Finding the center of mass of a piecewise function

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To find the center of mass (COM) of a piecewise function, the shape can be divided into simpler components like parallelograms and triangles. While calculating the COM for each shape is possible, integrating the equations directly may provide a more accurate result. The discussion emphasizes that even if the shapes are not perfect parallelograms, they can still be treated as point masses for COM calculations. There is some confusion regarding how integration relates to finding the COM versus just calculating area. Clarification is needed on the necessary multipliers for the integration process to accurately determine the COM.
JorkThePork
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Homework Statement
Find the center of mass of the "lightning bolt" shaped piecewise function as shown below (or in the desmos project: https://www.desmos.com/calculator/g6crwsecp1)
Relevant Equations
Xcm = A^-1 * ∫ a b x * (f(x) - g(x)) d x

Ycm = A^-1 * ∫ a b .5((f(x)^2 - g(x)^2) d x
desmos-graph.png
I understand that I can divide this shape into a few parallelograms and a triangle and calculate the center of mass of each, but am confused as to what I should do after that. My physics teacher also wants us to use integrals, but I'm assuming I can calculate the COM of each parallelogram and triangle by simply integrating.
 
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You can find the CM of the two parallelograms and the half-parallelogram which will give you 3 point masses at the locations of these CMs. Then you can find the CM of the 3 point masses the usual way.
 
kuruman said:
You can find the CM of the two parallelograms and the half-parallelogram which will give you 3 point masses at the locations of these CMs. Then you can find the CM of the 3 point masses the usual way.
the two seemingly big parallelograms actually are not parallelograms because of how I designed the lightning bolt (the equations I used are here https://www.desmos.com/calculator/g6crwsecp1). However, I could break the shape up into 2 big parallelograms, 2 very thin parallelograms, and a triangle. I'm assuming I could still then find the CM of the 5 point masses the usual way.
 
JorkThePork said:
the two seemingly big parallelograms actually are not parallelograms because of how I designed the lightning bolt (the equations I used are here https://www.desmos.com/calculator/g6crwsecp1). However, I could break the shape up into 2 big parallelograms, 2 very thin parallelograms, and a triangle. I'm assuming I could still then find the CM of the 5 point masses the usual way.
Yes, you can carve it up into simple non overlapping shapes, but since you have equations why not just integrate them and add the (signed) results?
 
haruspex said:
Yes, you can carve it up into simple non overlapping shapes, but since you have equations why not just integrate them and add the (signed) results?
Sorry, I’m a little confused how this would give me the center of mass. wouldn’t this just give the area?
 
JorkThePork said:
Sorry, I’m a little confused how this would give me the center of mass. wouldn’t this just give the area?
So what do you need to multiply the expressions by before integrating?
 
Last edited:
Thread 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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