Notes_07_01
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Mass Moment of Inertia Quiz
value
units
answer
mass m
_______________
_______________
_______________
length L
_______________
_______________
_______________
width w
_______________
_______________
_______________
thickness t
_______________
_______________
_______________
density 
_______________
_______________
_______________
mass moment J
_______________
_______________
_______________
1 slug = 32.174 lbm
Notes_07_01
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Mass Moment of Inertia Review
m
T
a
m
F

r
T=Fr
F=mr
F=ma
T = m r2 
a=r
J
J = m r2
m
b
JG 
1
mb 2
12
m/2
m/2
b/4
b/4
2
JG 
2
mb
mb
1
2
      mb
2 4
2 4
16
Notes_07_01
m/4
m/4
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m/4
b/8
m/4
b/8
3b/8
3b/8
2
2
2
2
m  3b 
mb
mb
m  3b 
20
1
JG             
mb 2 
mb 2
4 8 
4 8
4 8
4 8 
256
12.8
m/6
m/6
m/6
m/6
b/12
m/6
m/6
b/12
3b/12
3b/12
5b/12
5b/12
2
JG 
2
2
2
2
2
m  5b 
m  3b 
m b 
m b 
m  3b 
m  5b 
70
1
mb 2 
mb 2
                 
6  12 
6  12 
6  12 
6  12 
6  12 
6  12 
864
12.343
2
mb 2
 m  b 
J G  2   12  3 2  5 2  ... 
2n 3
 n  2n 


n
 2i  1
2
i 1
2
for n  100, J 
1
mb 2
12.0012
Notes_07_01
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y
m
r
x
h

J G   m r
JG   t 
h/2

h/2

h/2
h / 2
JG   t
JG   t
JG 
h / 2
h / 2
2
  m x

b/2
b / 2
x
2
2
y
 y2
2

   t  x y x
2
y
2

b
dx dy
b/2
 1 3

  x  xy 2 
 dy   t
 3



b
/
2



h/2
h / 2
 1 3 1 2   1 3 1 2 
  b  by     b  by   dy
2
2
  24

  24
h/2
1 3
1
1 3
1 3
1

2
  t b 3 h  bh 3 
 b  by  dy   t  b y  by 
3
12
 12

 12
 h / 2
 12




1
1
 t b h b2  h 2  m b2  h 2
12
12

Notes_07_01
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y
x
h
z
t
b
y
x
h
z
t
b
2
J G   t 0
R
0
r 2 dr r d   t 0
2
R
0
r 3 dr d
y
R
2
0
JG   t 
1
1 4
4 2
 r  d   t R 0 d
4
 4 0
x
1

2
J G   t R 4 0   t R 4
4
2
JG 
1
m R2
2
R
z
t
Notes_07_01
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Radius of Gyration
Radius of gyration k is defined as an equivalent radius at which the entire mass of an object
would be concentrated if it were a thin ring. It is a simple way to express mass moment that
helps visualize an equivalent size.
JG  m k2
k
JG
m
As an example, for a long thin rod
JG 
1
m b2
12
As an example, for a circular disk
JG 
1
m R2
2
k
k
1
b  0.2887 b
12
1
R  0.7071 R
2
Notes_07_01
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Mass Moment of Inertia for Disk with Slots
The disk shown below has two rectangular slots. The disk is 2.5 cm thick and the slots go
completely through the disk. The density is 7.8 g/cm3.
Determine the mass, centroidal mass moment of inertia and the radius of gyration of the disk
with slots.
1.298 kg
m ____________________
2
16.21 kg.cm
JG ____________________
1 cm
d
G
3.534 cm
rG ____________________
rC = 5 cm
mC =  r2 t  = 1.532 kg
JC = mC rC2 / 2 = 19.144 kg.cm2
bS = 6 cm, hS = 1 cm, dS = 2.5 cm
mS = bS hS t  = 0.117 kg
JS = mS (bS2 + hS2) / 12 = 0.361 kg.cm2
m = mC - 2 mS = 1.298 kg
JG = JC - 2 JS - 2 mS dS2 = 16.21 kg.cm2
rG = sqrt( JG / m ) = 3.534 cm
steel
What material was used to manufacture this disk? __________________________
Comment on the manufacturability of this design.
square corners on slots are difficult to machine