MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
Physics 8.01
IC_W09D1-2 Moment of Inertia Uniform Rod Solution A thin uniform rod of mass M and length L is mounted on an axis passing through the center of
the rod, perpendicular to the plane of the rod.
a) Calculate the moment of inertia about an axis that passes perpendicular to the rod through the
center of mass of the rod.
b) Calculate the moment of inertia about an axis that passes perpendicular through the end of the
rod.
Solution:
a) The axis is through the center, so take the point S to be the center of mass of the rod. Choose
the origin at the center of mass and the x -axis oriented along the rod, positive to the right in the
figure. Denote the length of a small element of the rod by
(1)
dl = dx .
Since the rod is uniform, the mass per unit length is a constant,
dm
mtotal
M
!=
=
=
.
dl Length L
(2)
Therefore the mass in the infinitesimal length element as given in Equation (1), is given by
M
dm = ! dx =
dx .
(3)
L
When the rod rotates, the small element traces out a circle of radius r! , dm = x ; that is, the
distance from the center of mass is the perpendicular distance from the axis.
The moment of inertia integral is now an integral with the coordinate x varying from x = ! L / 2
to x = L / 2 . The integral is then
I cm =
# (r ) dm =
2
! , dm
body
M x3
=
L 3
=
L/2
=
"L/ 2
M
L
#
L/2
"L/ 2
x 2 dx
M ( L / 2)3 " (" L / 2)3
L
3
(4)
1
M L2 .
12
b) In order to calculate the moment of inertia through an axis passing through the end of the rod,
we choose our origin at the end of the rod and we only need to adjust the limits of the integral in
Eq. (4):
I end =
M
" ( r ) dm = L "
2
! , dm
body
M x3
=
L 3
L
0
L
0
1
= M L2
3
x 2 dx
.
(5)
The axis passing through the center of mass is a perpendicular distance dend , cm = L / 2 from the
axis passing through the end. We can use the parallel axis theorem to find the moment of inertia
about the end,
2
I end = I cm + M dend
.
, cm
which becomes
(6)
I end =
agreeing with our result in Eq. (5).
1
1
1
M L2 + M L2 = M L2 .
12
4
3
(7)