sign in sign up
<<
Home , Coordinate system

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 • Both expressions would be a great deal simpler if the inertia tensor was diagonal. That is, if: = δ Iij Ii ij or

»I1 0 0 ÿ … Ÿ I = 0 I 0 … 2 Ÿ

…0 0 I3 ⁄Ÿ • Then we could write = ω = δ ω = ω Li ƒ Iij j ƒ Ii ij j Ii i j j

= 1 ω ω = 1 δ ω ω = 1 ω 2 Trot ƒ Iij i j ƒ Ii ij i j ƒ Ii i 2 i, j 2 i, j 2 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 ) is diagonal – These are called the principal axes of the body – All the rotational problems you did in first-year dealt with about a principal axis – that’s why the looked simpler. • If a body is rotating solely about a principal axis (call it the i axis) then: = ω = Li Ii i , 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 = ω • But the general relation L i ƒ I i j j also holds. So, j = ω = ω + ω + ω L1 I 1 I11 1 I12 2 I13 3 = ω = ω + ω + ω L2 I 2 I21 1 I22 2 I23 3 = ω = ω + ω + ω L3 I 3 I31 1 I32 2 I33 3 • Rearranging the equations gives: ( − )ω + ω + ω = I11 I 1 I12 2 I13 3 0 ω + ( − )ω + ω = I21 1 I22 I 2 I23 3 0 ω + ω + ( − )ω = I31 1 I32 2 I33 I 3 0 • fact: ω – We can consider this as a system of equations for the i – Such a system has a solution only if the determinant of the coefficients is zero • In other words, we need: − I11 I I12 I13 − = I21 I22 I I23 0 − I31 I32 I33 I • The determinant results in a cubic 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 – 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 ω 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 ω ω ω – 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 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 ÿ »1 1 0ÿ = … 2 2 Ÿ= 2 … Ÿ I m …2b 2b 0 Ÿ 2b m …1 1 0Ÿ … 0 0 4b2 ⁄Ÿ …0 0 2⁄Ÿ • 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⁄ÿ = 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: ω + ω = 1 2 0 ω + ω = 1 2 0 ω = 4 3 0 • So the ratio of the components in our coordinate system when the object it rotating about the principal axis with I = 0 is: ω ω ω = − 1 : 2 : 3 1: 1: 0 meaning the axis is defined by the vector: = − r ex ey In other words, along the axis of the dumbbell If an object has an axis of , that axis is always a principal axis • What about the other principal axes? – The axes associated with I = 4mb2 are: −ω + ω = 1 2 0 ω −ω = 1 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 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 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: = * Iij Iij • 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 will come up again in Quantum Mechanics Principal Moments ↔ Eigenvalues Principal Axes ↔ Eigenfunctions