Page 205 - Intro to Tensor Calculus

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Employing the e − δ identity the equation (2.2.38) can be simplified to the form

N
1 X (α) (α) (α) (α)
T = m (α) ω mω m x k x k − ω n ω k x k x n .
2
α=1
Define the second moments and products of inertia by the equation
N
X (α) (α) (α) (α)
I ij = m (α) x x δ ij − x x (2.2.39)
k k i j
α=1
and write the kinetic energy in the form
1
T = I ij ω i ω j . (2.2.40)
2
Similarly, the angular momentum of the system of particles can also be represented in terms of the
second moments and products of inertia. The angular momentum of a system of particles is defined as a
summation of the moments of the linear momentum of each individual particle and is

N N
X (α) (α) X (α)
H i = m (α) e ijk x j v k = m (α) e ijk x j e kmn ω m x (α) . (2.2.41)
n
α=1 α=1
The e − δ identity simplifies the equation (2.2.41) to the form

N
X (α) (α) (α) (α)
m (α) x x δ ij − x x = ω j I ji . (2.2.42)
H i = ω j
n n j i
α=1
The equations of motion of a rigid body is obtained by applying Newton’s second law of motion to the
system of N particles. The equation of motion of the αth particle is written

(α) (α)
m (α) ¨x = F . (2.2.43)
i i
Summing equation (2.2.43) over all particles gives the result

N N
X (α) X (α)
m (α) ¨x = F . (2.2.44)
i i
α=1 α=1
This represents the translational equations of motion of the rigid body. The equation (2.2.44) represents the
rate of change of linear momentum being equal to the total external force acting upon the system. Taking
(α)
the cross product of equation (2.2.43) with the position vector x produces
j
(α) (α) (α) (α)
m (α) ¨x t e rst x s = e rstx s F t

and summing over all particles we find the equation
N N
X (α) (α) X (α) (α)
m (α) e rst x ¨ x = e rst x F . (2.2.45)
s t s t
α=1 α=1