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# CDIM11 - CHAPTER 11 Dynamics of Rigid Bodies 11-1 The...

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CHAPTER 11 Dynamics of Rigid Bodies 11-1. The calculation will be simplified if we use spherical coordinates: sin cos sin sin cos x r y r z r θ φ θ φ θ = = = (1) z y x Using the definition of the moment of inertia, ( ) 2 ij ij k i j k I r x x x ρ δ dv = (2) we have ( ) ( ) ( ) 2 2 33 2 2 2 2 cos cos I r z dv r r r dr d d ρ ρ θ θ = = φ (3) or, ( ) ( ) 1 2 4 2 33 0 1 0 5 1 cos cos 4 2 5 3 R I r dr d R + = = π d ρ θ θ πρ φ (4) 353

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354 CHAPTER 11 The mass of the sphere is 3 4 3 M = R π ρ (5) Therefore, 2 33 5 I MR 2 = (6) Since the sphere is symmetrical around the origin, the diagonal elements of { I } are equal: 2 11 22 33 2 5 I I I MR = = = (7) A typical off-diagonal element is ( ) ( ) 12 2 2 2 sin sin cos cos I xy dv r r dr d ρ d ρ θ φ φ θ = = − φ (8) This vanishes because the integral with respect to φ is zero. In the same way, we can show that all terms except the diagonal terms vanish. Therefore, the secular equation is 11 22 33 0 0 0 0 0 0 I I I I I I 0 = (9) From (9) and (7), we have 2 1 2 3 2 5 I I I MR = = = (10) 11-2. a) Moments of inertia with respect to the x axes: i x 3 = x 3 R h CM x 1 x 1 x 2 x 2 It is easily seen that for i j . Then the diagonal elements become the principal moments I , which we now calculate. 0 ij I = ii I i The computation can be simplified by noting that because of the symmetry, . Then, 1 2 I I I = 3 ( ) 2 2 2 1 2 1 2 3 1 2 2 2 2 I I x x x d I I v ρ + = = = + + (1)
DYNAMICS OF RIGID BODIES 355 which, in cylindrical coordinates, can be written as ( ) 2 2 2 1 2 0 0 0 2 2 h Rz h d dz r z rd π I I r ρ φ = = + (2) where 2 3 M M V R ρ π = = h (3) Performing the integration and substituting for ρ , we find ( ) 2 1 2 3 4 20 I I M R h = = + 2 (4) 3 I is given by ( ) 2 2 2 3 1 2 I x x dv r rdr d dz ρ ρ = + = φ (5) from which 2 3 3 10 I MR = (6) b) Moments of inertia with respect to the x i axes: Because of the symmetry of the body, the center of mass lies on the 3 x axis. The coordinates of the center of mass are (0 0 ,0, ) z , where 3 0 3 4 x dv z h dv = = (7) Then, using Eq. (11.49), 2 ij ij ij i j I I M a a a δ = (8) In the present case, and 1 2 0 a a = = ( ) 3 3 4 a = h , so that 2 2 1 1 2 2 2 2 2 3 3 9 3 1 16 20 4 9 3 1 16 20 4 3 10 I I Mh M R h I I Mh M R h I I MR = = + = = + = 2 2 11-3. The equation of an ellipsoid is 2 2 2 3 1 2 2 2 2 1 x x x a b c + + = (1)

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356 CHAPTER 11 which can be written in normalized form if we make the following substitutions: 1 2 3 , , x a x b x c ξ η ζ = = = (2) Then, Eq. (1) reduces to 2 2 2 1 ξ η ζ + + = (3) This is the equation of a sphere in the ( ξ , η , ζ ) system. If we denote by dv the volume element in the system and by d τ the volume element in the ( ξ , η , ζ ) system, we notice that the volume of the ellipsoid is i x 1 2 3 4 3 V dv dx dx dx abc d d d abc d abc ξ η ζ τ π = = = = = (4) because d τ is just the volume of a sphere of unit radius.
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CDIM11 - CHAPTER 11 Dynamics of Rigid Bodies 11-1 The...

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