1.
Rotational Inertia of Geometrical Bodies
(a)
Annular cylinder about its central axis
Let 2R be the outer radius of the annular cylinder and 1R
be its inner radius, and l be its length. Let ! be its
density.
We will calculate expression for the rotational inertia by
integrating with variable r, the radial distance measured
from the axis.
Mass of annular cylinder is given by the integral
2
1
2 22 1
2
( ).
R
R
M rl dr
l R R
" !
" !
#
# $
%
Its rotational inertia is given by the integral 2
1
2
4 412 12
2 2 2 1
2 ,
=
1 ( ).2
R
R
annular cylinder
I rl dr r
l R R
I M R R
" !
" !
# &
' ($) *
# +
%
(b)
Solid cylinder (or ring) about central axis
Let the radius of the cylinder be R and its mass M. We
can obtain its rotational inertia I from the formula for the
rotational inertia of an annular cylinder by substituting
1 20 and .R R R# #
We have
2
1 .2solid cylinderI MR#
(c)
Solid disk of width h,
Let R be the radius, h, thickness and ! be the density of
the disk. For calculating the rotational inertia about the
axis as shown in the figure we choose angular variable -
measured from the vertical direction, and consider an
infinitesimal box of length dx, height dy and width h, .
The moment of inertia can be found by integrating
sin2
0 0
4 .R R
I dy h x dx-
!# ,% %
As sin ,y R -#
cos .dy R d- -# $
Therefore, 2 sin
2
0 0
24 4
0
4
4
4 sin ,
4 sin ,34 3 ,3 16
.4
R
I R d x h dx
h R d
h R
R h
" -
"
- - !
! - -
"!
" !
# & ,
# ,
# , &
# & ,
% %
%
But the mass of the disk is
02
2
.
R
M r h dr
R h
" !
" !
# ,
# ,
%
Thus the moment of inertia of a thin disk of mass M is
2
1 .4thin diskI MR#
We will use the parallel axis theorem for finding the
rotational inertia of a thin disk about an axis parallel to
the vertical axis passing through its centre.
This gives
2.O CMI I Mh# +
Using the expression of rotational inertia of a thin disk,
we have
4 2 2 disk .
4O thinI R h R hh" ! " !# , + ,
(d)
Cylinder about axis through its CM
We will use this result for calculating the rotational
inertia of a solid cylinder of length L, radius R, and mass
M about a vertical axis passing through its centre of
mass.
2 24 2 2
0 0
4 2 3
12 4
1 1 .4 12
L L
I R dh R h dh
R L R L
! " ! "
! " ! "
' (# & +. /
. /) *
# & + &
% %
Mass of the cylinder M is 2 .M R L" !#
We thus find that the rotational inertia of a cylinder about
axis as shown in the figure is
2 21 1 .4 12cylI MR ML# & +
(e)
Thin rod about an axis through its centre
Rotational inertia of a thin rod of length L and mass M
about an axis passing through its centre can be obtained
from the above result by putting in it R =0. We get
21 .12thinrodI ML#
(f)
Thin rod about axis at one of its ends
By applying parallel axis theorem and using the
expression of rotational inertia of a thin rod about axis
through its CM, we get
2 2 at end
2
1 ( ) ,2121 .3
thinrod axisLI ML M
ML
$ # & +
# &
(g)
Thin spherical shell about any diameter
Let radius of the shell be r, its thickness r, and ! be its
density. Using spherical polar coordinates and measuring
distance from the polar axis, we have
0 12
22
0 0
4 3
0
4
sin sin ,
2 sin ,
8 .3
thin spherical shellI d r r d r
r r d
r r
" "
"
- ! - 2 -
! " - -
" !
# ,
# , & &
# & ,
% %
%
Mass of a spherical shell of radius r, thickness r, and
density ! is 24 .M r r" !# ,
Using this expression for M, the rotational inertia of a
thin spherical shell of radius r can be expressed as
2
2 .3thin spherical shellI Mr# &
(h)
Rotational inertia of a solid sphere
With this result we obtain next the rotational inertia of a
solid sphere of radius R about any diameter.
Integrating the thin spherical shell expression with
respect to r from 0 to R, we get
4
0
5
8 ,3
8 .15
R
sphere
sphere
I r dr
RI
" !
"
# &
#
%
Mass of a homogeneous sphere of radius R and density
! is 34 .
3RM "
#
We thus find
22 .5sphereI MR#
(i)
Rotational inertia of a thin slab
Let the length of the slab be a, its width be b, its
thickness be c, and its density be ! . Its rotational
inertia about an axis perpendicular to its plane and
passing through its centre of mass can be calculated by
integrating the following expression;
0 1
0 1
2 22 2
2 2
2 2
,
.12
a b
a b
I c dx x y dy
c ab a b
!
!$ $
# , +
,# +
% %
Mass of the slab is .M c ab!# ,
We thus find
2 2
1 ( ).12thin slabI M a b# +
