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The Tensor of Inertia C Alex R Lecture 14 Wedneday - February 22, 2006 Written or last updated: February 23, 2006 P442 – Analytical Mechanics - II The Tensor of Inertia c Alex R. Dzierba We will work out some specific examples of problems using moments of inertia. Example 1 y R M m = M/4 x A Figure 1: A disk and a point mass Figure 1 shows a thin uniform disk of mass M and radius R in the x, y plane. The z axis is out of the paper. A point mass m = M/4 is attached to the edge of the disk as shown. The inertia tensor of the disk alone about its center of mass is: 1 0 0 MR2 0 1 0 4 0 0 2 (A) Find the inertia tensor of the disk alone about the point A. (B) Find the inertia tensor of the point mass alone about the point A. (C) For the combined tensor from the above two parts – find the principal moments. SOLUTION First look at the disk alone. When we go from the center of mass of the disk to point A the moment of 2 2 2 2 inertia about the z axis goes from Izz = MR /2 to MR /2 + MR = 3MR /2 by the parallel axis theorem. The moment of inertia about the y axis clearly does not change when we go from the center of mass of the 2 disk to point A so that remains as Iyy = MR /4. But the perpendicular axis theorem holds for a lamina so 2 Ixx + Iyy = Izz or Ixx = 5MR /4. Hence the tensor for the disk alone is: 1 5 0 0 MR2 I = 0 1 0 (1) disk 4 0 0 6 Now let’s treat the point mass of mass m. The moment of inertia about the x axis is mR2 as is the moment of inertia about the y axis. The moment of inertia about the z axis is 2mR2. The products of inertia involving 2 z are zero and Ixy = Iyx = −mR . But since m = 5M/4, for the point mass alone the tensor is: 1 −1 0 MR2 I = −1 1 0 (2) mass 4 0 0 2 The tensor for the total system is then: 6 −1 0 MR2 I = −1 2 0 (3) total 4 0 0 8 To find the eigenvalues we solve the characteristic equation: 6 − λ −1 0 −1 2 − λ 0 = 0 (4) 0 0 8 − λ The roots are: √ √ λ1 = 8 λ2 = 4 − 5 λ3 = 4 + 5 leading to the principal moments of inertia given by: MR2 I = λ (5) i 4 i The eigenvectors corresponding to the eigenvalues are: 0 eˆ1 = 0 1 √ 1 −2 + 5 eˆ2 = p √ 1 10 − 4 5 0 2 √ 1 −2 − 5 eˆ3 = p √ 1 10 + 4 5 0 You can use these three vectors to form the columns of a 3 × 3 matrix that is the rotation matrix that will take you to the principal axes. Example 2 A uniform solid cube of mass M and side a is oriented with its edges along the x, y, z axes of a coordinate system whose origin coincides with one of the corners of the cube. First, find the moment of inertia tensor. z y a x Figure 2: A cube. 3 We’ll assume that the mass density is ρ = M/a . Let’s start with Ixx: Z a Z a Z a 2 2 2 5 2 2 Ixx = dx dy dzρ(y + z ) = ρa = Ma (6) 0 0 0 3 3 It is clear from symmetry that Ixx = Iyy = Izz. The element: Z a Z a Z a 1 5 1 2 Ixy = −ρ xdx ydy dz = − ρa = − Ma (7) 0 0 0 4 4 and from symmetry Ixy = Iyz = Ixz. Thus we have: 8 −3 −3 Ma2 I = −3 8 −3 (8) 12 −3 −3 8 Let’s determine the direction of the angular momentum vector as given by L = Iω for the case where ω is oriented along the x axis. 3 8 −3 −3 ω 8 Ma2 Mωa2 L = −3 8 −3 0 = −3 (9) 12 12 −3 −3 8 0 −3 Clearly ω and L are not collinear. Suppose, instead, the cube is rotating about its main diagonal. Then: 8 −3 −3 1 2 Ma2 ω Mωa2 ω L = √ −3 8 −3 1 = √ 2 (10) 12 12 3 −3 −3 8 1 3 2 In this case, ω and L are collinear. What happens if we move the origin of coordinates to the center of the cube? Z a/2 Z a/2 Z a/2 2 2 1 2 Ixx = dx dy dzρ(y + z ) = Ma (11) −a/2 −a/2 −a/2 6 and again Ixx = Iyy = Izz. Also, from the symmetry of the problem, Ixy = Iyz = Ixz = 0. So for the cube, with the origin at the center of the cube the moment of inertia tensor is: 1 0 0 Ma2 I = 0 1 0 (12) 6 0 0 1 and it is easy to see that for any angular velocity vector ω that for this inertia tensor: Ma2 Ma2 L = Iω = 1ω = ω (13) 6 6 where is it understood that 1 is the unit matrix. So for any ω the angular momentum L and ω are collinear. 4.
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