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

This preview shows pages 1–5. Sign up to view the full content.

CHAPTER 11 Dynamics of Rigid Bodies 11-1. The calculation will be simplified if we use spherical coordinates: sin cos sin sin cos xr yr zr θ φ = = = (1) z y x Using the definition of the moment of inertia, () 2 ij ij k i j k Ir x x x ρδ d v =− (2) we have ( ) ( ) 22 33 22 2 2 cos cos Ir z d v rr r d r d d ρ θθ (3) or, ( ) 12 42 33 01 0 5 1c o s c o s 4 2 53 R d r d R + =⋅ ∫∫ π d πρ (4) 353

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
354 CHAPTER 11 The mass of the sphere is 3 4 3 M = R π ρ (5) Therefore, 2 33 5 IM R 2 = (6) Since the sphere is symmetrical around the origin, the diagonal elements of { I } are equal: 2 11 22 33 2 5 III M R === (7) A typical off-diagonal element is ( ) () 12 22 2 sin sin cos cos Ix y d v rr d r d d θφ φ θ =− (8) This vanishes because the integral with respect to 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 00 II 0 = (9) From (9) and (7), we have 2 123 2 5 M R (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, 12 =≠ 3 ( ) 222 3 1 2 2 xxxd v + == = ++ (1)
DYNAMICS OF RIGID BODIES 355 which, in cylindrical coordinates, can be written as ( ) 2 22 12 00 0 2 2 hR z h d d z r z r d π II r ρ φ == + ∫∫ (2) where 2 3 MM VR h (3) Performing the integration and substituting for , we find ( ) 2 3 4 20 M R h + 2 (4) 3 I is given by ( ) 2 31 2 Ix x d v r r d r d d z =+= (5) from which 2 3 3 10 IM R = (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 xd v z h dv (7) Then, using Eq. (11.49), 2 ij ij ij i j IIM a a a δ =− (8) In the present case, and 0 aa ( ) 3 34 a = h , so that 11 2 33 93 1 16 20 4 16 20 4 3 10 M h M R h M h M R h M R  = +   = + 2 2 11-3. The equation of an ellipsoid is 2 3 222 1 x xx abc + += (1)

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
356 CHAPTER 11 which can be written in normalized form if we make the following substitutions: 12 3 ,, xa xb xc ξ η ζ = == (2) Then, Eq. (1) reduces to 222 1 ξη + += (3) This is the equation of a sphere in the ( , , ) system. If we denote by dv the volume element in the system and by d τ the volume element in the ( , , ) system, we notice that the volume of the ellipsoid is i x 123 4 3 Vd vd x d x d xa b c d d d abc d abc ξηζ τπ = ∫∫ (4) because d is just the volume of a sphere of unit radius. The rotational inertia with respect to the passing through the center of mass of the ellipsoid (we assume the ellipsoid to be homogeneous), is given by 3 -axis x ( ) ( 22 31 2 M Ix x d v V M abc a b d V =+ ) (5) In order to evaluate this integral, consider the following equivalent integral in which z = r cos θ : ( ) 21 00 0 2 2 sin cos sin 2 35 4 15 R azd v az rd r r dd ad d r d a a = = = =× ×× = ππ θθφ φθ π 4 r (6) Therefore, () ( ) 2 4 15 abd a b 2 + (7) and ( ) 3 1 5 IM a b (8)
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 11/03/2008 for the course PHYS 3210 taught by Professor Ford during the Fall '08 term at Colorado.

### Page1 / 44

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

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online