Download Lecture 34: Principal Axes of Inertia

Lecture 34: Principal Axes of Inertia
• We’ve spent the last few lectures deriving the general
expressions for L and Trot in terms of the inertia tensor
• Both expressions would be a great deal simpler if the
inertia tensor was diagonal. That is, if:
I ij = I iδ ij
or
I1
0
0
I= 0
0
I2
0
0
I3
• Then we could write
I ijω j =
Li =
j
1
Trot =
2
I iδ ijω j = I iωi
j
i, j
1
I ijωiω j =
2
i, j
1
I iδ ijωiω j =
2
I iωi2
i
• We’ve already seen that the elements of the inertia tensor
transform under rotations
• So perhaps we can rotate to a set of axes for which the
tensor (for a given rigid body) is diagonal
– These are called the principal axes of the body
– All the rotational problems you did in first-year physics dealt
with rotation about a principal axis – that’s why the
equations looked simpler.
• If a body is rotating solely about a principal axis (call it the
i axis) then:
Li = I iωi , or L = I i
• If we can find a set of principal axes for a body, we call the
three non-zero inertia tensor elements the principal
moments of inertia
Finding the Principal Moments
• In general, it’s easiest to first determine the principal
moments, and then find the principal axes
• We know that if we’re rotating about a principal axis, we
have:
L=I
A principal moment
I ijω j also holds. So,
• But the general relation Li =
j
L1 = I ω1 = I11ω1 + I12ω 2 + I13ω3
L2 = I ω 2 = I 21ω1 + I 22ω 2 + I 23ω3
L3 = I ω3 = I 31ω1 + I 32ω 2 + I 33ω3
• Rearranging the equations gives:
( I11 − I ) ω1 + I12ω 2 + I13ω3 = 0
I 21ω1 + ( I 22 − I ) ω 2 + I 23ω3 = 0
I 31ω1 + I 32ω 2 + ( I 33 − I ) ω3 = 0
• Linear algebra fact:
– We can consider this as a system of equations for the ωi
– Such a system has a solution only if the determinant of the
coefficients is zero
• In other words, we need:
I11 − I
I12
I 21
I 31
I 22 − I
I 32
I13
I 23 = 0
I 33 − I
• The determinant results in a cubic equation for I
• The three solutions are the three principal moments of
inertia for the body (one corresponding to each principal
axis)
• And this brings us the resolution of the apparent
contradiction between freshman-level physics, in which
there were three moments of inertia, and this course, where
we needed 6 numbers
– In the earlier course, only rotations about principal axes were
considered!
Finding the Principal Axes
• Now all that’s left to do is find the principal axes. We do
this by solving the system of equations for ωi
– Using one of the possible values of I – call it I1
– This will give the direction of the first principal axis
• It turns out that we won’t be able to find all three
components
– But we can determine the ratio ω1 : ω 2 : ω 3
– And that’s enough to figure out the direction of the first
principle axis (in whatever coordinate system we’re using)
Example: Dumbbell
• Consider the same dumbbell that appeared last lecture, and
define the coordinate system as follows:
( −b, b,0)
m
m
2b 2
I = m 2b 2
0
2b 2
2b 2
0
(b, −b,0)
0
1 1 0
0 = 2b 2 m 1 1 0
4b 2
0 0 2
• So the equation we need to solve is:
1− I
1
0
1 1− I
0 =0
0
0
2−I
( 2 − I ) (1 − I )2 − 1 = 0
(2 − I )[I 2 − 2I ] = 0
I (2 − I ) ( I − 2) = 0
I = ( 0, 2 or 2 ) × 2mb2
• Let’s find the principal axis associated with I = 0:
ω1 + ω 2 = 0
ω1 + ω 2 = 0
4ω3 = 0
• So the ratio of the angular momentum components in our
coordinate system when the object it rotating about the
principal axis with I = 0 is:
ω1 : ω 2 : ω3 = 1: −1: 0
meaning the axis is defined by the vector:
r = ex − e y
In other words, along the axis of the dumbbell
If an object has an axis of symmetry, that axis is
always a principal axis
• What about the other principal axes?
– The axes associated with I = 4mb2 are:
−ω1 + ω 2 = 0
ω1 − ω 2 = 0
0=0
• There’s not much information in those equations!
– For example, the z component could be anything
• This means that any two axes perpendicular to the axis of
the dumbbell can be taken as principal axes
• Note that the principal axes one finds can depend both on
the shape of the body and on the point about which it’s
rotating
– Look over Examples 11.5 and 11.6 in the text
– They show that one finds different principal axes for a
uniform cube depending on whether it’s rotating about a
corner or about the center of mass
When Can We Find Principal Axes?
• We can always write down the cubic equation that one
must solve to determine the principal moments
• But if we want to interpret these as physically meaningful
quantities, the roots of that cubic have to be real
– Recall that in general, cubics can have two complex roots
• Fortunately, we’re not in the general case here
• The inertia tensor is both real and symmetric – in
particular, it satisfies:
I ij = I ij*
• Matrices that satisfy this restriction are called Hermitian
• For such matrices, the principal moments can always be
found, and they are always real (see proof in text)
This mathematics will come up again in Quantum Mechanics
Principal Moments ↔ Eigenvalues
Principal Axes ↔ Eigenfunctions