§ 7.6 - Moments, Centers of Mass and CentroidsThe Idea of Center of Mass as a Balance Point

If we took a meter stick or yard stick, we could balance it by placing our finger under its center. Forthe meter stick, that would be the 50-cm point. For the yard stick, it would be the 18-inch point. Thatpoint is called the center of mass and in addition to being the balance point, it has some usefulproperties. For example, if we threw it in the air, the stick would rotate about its center of mass, andits center of mass would follow a parabolic trajectory, just as if we had thrown a ball.

You probably first were exposed to this idea as a small kid balancing on a seesaw. If kids of equalweight sit on opposite ends and the fulcrum is in the middle, the seesaw balances. If one kid isheavier than the other, then the balance point (center of mass) is closer to the heavier kid. In fact, ifthe heavier kid is twice the weight (or there are two kids on one side), then the fulcrum should beplaced twice as far from the lighter kid as from the heavier kid. To balance, the product of thedistance from the fulcrum and the weight of the kid has to be the same on both sides. This isillustrated in the two diagrams below.

Finding the Center of Mass or Balance Point in One Dimension

The position of the center of mass can be found by understanding that it’s the point at which we canoften consider all the mass to be concentrated. That’s why if we support the center of mass, wesupport the whole object. The position of the center of mass is often designated by “x-bar”, and ifwe multiply the total mass of a group of objects by x-bar, it should be equivalent to adding up all theproducts of the individual masses times their respective distance from some reference point such asthe position of the left-most mass or maybe the origin of a coordinate system.

Solving for x-bar

Applying this idea to the top diagram, above, and measuring distances from the left-most mass:

from the left-most mass, which is exactly where we placed the fulcrum to achieve balance.

Applying this idea to the bottom diagram, above, and measuring distances from the left-most mass:

from the left-most mass, which again is exactly where we placed the fulcrum to achieve balance.

Example of finding the center of mass of a linear collection of several masses

from the left-most mass, which is exactly where the fulcrum should be placed to achieve balance.

Finding the Center of Mass of a Continuous Object

When we move from the discrete case with individual masses, as shown above, to the continuouscase of a more or less one-dimensional object with mass which may vary along its length (such as abaseball bat, for example), we imagine the object being cut into an infinite number of infinetesimallythin slices of mass dm, and we change the summations to integrals.

A real baseball bat is thicker at one end, making it more massive at that end, and moving the centerof mass closer to the thicker end. Imagine, instead, a hypothetical bat of uniform thickness whosedensity increases as we move along its length.

Then the mass dm of a thin slice dx (perpendicular to the length) would be

If we assume the bat is 3 feet long, then its center of mass would be at

from the left end.

Finding the Center of Mass of a Triangle in Two Dimensions

Suppose we have a thin, flat triangle (a thin flat shape is called a lamina) of uniform density, and wewant to find its center of mass.

To find the y-coordinate of the center of mass we need to take horizontal slices, as shown in the toptriangle diagram above. To find the x-coordinate of the center of mass we need to take verticalslices, as shown in the bottom triangle diagram above.

In two dimensions, the mass of a horizontal slice is the density times the area of the slice

So y-bar is

Note that the density cancels out as long as it’s constant and can be taken out of the integrals.

The integrals we’ve written are over y, but they contain an x, so we need to know how x and y arerelated, that is, we need to find the equation of the hypotenuse of the triangle.

Solving for x gives

So y-bar is

Similarly, the mass of a vertical slice is

and x-bar is

The center of mass of the triangle is shown by the intersecting dashed lines in the figure below.

Finding the Center of Mass of a Parabolic Section in Two Dimensions

Find the center of mass of the section of the parabola shown below.

From the fact that the parabola is symmetric with respect to the y-axis, we can see that the x-coordinate of the center of mass is on the y-axis

The y-coordinate of the center of mass can be found by taking horizontal slices, as we did with thetriangle previously. In this case, the mass dm of the horizontal slice is

and from the equation for the parqbola, we can solve for x

Let u = 4-y and du = -dy.

Finding the Center of Mass of a Parabolic Section Cut by a Line Two Dimensions

Find the center of mass of the section of the parabola shown in yellow below.

To find x-bar, we look at vertical strips with that extend from the straight line up to the parabola, asshown in the figure

so dm

and x-bar is

To find y-bar, we look at horizontal strips with that extend from the straight parabola on the left to theline on the right for y going from 0 to 3, and from the left side of the parabola to the right side of theparabola for y going from 3 to 4, as shown in the figure

so dm for y going from 0 to 3 is

and dm for y going from 3 to 4 is

and y-bar is

The center of mass of the parabola section is shown by the intersecting lines and the red dot in thefigure below.

Finding the Volume of a Figure of Rotation Using the Theorem of Pappus

The Theorem of Pappus says that you can find the volume of a figure of rotation by multiplying thecross-sectional area times 2π times distance from the axis of rotation to the center of mass of thecross-sectional area.

