Download Inertia Tensor - Lecture Notes | Classical Mechanics | PHYS 3210 and more Study notes Mechanics in PDF only on Docsity! PHYS 321 - Classical Mechanics Lecture 28, 8 November 2002 Inertia Tensor In a non-inertial reference frame, the kinetic energy of any body is represented by the equation 2 1 1 2 N k k k k dr T m r dt= ′ ′= + Ω × ∑ r r r . (1.1) For a rigid body in a rotating frame, the magnitude of kr′ r remains constant, implying that the time derivative of kr′ r is zero. The resulting energy equation for a rigid body in a rotating frame is ( ) ( ) 1 1 2 N k k k k T m r r = ′ ′= Ω× ⋅ Ω×∑ r rr r (1.2) This expression is simplified through the use of several properties of the cross product: (Ω x r'k) (Ω x r'k) = (r'k x Ω) (r'k x Ω) = r'k Ω x (r'k x Ω) (1.3) Using the vector triple product Ω x (r'k x Ω) = r'k(Ω Ω) - Ω( r'k Ω) ], (Ω x r'k) (Ω x r'k) = r'k2(Ω Ω) - ( r'k Ω)( r'k Ω) (1.4). Substituting equation 1.4 into 1.2 yields ( )( ) ( )( ) 1 1 2 N k k k k T m r r r r = ′ ′ ′ ′= ⋅ Ω⋅Ω − ⋅Ω ⋅Ω ∑ r r r rr r r r (1.5) The radial vector r'k is represented by the sum of all of the components of the vector in all possible directions: r'k = ∑ = 3 1u r'k µ(êµ) (1.6) where êµ represents the unit vector in the µ direction. The variable µ is limited to the values 1, 2, 3 due to the fact that there are 3 perpendicular directions in Cartesian coordinates (e.g. x, y, z). Using relation 1.6, 3 3 1 1 k k kr r r µ ν µ ν µ= ν= ′ ′ ′⋅ Ω = Ω ≡ Ω∑ ∑ rr (1.7) Substituting these expressions into equation 1.5 and rearranging the summations (which is possible due to the fact that all of the indices of summation are finite [note that even the summation going from 1 to N is finite as N represents the number of all particles that compose a rigid body and is on the order of magnitude of Avogadro's Number, ~1023]) produces ( ) 1 1 2 N k k k k T m r r r rµ νµ µν ν µ ν = ′ ′ ′ ′ = Ω ⋅ δ − Ω ∑∑ ∑ r r (1.8) where δµν is the Kronecker delta: 1 when µ = ν and 0 when µ ≠ ν. An alternative representation of the energy of the rigid body in a rotating frame is given by T = (1/2) ∑ = 3 1u ∑ = 3 1v (Ωµ) Iµν (Ων) (1.9) where Iµν is termed the inertia tensor. Comparing equations 1.8 and 1.9, the inertia tensor is Iµν = ∑ = N k 1 mk(r'k 2δµν - r'kµ r'kν) (1.10). Equation 1.10 implies, for a continuous mass distribution, Iµν = dm(r'k 2δµν - r'kµ r'kν) (1.11). There are two important observations to note about the inertia tensor: 1) the inertia tensor is real; 2) the inertia tensor is symmetric (3 x 3 matrix) which indicates that a rotation matrix (3 x 3 orthogonal matrix) can diagonalize the inertia tensor [the inertia tensor can be transformed by rotations]. Example: Compute the inertia tensor of a rod extended along the y-axis from - L/2 to L/2. Solution: All 9 of the elements of the 3 x 3 matrix are computed using equation 1.11 with r'k 2 = x2 + y2 + z2. Ixy = µ' ∫ b a dx ∫− 2/ 2/ L L dy ∫ b a dz δ(x)δ(z)[-xy] = 0
