8 ROTATIONAL MOTION 8.6 Moment of inertia

q

̧

2π σ dσ

̧b

2π σ dσ

h i

0

ring

= M b (^2)
ring
axis
M
b
I = 2 M b 2
original axis new axis
Figure 78: An application of the parallel axis theorem.
where the integrals are taken over the area of the disk, and the redundant z-
integration has been suppressed. Let us divide the disk up into thin annuli. Con-
sider an annulus of radius σ = x^2 + y^2 and radial thickness dσ. The area of this
annulus is simply 2π σ dσ. Hence, we can replace dx dy in the above integrals by
2π σ dσ, so as to give
The above expression yields
b 3
I = M^0. (8.50)
0
2 π σ^4 /4
b
I = M 0
[2 π σ^2 /2]b

1
M b^2. (8.51)
2
Similar calculations to the above yield the following standard results:
The moment of inertia of a thin rod of mass M and length l about an axis
passing through the centre of the rod and perpendicular to its length is
I =
1
M l^2.
12
The moment of inertia of a thin rectangular sheet of mass M and dimensions
a and b about a perpendicular axis passing through the centre of the sheet
is
I =
1
M (a^2 + b^2 ).
12
axis (^) I
M
b