The Parallel Axis Theorem
27 October, 2005
Mark Halpern
Our textbook has an equation which it calls the parallel axis theorem in Chapter 10. Let
CM Ixx be the moment of inertia of an object for rotations about the centre of mass of the object
and pointing in the xdirection. Remember that we will use a direction to describe a rotation; it
is the orientation of the axis of rotation and as a matter of convention we choose the end of the
axis to look towards, and the end to look from, so that the rotation appears clockwise to us. I
have chosen the xaxis here completely arbitrarily, but I do not want to write down a moment
of inertia without including a direction.
[ Check back to Section 9.5 if you want to review what we mean by centre of mass.]
But let’s imagine that this object is rotating about an axis parallel to the x axis but which does
not pass through the centre of mass but at closest approach is a distance Daway. For example,
imagine your physics book rotating about its corner rather than its centre. In this case, as the
text works out, the moment of inertia is
Ixx =CM Ixx +M D2.10.18 (1)
This is a very useful theorem, especially if someone nails the corner of your textbook to a table.
It provides a recipe for the angular momentum of an object whose motion and rotation have a
very particular relation so that one point on the object sits still. But there is another expression
which is also called the parallel axis theorem, one which we used in class to calculate an angular
momentum of a sliding block. It is a recipe for finding the angular momentum of an object which
is moving and rotating but with no restriction on the relation between motion of the centre of
mass and rotation about the centre of mass. The earth spins and orbits the sun, but it does not
move in a way that some spot sits still. Equation (10.18) is not directly useful.
We will work this out for a very particular case, but the result is general. Imagine a dumbell
consisting of two equal point masses, each with mass m, separated by a massless rod of length
2R. Imagine that the centre of mass of this object is at x=D, that the dumbell is spinning
about its centre of mass with angular frequency ωCM and that the velocity of the centre of mass
is vy. The angular momentum of the dumbell about the origin can be found from the sum of the
angular momenta of the two end masses. The mass nearest the origin is located at x=D−R
and has velocity ~v1= (vy−ωCM R)ˆy. Its angular momentum is
~
L1=~r ×~p = (D−R)m(vy−ωC M R)ˆz. (2)
The angular momentum of the second mass is
~
L2=~r ×~p = (D+R)m(vy+ωC M R)ˆz. (3)
so the sum is ~
L=~
L1+~
L2= 2Dmvyˆz+ 2mR2ωCM ˆz(4)
1
