I
Moment of Inertia
I
p. 1
2
1. Point Mass ( I = mr )
Let's use the simple version of torque (where r sin   r  r ) :   rF
From 'F = ma' we obtain: rF  r  ma (now using at  r , we get...) rF  r  m  r 
Finally we have a relationship between torque and angular acceleration:
  mr 2   and if we call mr2, the 'moment of inertia, I' then we have our
Fnet = ma in its rotational form:  net  I
2. Lotsa point masses ( I   mi ri2 ) That's a Greek sigma for 'S' (summation)!
i
3. Continuous Mass Distribution ( I   r 2dm ) Just think of the integral symbol as an
elongated 'S' for summation!
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−4. Hoop or Shell of mass, M, and radius, R ( I = MR2 )
I   r 2dm = ? Well, since all the little masses ( dm's ) are all at a distance of r = R,
we can 'slide' out R2 as a constant to get: I  R 2  dm  R 2  M (since the dm's add to M)
HOOP
VS
SOLID CYLINDER ( I = ½ MR2 )
End View
Side View
HOOP
I = MR2
5. Solid Disk or Solid Cylinder of mass, M, and radius, R, (any length!) ( I = ½ MR2 )
The plan here is to sum a lot of hoops with radius, x. (See inside ring above.)
So… I   r 2dm =  x 2 dm (where we will integrate from x = 0 to x = R)
For each x or ring, mass or 'dm' varies with its area (circumference, 2 r , times width, dx)
M
So replace 'dm' with ' 2 x  dx ' and replace ' ' with
(see box info below) to get:
 R2
 M  3
I =  x 2dm   x 2 2 x dx  2 
M
M
  x dx

Area Density:  
  R2 
Area  R 2
4 R
R
M
2M
2M x
Also  dA  dm   2 x  dx
I = 2  x3dx  2
= 2 R 4  04
R x 0
R 4 0 2R

Finally we get: I = 1 2 MR 2

whew! (I mean Q.E.D)
I
Moment of Inertia I
p. 2
1 ML2 ) rotation axis through it's c.m.
6. Rod of mass, M, and length, L ( ICM = 12
Here the plan is to integrate from x = 0 to x = L/2 (from the center out) and double
our answer to get the I for the other half of the rod.
I1/2 =  r 2dm  x 2dm
M
Linear Density:  
Doubling and substituting for 'dm' we get:
L
L
L
3 2
2
2M x
Also  d  dm   dx
M 
2 I1/2 = I = 2 x 2 dx 2    x 2dx 
L 3 0
 L  x 0
Finally, we get:
2M  L3 03 
1 ML2
  = Icm  12

L  38 3 
x = L/2
x=0
7. Parallel Axis Theorem – When rotating any mass about a line parallel to a line through
its center of mass: I = Icm + MD2 (where D is the distance between the two lines)
x=L
c.m.
D = L/2 (distance from c.m.)
x=0
8. Rod of mass, M, and length, L, rotated about a line at one of its ends (parallel lines...)
( I  13 ML2 )
Method 1/ Using the Parallel Axis Theorem, I =
So we get: I =
1 ML2
12
1 ML2 +
12
MD2 =
1 ML2 +M(L/2)2
12
 14 ML2  I  13 ML2
Method 2/ Integrating (we need the practice!) I =  r 2dm  ?
This time we'll integrate from x = 0 to x = L (no doubling here!)
So we obtain: I =  x 2dm   x 2 dx   x 2  ML  dx
M
Linear Density:  
L
L
L
x3
2
M
M
= L  x dx  L
= I  13 ML2
Also  d  dm   dx
3 0
0
I
Moment of Inertia I
p. 3
9. Hollow Cylinder, mass M and with an outer (R2) and an inner (R1) radius.
Same as the 'Solid Cylinder'.
M
M
Just integrate from x = R1 to x = R2

Area Density:  
2
Area  R2   R12
Also  dA  dm   2 x  dx
I =  r 2dm   x 2 dA   x 2 2 x dx


 2 M  R2 3
 2M  x 4
M
3
x dx =  2
=
2  x dx =  2
2 
2
2
2
R

R
 R2  R1  x  R1
  R2   R1 
 2
1  4
=


1 M
R22  R12 R22  R12
2
2
2 R2  R1


1
I  M R22  R12
2

R2
 12
R1

M
R24  R14
2
2
R2  R1
(This is just an exercise!)
R1
R2
b
a


1 2
b
12
 x 2 dx
1
M a 2  b2
12
Here, we'll consider long rods with length, L = b, and integrate from x = 0 to x = a/2
and then double our result to get the other half of the plank or plate!
For each 'dx' there will be a rod with area, bdx .
M
M
Area Density:  

