Fidning the center of mass of an incomplete circle

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Discussion Overview

The discussion revolves around finding the x-coordinate of the center of mass of a homogeneous rod bent into the shape of a circular arc with a specified radius. The problem involves mathematical reasoning and integration to determine the center of mass, particularly focusing on the implications of using polar coordinates.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant calculated the mass "M" and the moment "My" but found their initial result for the x-coordinate of the center of mass to be incorrect.
  • Another participant suggested drawing a diagram and clarifying the relationship between polar coordinates and Cartesian coordinates, prompting a recalculation of the moment.
  • A later post indicated a different result for the x-coordinate, suggesting it to be around 21 mm, while also referencing a formula from Wikipedia that evaluates to approximately 32 mm.
  • Another participant proposed setting up a coordinate system with the open end of the semi-circle facing left and suggested integrating moments to find the correct x-coordinate, arriving at a value of 32.47 mm.

Areas of Agreement / Disagreement

Participants have not reached a consensus on the correct x-coordinate of the center of mass, with differing results presented (around 21 mm versus approximately 32 mm). The discussion remains unresolved regarding the correct approach and final answer.

Contextual Notes

There are unresolved assumptions regarding the integration limits and the application of polar coordinates in the calculations. The dependence on different interpretations of the setup may affect the results.

Jbray
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Homework Statement



Locate the x coordinate of the center of mass of the homogeneous rod bent into the shape of a circular arc. Take r = 170 .

The arc goes from (-5/6) to (5/6)pi (counterclockwise). It has a radius of 170mm.

Homework Equations



x=rcosθ, y=rsinθ, dL=r*dθ

The Attempt at a Solution



I found "M" by integrating "170 dθ" from (-5/6)pi to (5/6)pi. This gave me 890.12.I found "My" by integrating "170 (cosθ) 170 dθ" from (-5/6)pi to (5/6)pi. This gave me 170^2 or 28900.

I used My/M to find the x coordinate of the center of mass as 28900/890.12 or 32.468. However this is incorrect.
 
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Draw a picture of the bent rod. Include the coordinate axes. If you have polar coordinates r and theta, are they the same as x and y? What are x and y expressed in terms of r and theta? Recalculate your moment based on your findings.
 
Disregard previous post.
 
Jbray said:
I used My/M to find the x coordinate of the center of mass as 28900/890.12 or 32.468. However this is incorrect.
What is the correct answer? I seem to get around 21 mm for x (but I did need wolfram alpha ).

EDIT: The formula from Wikipedia evaluates to ≈ 32 mm.
 
Last edited by a moderator:
Set up xy coord. system so the semi-circle's open end faces left.
Draw an axis parallel to the y-axis at x = x_0 somewhere between x = 0 and x = R.
Now sum (integrate) moments to the left & to the right and equate to zero. Express x in polar coordinates and integrate from theta = -5pi/6 to +5pi/6. Solve for x_0.

(I get x_0 = 0.191R = 32.47 mm.)
 

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