solution manual if you'll notice. Either way it shouldn't affect the result of Xc where the x-axis is, since we're only looking at the distance to the y-axis while calculating Xc.

Btw, if you had finished your previous calculation with the minus sign you already mentioned, you should have the same result as the solution manual, except for a difference of 20...

I'll check that soon, I want to see if I get the same result as the manual by using the same axis. I'm not sure what's wrong with my equation
 
  • #17
Femme_physics said:
Haha, you've turned to tiny-tim ILS ;D

Ahem,

Good morning, I Like Serena! :wink:

Yes, well, tiny-tim is one of my great examples here on PF! :smile:
I feel more people should follow his smiley approach in helping people.
Btw, he has given me a concise guide book on how to use smileys! :approve:
In time I'll develop a more personal touch again.


Femme_physics said:
Actually, the x coordinate is the same as the solution manual if you'll notice. Either way it shouldn't affect the result of Xc where the x-axis is, since we're only looking at the distance to the y-axis while calculating Xc.

Nope. You'll see for instance that the solution manual has an x3 of 60 mm.

And you're right that the result is unaffected by the location of the x axis.
But all x coordinates change, when you change the location of the y axis. :wink:
 
  • #18
Heh @ tiny-tim comment

Yes, he's great with his smiley's approach :)

Nope. You'll see for instance that the solution manual has an x3 of 60 mm.

Yes, I also used a distance of 60mm. It's the distance from the 20mm shape to the y axis.
 
  • #19
Femme_physics said:
Yes, I also used a distance of 60mm. It's the distance from the 20mm shape to the y axis.

Funny, I don't see that in the calculation you posted.
Did you post the wrong webcam photo?
 
  • #20
Oh, and I just saw you have the wrong x coordinate for the half circle segment.

In the problem you posted, you can see that they gave:
x_C = \frac {2 R} {\pi}
But that is not what you used... :smile:
 
  • #21
So they've defined my Y axis for me at the center of gravity of the circle?
 
  • #22
Femme_physics said:
So they've defined my Y axis for me at the center of gravity of the circle?

As yet I haven't seen anything that would define the Y axis and I'm still wondering if the Hebrew text says anything about it. If not then the choice would be arbitrary.

What they gave (I presume) is how far the center of mass of a half circle is off its center.
 
  • #23
I translated everything word by word, except their little comment. But it just said "a half circle center of gravity coordinate is Xc = 2R/pi "

*shrugs* I have all my formulas in the formulas page.
What they gave (I presume) is how far the center of mass of a half circle is off its center.


I'm not sure what you mean :-/
 
  • #24
Femme_physics said:
I'm not sure what you mean :-/

A full circle has its center of mass in the center of the circle.
A half circle has its center of mass at a certain offset from the center of the circle.
 
  • #25
A half circle has its center of mass at a certain offset from the center of the circle.

That should be 4R/3pi according to my formulas page
 
  • #26
Femme_physics said:
That should be 4R/3pi according to my formulas page

Is that for a thin wire circle or for a circle disk?
 
  • #27
Hmm... OH! A half circle disc has the formula of 2R/pi?!?

My godness. You sure? That radically changes everything!
 
  • #28
Femme_physics said:
Hmm... OH! A half circle disc has the formula of 2R/pi?!?

My godness. You sure? That radically changes everything!

Yeah! I know! :smile:
Now you have to learn everything from scratch again!

Btw, I hope you just meant "half circle wire" instead of "half circle disc"! :wink:
 
  • #30
Femme_physics said:
Ah, darn it. What am I missing this time? (this is neverending!)

You took the half circle offset from the left side of the circle, instead of from the center of the circle.
That would be correct in your original choice of the coordinates (with the center of the circle as the origin), but with the origin at the leftmost point of the object, you need to compensate. ;)
 

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