What is the Distance from C to C' for a Circular Pizza with a Piece Removed?

[jessedevin]

Homework Statement

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.)

Homework Equations

The Attempt at a Solution

I started off with taking the smaller circle as the negative mass.
x_CG = (m₁ x₁ - m₂ x₂) / (m₁ - m₂)
But I do not know what to do after this. Please help asap. Thanks.

 

[Doc Al]

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

 

[jessedevin]

I know you are suppose to relate it to desnity or thickness but i don't 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.

 

[Doc Al]

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?

 

[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, that's just how they turned out.

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

 

[physics girl phd]

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, that's just how they turned out.

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

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.

 

[jessedevin]

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.

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

Sooo...
X_CG = (m1 x1 - m2 x2) / (m1 - m2)
X_CG = -($$\sigma$$$$\pi$$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!

 

[Doc Al]

Excellent!

 

[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?

 

[Doc Al]

Can I ask why x1 is taken to be zero?

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