Finding the Center of Mass of a Meter Stick with Variable Mass Density

AI Thread Summary
The discussion focuses on finding the center of mass of a meter stick with a variable mass density defined by the equation ρ(x) = 0.800(1 + 0.00250x) grams/cm³. Participants clarify that the solution involves using two integrals: one for the mass, ∫ ρ dx, and another for the moment of the mass, ∫ xρ dx. The volume element is specified as dV = Adx, where A represents the cross-sectional area. The conversation emphasizes understanding the derivation of these integrals to solve the problem effectively. Overall, the thread aims to aid comprehension of the mathematical approach necessary for determining the center of mass in this scenario.
XwakeriderX
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Homework Statement



A meter stick has constant thickness and width , but the material that the stick is constructed from is very strange ... it has a variable mass density that is given by, ρ(x) = 0.800(1 + 0.00250x) grams/cm3 where x is measured in cm. Find the center of mass of the meter stick, as measured from it's left end.

The problem shown asks for the center of mass of a meter stick with a variable mass density, the picture is the solution to the problem! where did this formula come from??


Homework Equations


Xcm=(M1X1+m2x2)/(M1+m2)


The Attempt at a Solution


The hint says to use two simple integrals using the volume element dV=Adx
Can someone please help me understand where they got these 2 integrals from?
 

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Hi XwakeriderX! :smile:
XwakeriderX said:
The hint says to use two simple integrals using the volume element dV=Adx
Can someone please help me understand where they got these 2 integrals from?

The first is the integral of the mass, ∫ ρ dx

The second is the integral of the moment of the mass, ∫ xρ dx :wink:
 
Ahh i see now thanks!
 
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