Aliasa
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Homework Statement
Find the center of mass of a uniform sheet in the form of a circular disc
with a hole bounded by the equations x^2 + y^2 ≤ 1 and (x - 1/2)^2 + y^2 ≥ (1/16).
The problem involves finding the center of mass of a uniform circular disc with a hole, defined by specific mathematical inequalities. The equations provided describe the boundaries of the disc and the hole within it.
There is a mix of interpretations regarding the wording of the question, with some participants expressing confusion over the definitions provided by the equations. Others assert that the equations define a precise shape, leading to a debate about the nature of the center of mass. Some participants have indicated they believe they have solved the problem, while others are still grappling with its complexities.
Participants note that the question may be poorly worded, leading to misunderstandings about the boundaries and definitions of the shapes involved. There is also mention of varying difficulty levels in relation to other problems in the assignment.
Aliasa said:That is all the info given.
SteamKing said:Can you find the area of the figure?
Yes, but can you find the area? This is a disk with a hole in it. It would be a good idea to at least draw a picture of the "hole bounded by the equations x^2 + y^2 ≤ 1 and (x - 1/2)^2 + y^2 ≥ (1/16)." Those boundaries are circles. Can you graph the circles?Aliasa said:That is all the info given.
You are misunderstanding the question. The metal template is very precisely defined by the two equations; there is no range of possible shapes and sizes. The centre of mass is an exact point. The two equations together define the metal surface, the hole is where there is metal missing.Aliasa said:It turns out the question is horribly worded. The sheet is bounded by the former equation, while hole by the latter -_-. Since the equations are inequality, by taking density into account, I feel the center of mass should be a range too. The hole can have a max radius of .5, sheet, 1. Hole can have a min radius of 1/4, when the sheet can have a minimum of 3/4.
Aliasa said:If that is the case then the question is trivial. I have solved it if those equations represent what you say. But doing that only takes me 5 minutes or less, which I can't understand. All the other questions on the assignment take in excess of 3 hours.
Yes. That's correct. Nice job.Aliasa said:The answer is (-1/30,0) btw.