Finding the Center of Mass (1) for several objects.ppt

Define center of mass, it is not my work. thanks for the original author

1. Center of Mass
of a
Solid of Revolution

2. See-Saws
We all remember the fun
see-saw of our youth.
But what happens if . . .

3. Balancing Unequal Masses
Moral
Both the masses and their positions affect
whether or not the “see saw” balances.

4. Balancing Unequal Masses
Need:
M
1d
1= M
2d
2
M
1
M
2
d
1 d
2

5. 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.

6. 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!

7. In other words. . .
(We still) need:
M
1d
1= M
2d
2
M
2
d
1 d
2
M
1

8. 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

9. Mathematical Setting
And place the objects in the coordinate system. . .
0
M
2
M
1
M
3
M
4
d
2d
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!)

10. Mathematical Setting
And place the objects in the coordinate system. . .
0
M
2
M
1
M
3
M
4
d
2d
1
d
3 d
4
Place the fulcrum at some coordinate .
is called the center of massof the system.x x x

11. Mathematical Setting
And place the objects in the coordinate system. . .
0
M
2
M
1
M
3
M
4
d
2d
1
d
3 d
4
In order to balance 2 objects, we needed:
M
1d
1= M
2d
2 OR M
1d
1-M
2d
2=0
For a system with nobjects we need:x 1 1 2 2 3 3
( ) ( ) ( ) ( ) 0
nn
M d x M d x M d x M d x        

12. Finding the Center of Mass of the System1 1 2 2 3 3
( ) ( ) ( ) ( ) 0
leads to the following set of calculations
nn
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 . . .

nn
n
M d M d M d M d
x
M M M M
   

   

13. The Center of Mass of the System1 1 2 2 3 3
1 2 3
nn
n
M d M d M d M d
x
M M M M
   

   
In the expression
The numerator is called the
first momentof the system
The denominator is the
total massof the system

14. The Center of Mass of a Solid of
Revolution.
Some preliminary remarks:
–I will ask you to believe the following (I think)
plausible fact:
Due to the circular symmetry of a solid of revolution,
the center of mass will have to lie on the central
axis.
–In order to approximate the location of this
center of mass, we “slice” the solid into thin
slices, just as we did in approximating
volume.

15. The Center of Mass of a Solid of
Revolution
We can treat this
as a discrete, one-
dimensional center
of mass problem!

16. Approximating the Center of Mass
of a Solid of Revolution 1 1 2 2 3 3
1 2 3
nn
n
M d M d M d M d
x
M M M M
   

   
What is the mass of each “bead”?
•Assume that the solid is made of a single material so
its density is a uniform throughout.
•Then the mass of a bead will simply be times its
volume.

17. Approximating the Center of Mass
of a Solid of Revolution

18. Approximating the Center of Mass
of a Solid of Revolution 2
2
d volume
d mass d volume
Rh
Rh

  


f
R
hi
x 1i
x
 i
x  
11
,
ii
x f x
  
 
2
i1
2
i1
So . . .
d volume ( )
d mass d volume ( )
ii
ii
f x x
f x x

  



  

19. Summarizing:
The mass of the i
th
bead is
The position of the i
th
bead is
Approximating the Center of Mass
of a Solid of Revolution 
2
i1
d mass d volume ( ) .
ii
f x x  

          
     
 
 
1 1 2 2 3 3
1 2 3
2 2 2 2
0 0 1 1 2 2 1 1
2 2 2 2
0 1 2 1
2
11
1
2
1
1
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
()
()
nn
n
nn
n
n
ii
i
n
i
i
M d M d M d M d
x
M M M M
f x x f x x f x x f x x
f x f x f x f x
x f x
fx
   
   








   

   
   

   


 1
.
ii
dx



20. The Center of Mass of a Solid of
Revolution       
     
 
 
1 1 2 2 3 3
1 2 3
2 2 2 2
0 0 1 1 2 2 1 1
2 2 2 2
0 1 2 1
2
11
1
2
1
1
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
()
()
nn
n
nn
n
n
ii
i
n
i
i
M d M d M d M d
x
M M M M
f x x f x x f x x f x x
f x f x f x f x
x f x
fx
   
   








   

   
   

   



Both the numerator and
denominator are
Riemann sums. As we
subdivide the solid more
and more finely, they
approach integrals.
. . . And the fraction
approaches the center of
mass of the solid!

21. The Center of Mass of a Solid of
Revolution

2
2
()
()
b
a
b
a
x f x dx
x
f x dx





where aand bare the endpoints of the region over which the solid is
“sliced.”
In the limit as the number of “slices” goes to infinity, we get the
coordinate of the center of mass of the solid . .

22. The derivation is more or less the same, except that when we compute
the area of the little cylinder, we get
as we did when we computed the volume of a solid of revolution.
So the coordinate of the center of mass will be:
If the cross sections are
“washers”. . . 
 
22
22
( ) ( )
( ) ( )
b
a
b
a
x f x g x dx
x
f x g x dx







where aand bare the endpoints of the region over which the solid is
“sliced.”   
22
i 1 1
d volume ( ) ( )
i i i
f x g x x

  