# Chapter 7 Extra Topics Crater Lake, Oregon

```Chapter 7 Extra Topics
Crater Lake, Oregon
Photo by Vickie Kelly, 1998
Greg Kelly, Hanford High School, Richland, Washington
Fg
Centers of Mass:
d
Torque
is a function
of force and distance.
Lake Superior, Washburn,
WI
Photo by Vickie Kelly, 2004
(Torque is the tendency of a system to rotate about a point.)

If the forces are all gravitational, then torque 
m1 g
 mgx
m2 g
If the net torque is zero, then the system will balance.
Since gravity is the same throughout the system, we could
factor g out of the equation.
M O   mk xk
This is called the

If we divide Mo by the total mass, we can find the center
of mass (balance point.)
M O   mk xk
MO
x

M
x m
k
k
m
k

x
MO

M
x m
k
For a thin rod or strip:
k
d = density per unit length
m
(d is the Greek letter delta.)
k
M O   x  d  x  dx
b
a
M   d  x  dx
b
mass:
center of mass:
a
MO
x
M
For a rod of uniform density and thickness, the center of
mass is in the middle.

For a two dimensional shape, we need two distances to
locate the center of mass.
y
strip of mass dm
x  distance from the y axis to
the center of the strip
x
y  distance from the x axis to
y
x
the center of the strip

 y(pronounced
dm
Moment about x-axis: M xx tilde
Center
mass:
ecksoftilda)

Moment about y-axis: M y  x dm

Mass: M  dm
x
My
M
Mx
y
M

For a two dimensional shape, we need two distances to
locate the center of mass.
y
For a plate of uniform thickness
and density, the density drops out
of the equation when finding the
center of mass.
x
y
x
Vocabulary:
center of mass = center of gravity = centroid
constant density d
= homogeneous
= uniform

9
8
yx
2
Mx  
3
Mx  
3
0
7
6
0
1 2 2
x  x dx
2
1 4
x dx
2
3
M y   x  x 2 dx
0
3
M y   x3dx
0
5
4
3
2
x
1
.5x
0
1
2
2
3
xx
coordinate of
1 2
centroid
y  =x
(2.25, 2.7)
2
1 53
Mx  x
10 0
1 43
My  x
4 0
243
Mx 
10
81
My 
4
1 33
M   x dx  x  9
0
3 0
243
81
Mx
27
My
10
9
4
y


x


M
9
10
M
9 4
3
2

Note: The centroid does not
have to be on the object.

If the center of mass is obvious, use a shortcut:

square

rectangle

circle

h
3
b
3
right triangle 
Theorems of Pappus:
When a two dimensional shape is rotated about an axis:
Volume = area . distance traveled by the centroid.
Surface Area
= perimeter . distance traveled by the centroid of the arc.
Consider an 8 cm diameter
donut with a 3 cm diameter
cross section:
1.5
V  2 r  area
V  2  2.5    1.5 
V  11.25
2
2.5
2
V  111cm
3

We can find the centroid of a semi-circular surface by using
the Theorems of Pappus and working back to get the
centroid.
1 2
A  r
2
y
4 3
V=  r
3
1 2 4 3
2 y   r   r
2
3
4r
y
3

```