m23 Center of Mass (Centroid) (1)

# m23 Center of Mass (Centroid) (1) - MATH023 Center of...

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MATH023 Center of Mass ( Centroid )

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Objectives At the end of the period, you should be able to: Locate the centers of mass of simple regions and regions bounded by curves . Locate the centers of mass of solids of revolution .
See-Saws We all remember the fun see-saw of our youth. But what happens if . . .

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Balancing Unequal Masses Moral Both the masses and their positions affect whether or not the “see saw” balances.
Balancing Unequal Masses Need: M 1 d 1 = M 2 d 2 M 1 M 2 d 1 d 2

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Changing our Point of View The great Greek mathematician Archimedes said, “give me a place to stand and I will move the Earth,” meaning that if he had a lever long enough he could lift the Earth by his own effort.
In other words. . . We can think of leaving the masses in place and moving the fulcrum. It would have to be a pretty long see-saw in order to balance the school bus and the race car, though!

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In other words. . . (We still) need: M 1 d 1 = M 2 d 2 M 2 d 1 d 2 M 1
What happens if there are many things trying to balance on the see-saw ? Where do we place the fulcrum? Mathematical Setting First we fix an origin and a coordinate system. . . 0 1 -1 -2 2

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Mathematical Setting And place the objects in the coordinate system. . . 0 M 2 M 1 M 3 M 4 d 2 d 1 d 3 d 4 Except that now d 1 , d 2 , d 3 , d 4 , . . . denote the placement of the objects in the coordinate system, rather than relative to the fulcrum. (Because we don’t, as yet, know where the fulcrum will be!)
Mathematical Setting And place the objects in the coordinate system. . . 0 M 2 M 1 M 3 M 4 d 2 d 1 d 3 d 4 Place the fulcrum at some coordinate . is called the center of mass of the system. x x x

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Mathematical Setting And place the objects in the coordinate system. . . 0 M 2 M 1 M 3 M 4 d 2 d 1 d 3 d 4 In order to balance 2 objects, we needed: M 1 d 1 = M 2 d 2 OR M 1 d 1 - M 2 d 2 =0 For a system with n objects we need: x 1 1 2 2 3 3 ( ) ( ) ( ) ( ) 0 n n M d x M d x M d x M d x
Finding the Center of Mass of the System 1 1 2 2 3 3 ( ) ( ) ( ) ( ) 0 leads to the following set of calculations n n M d x M d x M d x M d x x 1 1 1 2 2 2 3 3 3 0 n n n M d M x M d M x M d M x M d M x Now we solve for . 1 1 2 2 3 3 1 2 3 n n n M d M d M d M d M x M x M x M x 1 1 2 2 3 3 1 2 3 n n n M d M d M d M d M M M M x 1 1 2 2 3 3 1 2 3 And finally . . . n n n M d M d M d M d x M M M M

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The Center of Mass of the System 1 1 2 2 3 3 1 2 3 n n n M d M d M d M d x M M M M In the expression The numerator is called the first moment of the system The denominator is the total mass of the system
Center of Mass Our main objective here is to find the point P on which a thin plate of any given shape balances horizontally as shown. This point is called the center of mass (or center of gravity) of the plate.

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Center of Mass The center of mass of the system is located at where m = Σ m i is the total mass of the system.
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