Finding the Center of Mass for a Uniform Plate

AI Thread Summary
To find the center of mass for a uniform plate with a length of L = 5.0 cm, the approach involves treating the plate as four rectangles and calculating their individual contributions to the center of mass. The total mass is calculated as 110 cm, and the x-coordinate of the center of mass is derived from the formula xcom = ((20(10)) + (30(-5)) + (20(5)) + (40(5))) / 110. It's advised to maintain the variable L throughout calculations to minimize errors, applying the specific value only at the end. The calculations confirm that the total mass and method are correct, emphasizing the importance of careful variable usage.
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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