We actually used this idea in the first week (§7.2) when we checked the volume of a torus (a circlerotated about an external axis) by multiplying the area of the circle times 2π times the distance fromthe axis of rotation to the center of the circle (which is the circle’s center of mass).

Using the Theorem of Pappus, we can say the volume of the torus is simply 2π times the distancefrom the axis of rotation to the center of mass (the center of the circle) times the cross-sectional areaof the circle (πr ).

This is the same answer we got using the shell method in §7.2, but note how much simpler it is usingthe Theorem of Pappus.

⋅ ∑ = ∑ ⋅x mi mi xi

=x ∑ ⋅mi xi∑ mi

= = = = 6 f tx ∑ ⋅mi xi∑ mi

(5 lb)(0 f t) + (5 lb)(12 f t)5 lb + 5 lb

60 f t − lb10 lb

= = = = 8 f tx ∑ ⋅mi xi∑ mi

(5 lb)(0 f t) + (10 lb)(12 f t)5 lb + 10 lb

120 f t − lb15 lb

= =x ∑ ⋅mi xi∑ mi

(3 lb)(0 f t) + (4 lb)(5 f t) + (2 lb)(8 f t) + (1 lb)(10 f t)3 lb + 4 lb + 2 lb + 1 lb

= = 4.6 f t46 f t − lb10 lb

=x ∫ x dm∫ dm

density = ρ = 1 + 0.6x lb/f t

dm = ρ ⋅ dx = (1 + 0.6x) dx lb

= = =x ∫ x dm∫ dm

x (1 + 0.6x) dx∫ 30

(1 + 0.6x) dx∫ 30

[ + 0.2 ]x2

2 x33

0

[x + 0.3 ]x2 30

= [ ] = 1.74 f t4.5 + 5.43 + 2.7

dm = ρ ⋅ x dy

= = =y ∫ y dm∫ dm

ρ ∫ y ⋅ x dyρ ∫ x dy

∫ y ⋅ x dy∫ x dy

y = 2 − x12

x = 4 − 2y

= = = f t f rom the bottomy y(4 − 2y) dy∫ 20

4 − 2y dy∫ 20

[2 − ]y2 23 y3 2

0

[4y − ]y2 20

23

dm = ρ ⋅ y dx

= = = = f t f rom the lef tx ∫ x ⋅ y dx∫ y dx

x(2 − x) dx∫ 40

12

(2 − x) dx∫ 40

12

[ − ]x2 16 x3 4

0

[2x − ]14 x2 4

0

43

= 0x

dm = ρ ⋅ 2x dy

x = 4 − y‾ ‾‾‾‾√

= = =y ∫ y dm∫ dm

y ⋅ 2x dy∫ 40

2x dy∫ 40

y dy∫ 40 4 − y‾ ‾‾‾‾√

dy∫ 40 4 − y‾ ‾‾‾‾√

= = = = = 1.6y − (4 − u) du∫ 04 u‾√− du∫ 0

4 u‾√

4 − du∫ 40 u‾√ u3/2

du∫ 40 u‾√

[ − ]83 u3/2 2

5 u5/2 40

[ ]23 u3/2 4

0

85

dm = ρ ⋅ [(4 − ) − (x + 2)]dx = ρ ⋅ [2 − − x]dxx2 x2

= = = =x ∫ x dm∫ dm

x(2 − − x) dx∫ 1−2 x2

2 − − x dx∫ 1−2 x2

2x − − dx∫ 1−2 x3 x2

2 − − x dx∫ 1−2 x2

[ − − ]x2 14 x4 1

3 x3 1−2

[2x − − ]13 x3 1

2 x2 1−2

= − = −0.512

dm = ρ ⋅ [(y − 2) − (− )] dy = ρ ⋅ (y − 2 + ) dy4 − y‾ ‾‾‾‾√ 4 − y‾ ‾‾‾‾√

dm = ρ ⋅ [( ) − (− )] dy = ρ ⋅ 2 dy4 − y‾ ‾‾‾‾√ 4 − y‾ ‾‾‾‾√ 4 − y‾ ‾‾‾‾√

= =y ∫ y dm∫ dm

y(y − 2 + ) dy + y(2 ) dy∫ 30 4 − y‾ ‾‾‾‾√ ∫ 4

3 4 − y‾ ‾‾‾‾√y − 2 + dy + 2 dy∫ 3

0 4 − y‾ ‾‾‾‾√ ∫ 43 4 − y‾ ‾‾‾‾√

=+[ − − (4 − y + (4 − y ]1

3 y3 y2 83 )3/2 2

5 )5/2 30 [− (4 − y + (4 − y ]8

3 )3/2 25 )5/2 4

3

+[ − 2y − (4 − y ]12 y2 2

3 )3/2 30 [− (4 − y ]2

3 )3/2 43

= = = 2.454592

125

2

Volume = 2π(5)(4π) = 40π2