# Center of Mass of Semi-Circle

1. Jul 30, 2007

### alfredska

I'm having difficulty in making two methods of the calculation of center of mass come out equivalently. One method is via intuition and the other is via standard formula.

1. The problem statement, all variables and given/known data
Find the center of mass of a semi-circular band of length L and mass M, and represent it in terms of $$\alpha R$$.

2. Relevant equations

3. The attempt at a solution

Can anyone track down the problem here?

Last edited: Jul 30, 2007
2. Jul 30, 2007

### mgb_phys

Do you have the equation for the area enclosed by a chord?
Calculate what chord is required to have an area of half the semicircle.

3. Jul 30, 2007

### alfredska

Chord?

I must not understand you. Area via chord formula? I'm using a path integral for the formula method, are you suggesting something is wrong there? Please elaborate.

Chord? As in a string laid out in an arbitrary manner?

4. Jul 30, 2007

### mgb_phys

A chord is a line drawn across 2 points i=on the circumference of a circle, the chord area is the area between the chord and the circumference ( which becomes a semi circle in the case where the line is the diameter )

The area of this chord area is quite easy to derive http://mathworld.wolfram.com/Chord.html, what you are trying to find is the distance of the chord from the centre such that the area is 1/4 the are of a circle.

5. Jul 30, 2007

### alfredska

Band

Realize this is a band of length L, in the shape of a semi-circle. I am not seeking the center of mass of solid half circle, rather the center of mass of a "rod" bent into a semi-circle.

6. Jul 30, 2007

### Dick

It is NOT correct that at the center of mass, mass above=mass below. Take a one dimensional example, point mass m at x=0, and mass 2m at x=1. We all know that the center of mass is at x=2/3. But mass above 2/3 is not equal to mass below 2/3. The mass needs to be weighted by its distance from the center.

7. Jul 30, 2007

### alfredska

Mass Above not equal to Mass Below

Dick,
I think you have explained why my intuitive method is incorrect. Splitting the mass in half as I have done, still leaves the upper portion further away from the CM than the lower half, and no weighting was done to take that into account. Thank you.

I am under the assumption that the formula method I presented is correct. Any verification by another member would be appreciated.

Thanks again.

8. Jul 30, 2007

### Dick

The other method looks fine to me.

9. Jul 30, 2007

### mgb_phys

Oops - sorry, that does make a difference!