# Fidning the center of mass of an incomplete circle

1. May 21, 2013

### Jbray

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

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.

2. Relevant equations

x=rcosθ, y=rsinθ, dL=r*dθ

3. 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.

2. May 21, 2013

### SteamKing

Staff Emeritus
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.

3. May 22, 2013

### SteamKing

Staff Emeritus
Disregard previous post.

4. May 23, 2013

### Staff: Mentor

What is the correct answer? I seem to get around 21 mm for x (but I did need wolfram alpha [Broken] ).

EDIT: The formula from Wikipedia evaluates to ≈ 32 mm.

Last edited by a moderator: May 6, 2017
5. May 23, 2013

### rude man

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.)