Finding the Center of Mass for a Uniform Plate

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SUMMARY

The discussion focuses on calculating the center of mass for a uniform plate, specifically addressing the x and y coordinates given a length (L) of 5.0 cm. The user proposes a method involving the decomposition of the plate into four rectangles and applying the formula for the center of mass. The correct total mass is confirmed as 110 cm, and the x-coordinate calculation is validated. The conversation emphasizes the importance of maintaining consistent variable representation throughout calculations to minimize errors.

PREREQUISITES
  • Understanding of center of mass calculations
  • Familiarity with rectangular lamina and moment of inertia
  • Basic algebra for manipulating equations
  • Knowledge of the formula ML^2/12 for moment of inertia
NEXT STEPS
  • Review the concept of center of mass for composite shapes
  • Study the derivation and application of the moment of inertia formula
  • Practice problems involving the calculation of center of mass for various geometries
  • Learn about error minimization techniques in mathematical calculations
USEFUL FOR

Students in physics or engineering courses, educators teaching mechanics, and anyone interested in understanding the principles of center of mass in uniform plates.

AnkhUNC
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Homework Statement


What are (a) the x coordinate and (b) the y coordinate of the center of mass for the uniform plate shown in Fig. if L = 5.0 cm?

http://edugen.wiley.com/edugen/courses/crs1650/art/qb/qu/c09/fig09_38.gif


Homework Equations





The Attempt at a Solution



So I think I should set this up as four different rectangles with a center particle mass. So this leads to Total = 110cm, xcom = ((20(10))+(30(-5))+(20(5))+(40(5))/110

Is this correct?
 
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You must consider the plates as rectangular lamina and first calculate their individual moment of inertia using ML^2/12.
 
AnkhUNC said:
Total = 110cm, xcom = ((20(10))+(30(-5))+(20(5))+(40(5))/110

Is this correct?

Hi Ankh! :smile:

First - common-sense - it's much easier if you use L throughout your calculations, and only put L = 5 right at the end - that way you're much less likely to make a mistake! :smile:

110 is correct.

hmm … you think I was kidding about mistakes? … try L^2 instead of L! :smile:
 

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