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Balancing unequal masses, Mathematical setting, Center of mass of a system
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
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.
Moral
Both the masses and their positions affect
whether or not the “see saw” balances.
4.
Balancing Unequal Masses
• If the heavy mass and the light mass are equidistant from the fulcrum, the
seesaw will not balance.
• The heavier object must be closer to the fulcrum than the lighter object.
5.
Balancing Unequal Masses
M1
M2
d1 d2
Need:
M1 d1 = M2 d2
6.
Changing our point of view
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. . .
M2
d1 d2
We (still) need:
M1 d1 = M2 d2
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. . .
-2 -1 0 1 2
9.
Mathematical Setting
And place the objects in the coordinate system. . .
M4
M2 M3
d1 d2 0 d3 d4
Except that now d1, d2, d3, d4, . . . 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. . .
M4
M2 M3
d1 d2 x 0 d3 d4
We want to place the fulcrum at some coordinate x that will balance the
x is called the center of mass of the system.
11.
Mathematical Setting
And place the objects in the coordinate system. . .
M4
M2 M3
d1 d2 x 0 d3 d4
In order to balance 2 objects, we needed:
M1 d1 = M2 d2 OR M1 d1 - M2 d2 =0
For a system with n objects we need:
M 1 (d1 x ) M 2 (d 2 x ) M 3 (d 3 x ) M n (d n x ) 0
12.
Finding the Center of Mass of the System
Now we solve for x .
M 1 (d1 x ) M 2 (d 2 x ) M 3 (d3 x ) M n (d n x ) 0
leads to the following set of calculations
M1d1 M1 x M 2 d 2 M 2 x M 3d3 M 3 x M n d n M n x 0
M 1d1 M 2 d 2 M 3d 3 M n d n M 1x M 2 x M 3 x M n x
M 1d1 M 2 d 2 M 3d3 M n d n M 1 M 2 M 3 M n x
And finally . . .
M d M 2 d 2 M 3 d3 M n d n
x 1 1
M1 M 2 M 3 M n
13.
The Center of Mass of the System
In the expression
The numerator is called the
first moment of the system
M 1d1 M 2 d 2 M 3d 3 M n d n
x
M1 M 2 M 3 M n
The denominator is the
total mass of the system
14.
The Center of Mass of a Solid of
Revolution.
For the goblet project,
you will need to
calculate the position
of the center of mass
of your goblet (which is
a solid of revolution!)
15.
The Center of Mass of a Solid of
Revolution.
Some preliminary remarks:
First, 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.
16.
The Center of Mass of a Solid of
Revolution.
Some preliminary remarks:
Next, 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.
17.
The Center of Mass of a Solid of
Revolution
We can treat this
as a discrete, one-
dimensional center
of mass problem!
18.
Approximating the Center of Mass
of a Solid of Revolution
M 1d1 M 2 d 2 M 3d3 M n d n
x
M1 M 2 M 3 M n
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.
19.
Approximating the Center of Mass
of a Solid of Revolution
20.
Approximating the Center of Mass
of a Solid of Revolution
R
h
d volume R 2 h
d mass d volume R 2 h
xi 1 , f xi 1
f
xi 1 xi So . . .
2
d volumei f ( xi 1 ) xi
2
d massi d volume f ( xi 1 ) xi
xi
21.
Approximating the Center of Mass
of a Solid of Revolution
Summarizing:
2
The mass of the i bead is th d massi d volume f ( xi 1 ) xi .
The position of the ith bead is di xi 1.
M 1d1 M 2 d 2 M 3d 3 M n d n
x
M1 M 2 M 3 M n
2 2 2
f ( x0 ) x0 x1 f ( x1 ) x1 x2 f ( xn 1 ) xn 1 xn
2 2 2
f ( x0 ) x1 f ( x1 ) x2 f ( xn 1 ) xn
n
2
xi 1 f ( xi 1 ) xi
i 1
n
2
f ( xi 1 ) xi
i 1
22.
The Center of Mass of a Solid of
Revolution
M 1d1 M 2 d 2 M 3 d3 M n d n
x
M1 M 2 M 3 M n
n
2
xi 1 f ( xi 1 ) xi 1
i 1
n
2
f ( xi 1 ) xi 1
i 1
Both the numerator and
denominator are
. . . And the fraction Riemann sums, and as
approaches the center of we subdivide the solid
mass of the solid! more and more finely,
they approach integrals.
23.
The Center of Mass of a Solid of
Revolution
In the limit as the number of “slices” goes to infinity, we get the
coordinate of the center of mass of the solid . .
n
2 b 2
xi 1 f ( xi 1 ) xi 1 x f ( x) dx
a
x i 1
n
2
x b 2
f ( xi 1 ) xi 1
i 1
f ( x) dx
a
where a and b are the endpoints of the region over which the solid is
“sliced.”
24.
If the cross sections are
“washers”. . .
The derivation is more or less the same, except that when we compute
the area of the little cylinder, we get
d volumei f (x 2
i 1 ) g ( xi 1 )
2
x ,
i
as we did when we computed the volume of a solid of revolution.
So the coordinate of the center of mass will be:
b 2 2
x f ( x) g ( x) dx
a
x b
f ( x) g ( x) dx
2 2
a
where a and b are the endpoints of the region over which the solid is
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