Homework Help: Pizza Center of Gravity

1. Nov 13, 2008

jessedevin

1. The problem statement, all variables and given/known data
A circular pizza of radius R has a circular piece of radius R/2 removed from one side. Clearly the center of gravity has moved from C to C' along the x-axis. Show that the distance from C to C' is R/6. (Assume that the thickness and density of the pizza are uniform throughout.)

2. Relevant equations

3. The attempt at a solution

I started off with taking the smaller circle as the negative mass.
xCG = (m1 x1 - m2 x2) / (m1 - m2)

Attached Files:

• p12-06.gif
File size:
7.5 KB
Views:
113
2. Nov 13, 2008

Staff: Mentor

So far, so good. How does m_1 compare with m_2? Measure x_1 and x_2 from the center; what are they?

3. Nov 13, 2008

jessedevin

I know you are suppose to relate it to desnity or thickness but i dont know how.
But I have no clue on how m1 relates to m2 from the CG or x1 relates to x2 from the CG. Could you explain or give a couple hints.

4. Nov 13, 2008

Staff: Mentor

For m1 versus m2, first consider how the radii of the two disks compare. Then use that to compare their areas.

For x1 & x2, consider the big disk to have its center at the origin. Where's the center of the cut out disk?

5. Nov 13, 2008

jessedevin

okay, so for m1, the radius is R, where in m2, the radius is R/2, which is stated in the problem.
So for m1, the area is $$\pi$$* R2
For m2, the are is $$\pi$$* (R/2)2=$$\pi$$* R2/4

Sorry the pi's look so weird, thats just how they turned out.

The center of the cut out disk is R/2. So what next?

6. Nov 13, 2008

physics girl phd

Once you have the area, you can just recognize that as long as the pie is uniform density, the masses are propotional to the area... give it an area density (say sigma). Them m1=sigma*a1...

So then you should be able to put m1, m2, x1 (=0) and x2 = R/2 (via the chosen system)... all in your equation in the original post.

7. Nov 13, 2008

jessedevin

So let me see if I got this straight
m1= $$\sigma$$$$\pi$$R2
m2=$$\sigma$$$$\pi$$R2/4
x1=0
x2=-R/2

Sooo....
XCG = (m1 x1 - m2 x2) / (m1 - m2)
XCG = -($$\sigma$$$$\pi$$R2/4*-R/2)/($$\sigma$$$$\pi$$R2-$$\sigma$$$$\pi$$R2/4)
R/6=(R3/8)/(R2-R2/4)
R/6=(R/8)/(3/4)
R/6=R/6

Thanks So much!

8. Nov 14, 2008

Staff: Mentor

Excellent!

9. Sep 28, 2011

Icetray

I hope it's okay that I am reviving this old thread but I was looking for a solution to this question. (: Can I ask why x1 is taken to be zero?

10. Sep 29, 2011

Staff: Mentor

x1 is the center of mass of the large disk. Since that's the point we're measuring from, x1 = 0.