Using #7 above, the parallel axis theorem:
Area ab
1 ML2 + Mx2 (where each rod is D=x away...)
I = 12
Also  dA  dm   b dx
2
2
1
each rod will have an I(or dI) of 12 b dm + x dm .
10. Rectangular Plate with dimensions 'a' and 'b' , I =
I1/2 =

1 b
12
2

 x dm =
2
a
2

x 0

1 b2
12

M 
 x 2  b dx =   b 
 ab 
 


a


a 3
3 2

M
x
 
M   1 2 a
M
2
1 2
I1/2 =   b   12 b x   =    12 b 2 
 (0  0)   
  a
3
3
 ab  
 a  
0


 b2 a 2 
1
I = 2I1/2 = 2M    = I  M a 2  b2
12
 24 24 
 


2
a3 
  ab
 

  24 24 

I
Moment of Inertia
I
p. 4
Time Out for Geometry! Area of a Frustum: A    R1  R2  L (L is the slant height)
Step 1/ What's the lateral area of a cone?
L.A. =  r (where is the slant height)
Proof: Cut along a slant height and unroll.

2 r
Notice the sector of a circle with radius, , and arc length, s = 2 r .
The area of a sector when  is measured in degrees is: Asector =

360
 R2

 2
2

R

A

R
sector
2
2 rad
 rad
When  is measured in radians the arc length is: s  rad 2 R  s  R
2
Using the figure above with s = 2 r = R  =  (since here R = )
2 r
 2  2 r  2
 Asector   r
we get:  
 Asector  R 
When  is measured in radians is: Asector =
rad
2
2
Step 2/ What's the area of a frustum of a cone? Area =   R1  R2  L (say R1 < R2)
The area is simply the big cone area minus the little cone area.
That formula is: Area =  R2 2   R1 1 (Different?!)
L
To show the two formulae are equal we'll use:
 's (ii)
R1
2

1
and by
1

R2
 R1 2  R2 1
Now we'll start with the first area formula and substitute
L with 2  1 to get: Area =   R1  R2  2  1  , now we'll 'foil' and...
get: Area =   R1 2  R1 1  R2 2  R2 1  , but by (ii) above R1 2  R2 1 = 0 (whoa!)
so our first formula becomes: Area =   R2 2  R1 1  =  R2 2   R1 1 (whoa!)
and that's our second area formula!
(i) L =
2
I
Moment of Inertia
I
p. 5
Now what was that frustum area all about? We'll actually use that A    R1  R2  L
to calculate the moment of inertia of a frustum-like piece of a thin spherical shell.
11. Thin Spherical Shell with mass, M, and radius, R ( I = 23 MR 2 )
Integrate from x = 0 to x = R and double our result.
M  M
Area Density:   Area
dy
dy
x
2
2
2
4 R 2
Also use: x + y = R  2 x  2 dx  0  dx   y
Also  dA = dm =  2 R  ds
Obtain slant height 'ds' with the Pythagorean Thm.
dA =   R1  R2  L    2r  ds
2
2
2
2
2
(ds)  (dx)  (dy)  ds  (dx)  (dy) or
where ds = L (slant height) and
2
we replace (R1+R2) with 2r.
ds  1  dy

dx
(used
for
arc
length)
and
dx
 
dA  2 y 1 
 
dy 2
dx
 dx where 'r' is replaced by 'y', giving us:
R
I1/2 =  r 2dm   y 2 dA =   y 2  2 y 1 
0
1
 
dy 2
dx
 
x
y
 1
 I  2I 12  2  2
2
 1
x2
y2
y 2  x2

y
2
 
dy 2
 dx
dx

R
y
and
 I 12  2

 y
R
M
4 R 2
3 R
 y  dx
0
 4M R   y 2dx  MR   R2  x2  dx  MR  R2 x  x3  0 =  I  23 MR 2
R
R
0
0
3
R
12. Solid Sphere with mass, M, and radius, R ( I  52 MR 2 )
At least the hard work has all been done above! Now we'll sum up (integrate) a lot of thin
spherical shells from x = 0 to x = R . (Note: We don't have to double our result this time.)
I   dI   23 dm  x 2 #10 (I  23 MR 2 ) and since dm   dV   4 x2dx...

I   23 x 2
I  2 M3
R

  4 x dx   
x5
5
2
R
 2 M3
R
0
That's All Folks!
M
4 R3
3

5
R
5

 83
R

x 4dx
x 0
 0  I  52 MR 2
M
Volume Density:   Volume

M
4  R3
3
Also:  dV  dm and with x = r
dV = 4 r 2 dx  4 x 2 dx