Lecture 34: Principal Axes of Inertia - University of Arizona

[Pages:11]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:

Iij = Ii ij

or

I1 0 0 I = 0 I2 0

0 0 I3

? Then we could write

Li = Iij j = Iiij j = Iii

j

j

Trot

=

1 2

i, j

Iiji j

=

1 2

i, j

Iiiji j

=

1 2

i

I

2

ii

? 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 = Iii , or L = Ii

? 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

? But the general relation Li = Iij j also holds. So,

j

L1 = I1 = I111 + I122 + I133 L2 = I2 = I 21 1 + I 22 2 + I 23 3 L3 = I3 = I 31 1 + I 32 2 + I 33 3

? Rearranging the equations gives:

( I11 - I )1 + I122 + I133 = 0 I 21 1 + ( I22 - I )2 + I 23 3 = 0 I 31 1 + I322 + ( I33 - I )3 = 0

? Linear algebra fact:

? We can consider this as a system of equations for thei

? Such a system has a solution only if the determinant of the coefficients is zero

? In other words, we need:

I11 - I I 21 I 31

I12 I22 - I

I 32

I13 I23 = 0 I33 - 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 i

? 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 1 : 2 : 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

(b, -b,0) m

2b2 2b2 0

110

I = m 2b2 2b2 0 = 2b2m 1 1 0

0 0 4b2

002

? 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 )[I2 - 2I] = 0

I (2 - I )(I - 2) = 0

I = (0, 2 or 2) ? 2mb2

? Let's find the principal axis associated with I = 0:

1 + 2 = 0 1 + 2 = 0 43 = 0

 ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download