University:
Michigan State University
Course:
ERG 160 | Success in Science, Technology, Engineering, and Mathematics
Academic year:

2021
Views:

263

Pages:

2
Author:

Madelyn Bradshaw

COSMOS: Complete Online Solutions Manual Organization System Chapter 9, Solution 179. (a) At the center of the cube have (using Figure 9.28) Ix = I y = Iz = 1 1 m a 2 + a 2 = ma 2 12 6 ( Now observe that symmetry implies ) I xy = I yz = I zx = 0 Using Equation (9.48), the equation of the ellipsoid of inertia is 1 2 2 1 2 2 1 2 2  ma  x +  ma  y +  ma  z = 1 6  6  6  or x2 + y 2 + z 2 = 6 = R2 ma 2 ( ) which is the equation of a sphere. Since the ellipsoid of inertia is a sphere, the moment of inertia with respect to any axis OL through the center O of the cube must always  1  be the same  R = .  I OL   ∴ I OL = 1 2 ma 6 (b) The above sketch of the cube is the view seen if the line of sight is along the diagonal that passes through corner A. For a rectangular coordinate system at A and with one of the coordinate axes aligned with the diagonal, an ellipsoid of inertia at A could be constructed. If the cube is then rotated 120° about the diagonal, the mass distribution will remain unchanged. Thus, the ellipsoid will also remain unchanged after it is rotated. As noted at the end of section 9.17, this is possible only if the ellipsoid is an ellipsoid of revolution, where the diagonal is both the axis of revolution and a principal axis. It then follows that I x′ = I OL = continued 1 2 ma 6 COSMOS: Complete Online Solutions Manual Organization System In addition, for an ellipsoid of revolution, the two transverse principal moments of inertia are equal and any axis perpendicular to the axis of revolution is a principal axis. Then, applying the parallelaxis theorem between the center of the cube and corner A for any perpendicular axis I y′ = I z ′ =  3  1 2 ma + m  a  6  2  2 or I y′ = I z′ = 11 2 ma 12 Note: Part b can also be solved using the method of Section 9.18. First note that at corner A Ix = Iy = Iz = 2 2 ma 3 I xy = I yz = I zx = 1 2 ma 4 Substituting into Equation (9.56) yields k 3 − 2ma 2k 2 + 55 2 6 121 3 9 m a k− ma =0 48 864 For which the roots are k1 = 1 2 ma 6 k2 = k3 = 11 2 ma 12