Calculating Center of Gravity After Pizza Cut: A Headache?

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Homework Help Overview

The discussion revolves around calculating the new center of gravity of a pizza after a smaller circular piece is removed. The original poster describes the setup involving a pizza with a defined radius and the removal of a smaller section, leading to a question about the distance between the original and new center of gravity.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the concept of treating the removed piece as having negative mass to find the new center of mass. There are discussions about using area and mass relationships in the calculations.

Discussion Status

Some participants have offered insights into methods for approaching the problem, including the negative mass concept. The original poster expresses appreciation for the suggestions and appears to be progressing in their understanding, though no consensus or final solution has been reached.

Contextual Notes

The original poster notes a typo in their question, indicating a focus on clarity in the problem setup. There may be assumptions regarding the uniformity of the pizza's mass distribution that are not explicitly stated.

skiboka33
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hi, here's my question...

A pizza has a center of gravity at C, in the middle of the pizza radius = R
A smaller circle of pizza is cut from the pizza of radius R/2 at the left side of the pizza so that one the diameter of the hole where the smaller piece was stretches from the edge of the pizza to the center...

This gives a new center of gravity, C', what is the distance, x, between C' and C...

thanks this question is really givin me headaches...
 
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and let me know if its confusing... (ps, take the word "one" out in "one the diameter", just a typo, lol, thanks again)
 
the old "negative mass" trick :-)

Treat the small pizza as a pizza with negative mass. Then find the center of mass of the two "pizzas" taken together: The normal pizza + the small pizza with negative mass.
 
You will also need the \sigma = \frac{mass}{area} trick.
 
:)

thanks guys i think i got it, didnt know about that negative mass trick!

C' = [(A1)(C) - (A2)(C-R/2) ] / (A1 - A2)

right?

which simplifies to

C' = C + R/6

C' - C = R/6 = Xcm

woot! thanks! :biggrin:
 

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