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# week10 - 10 Eigenvalue problems/complex numbers Read Boas...

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Unformatted text preview: 10 Eigenvalue problems/complex numbers Read: Boas Ch. 3, sec. 10-12, Ch. 2 10.1 Eigenvalues and eigenvectors Figure 1: Left: 3 masses for computation of inertia tensor. Right: imagine rotating cylinder around axis of symmetry, or around one rotated by an angle. Let’s start with a physical example which illustrates the kind of math we need to develop. A set of masses or a rigid body with some mass distribution can be characterized in its resistance to rotation about any axis by its inertia tensor I ij ≡ X k m k r 2 k- X k m k x k,i x k,j , (1) where k is an index that runs over a set of assumed discrete particles with mass m k . The indices i and j refer to Cartesian coordinates x, y, z , and r k is the distance of particle k from the origin, r k = q x 2 k, 1 + x 2 k, 2 + x 2 k, 3 (2) I need to convince you that this has something to do with what you learned in elementary physics about the moment of inertia associated with the rotation of a rigid body about some axis. First imagine there is a point mass m located at (a,0,0). There’s only one mass so that the sum over k has only one term. Check that I 11 = m · a 2- m · a 2 = m ( a- a ) = 0, I 22 = I 33 = m · a 2- m · · 0 = ma 2 . These correspond to the formulae you learned for the moment of inertia associated with rotating a point mass about x axis where it sits, a distance 0 away, or about the y or z axes, a distance a away. 1 In a more general situation as shown in the figure, we may have many masses indexed by k in (1). For the situation shown, let’s calculate the I ij explicitly: I 11 = (2 + 2 + 2)- (1 + 0 + 1) = 4 (3) I 12 =- (1 + 0 + 0) =- 1 , (4) etc. We find for the matrix or 2nd rank tensor I ij I = 4- 1- 1- 1 4- 1- 1- 1 4 . (5) Note I is symmetric due to its definition. mechanics: | L i = I | ω i (6) | τ i = I | α i , (7) where now the vectors | ω i and | α i have components which are angular velocities and accelerations, respectively, around each of three axes. Since there are off- diagonal elements of I ij , it’s hard to get much intuition for what it means; note for example that L x = I xx ω x + I xy ω y + I xz ω z . It is possible, however, to find a basis (rotate to a new set of coordinates) where the inertia tensor is diagonal, i.e. L x = I xx ω x , L y = I yy ω y , L z = I zz ω z . This should be intuitive to you as you are used to calculating the moments of inertia for highly symmetric situations, as the I for a cylinder rotated around its own axis, as shown in Fig. 1b. As also shown there, however, you can calculate the same thing for a rotation around a z axis, at a different angle with respect to the cylinder axis. The I ij is no longer diagonal, but some very general symmetric matrix for such coordinates. So the idea is: in some general case, let’s find a coordinate system where the tensor is diagonal. The elements on the diagonal will be the moments of inertia with respect to the three...
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week10 - 10 Eigenvalue problems/complex numbers Read Boas...

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