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Chapter 8 – Problems
Because of the time factor, I have decided to
work all 5 of the Chapter 8 problems in class
for you.
This is a tentative schedule of the time we
have left:
Date
Wednesday, Nov. 20
Friday, Nov. 22nd
Monday, Nov. 25th
Assignment
th
Wednesday, Nov. 27th
Monday, Dec. 2nd
Wednesday, Dec. 4th
Work 2 Chapter 8 problems in class
Work 3 Chapter 8 problems in class
Review Chapter 8 and receive Take
Home Test (3 problems due on
12/02/02)
3 In-class test problems
Go over test, do evaluations
Discuss Final Exam
The in-class final exam will be at your
scheduled time for your local campus.
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Textbook page 330, Problem 8.3
A solid uniform sphere of radius a has a
spherical cavity of radius a / 2 centered at
a point a / 2 from the center of the sphere.
Find the center of mass.
y

x
This problem is very difficult to work using
direct integration techniques. However, using
the additive nature of an integral it is an easy
problem.
Conceptually, a sphere with a hole in it can be
generated by looking at a solid sphere and
subtracting the smaller one from it. In
algebraic form this is:
Mass solid sphere X location of it’s center of
mass=mass of sphere with hole X location of
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it’s center of mass + mass of little sphere
(hole) X location of it’s center of mass.
4
Mass solid sphere =   R3
3
3
4  R 1 4
Mass of little sphere =       R 3
3 2 8 3
Mass of big sphere with hole =
7 4 3
 R
8 3
4
7
4
1 R
4
  R3  0    R3   x    R3  
3
8
3
8 2
3
This gives x 

1R
82  R
7
14
8
The location of the center of mass for the
R
y  0, z  0.
hollow sphere is at x  
14,
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Textbook page 330, Problem 8.5
Find the moment of inertia of the sphere
in Problem 8.3 about an axis passing
through the center of the sphere and the
center of the cavity.
Again, we have the same math problem,
so we take advantage of the symmetry.
This works because of the additive nature
of an integral. That is, since I   r 2 dm we
can say:
Isolid with hole  I solid  I"hole"
For a solid sphere, we know
2
I  MR 2
5
using this, we have
I solid with hole
2
2
 R
 M solid R 2  M hole  
5
5
2
3
2
2  4 3  2 2  4  R   R 
    R  R      
5 3
5  3  2   2 

2
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
I solid with hole 
24 5 
1
R 1  
53
 32 
2  4 5  31
 R   
5 3
 32
You look at the masses of the various
spheres:
4
Mass of Solid Sphere=   R3  M
3
Mass of little sphere (hole)
3
4  R 1
=      M
3 2 8
Mass of solid with hole =
7
M m
8
You have for I
2
31
I solid with hole  MR 2 
5
32
28 
31
  m R2 
57 
32

31
mR 2
70
 I solid with hole 
31 2
mR
70