Center of Mass

of a

Solid of Revolution

See-Saws

We all remember the fun see-saw of our youth.

But what happens if . . .

Balancing Unequal Masses

MoralBoth the masses and their positions affect

whether or not the “see saw” balances.

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.

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.

Balancing Unequal Masses

Need:M1 d1 = M2 d2

M1

M2

d1 d2

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!

In other words. . .

We (still) need:M1 d1 = M2 d2

M2

d1 d2

M1

What happens if there are many things

trying to balance on the see-saw?

Where do we place the fulcrum?

Mathematical SettingFirst we fix an origin and a coordinate system. . .

0 1-1-2 2

Mathematical Setting

And place the objects in the coordinate system. . .

0

M2

M1 M3

M4

d2d1d3 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!)

Mathematical Setting

And place the objects in the coordinate system. . .

0

M2

M1 M3

M4

d2d1d3 d4

We want to place the fulcrum at some coordinate that will balance the system.

is called the center of mass of the system.

x

x

x

Mathematical Setting

And place the objects in the coordinate system. . .

0

M2

M1 M3

M4

d2d1d3 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:

x

1 1 2 2 3 3( ) ( ) ( ) ( ) 0n nM 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 calculationsn nM d x M d x M d x M d x

x

1 1 1 2 2 2 3 3 3 0n n nM 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 3n n nM d M d M d M d M x M x M x M x

1 1 2 2 3 3 1 2 3n n nM 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 dx

M M M M

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 dx

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

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!)

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.

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.

The Center of Mass of a Solid of Revolution

We can treat this as a discrete, one-dimensional center of mass problem!

Approximating the Center of Mass of a Solid of Revolution

1 1 2 2 3 3

1 2 3

n n

n

M d M d M d M dx

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.

Approximating the Center of Mass of a Solid of Revolution

Approximating the Center of Mass of a Solid of Revolution

2

2

d volume

d mass d volume

R h

R h

f

R

h

ix

1ix ix

1 1,i ix f x

2

i 1

2

i 1

So . . .

d volume ( )

d mass d volume ( )

i i

i i

f x x

f x x

Summarizing:

The mass of the ith bead is

The position of the ith bead is

Approximating the Center of Mass of a Solid of Revolution

2

i 1d mass d volume ( ) .i if x x

1 1 2 2 3 3

1 2 3

n n

n

M d M d M d M dx

M M M M

1.i id x

2 2 2

0 0 1 1 1 2 1 12 2 2

0 1 1 2 1

( ) ( ) ( )

( ) ( ) ( )n n n

n n

f x x x f x x x f x x x

f x x f x x f x x

2

1 11

2

11

( )

( )

n

i i iin

i ii

x f x x

f x x

The Center of Mass of a Solid of Revolution

1 1 2 2 3 3

1 2 3

2

1 1 11

2

1 11

( )

( )

n n

n

n

i i iin

i ii

M d M d M d M dx

M M M M

x f x x

f x x

Both the numerator and denominator are

Riemann sums, and as we subdivide the solid more and more finely,

they approach integrals.

. . . And the fraction approaches the center of

mass of the solid!

The Center of Mass of a Solid of Revolution

2

2

( )

( )

b

ab

a

x f x dxx

f x dx

where a and b are 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 . .

2

1 1 11

2

1 11

( )

( )

n

i i iin

i ii

x f x xx

f x x

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”. . .

2 2

2 2

( ) ( )

( ) ( )

b

ab

a

x f x g x dxx

f x g x dx

where a and b are the endpoints of the region over which the solid is “sliced.”

2 2

i 1 1d volume ( ) ( ) ,i i if x g x x