Angular Momentum
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Consider
a system of particles. Define the mechanical angular momentum, Li,
of the ith particle as
where pi
º
mi
vi
.
Take the derivative with respect to time if Eq. (1) to obtain
where
the substitution
has been
made in Eq. (2). Define the total mechanical angular momentum, L,
as
Define
the torque, Ni
exerted on the ith particle relative to the point r = 0 as
Thus Eq.
(2) becomes
Taking
the derivative of Eq. (4) gives
Let the
force exerted on the ith particle be separated into two terms; the force
on the ith particle due to the jth particle, Fji,
and the external force, F(e)i,
exerted on the ith particle from outside the system of particles. Thus
Substituting
this expression for the force on the ith particle into Eq. (5) gives
The total
torque, N, is defined as
When the
expression for Ni
in Eq. (8) is substituted Eq. (9) we obtain
Define
the external torque, N(e)i,
on the ith particle as
Define
the total external torque, N(e),
as
With
these terms Eq. (10) may be expressed as
If the
weak form of Newton’s Third Law of action and reaction holds then
In this
case the second term on the right hand side of Eq. (14) is a sum of terms of the
form
where rij
=
ri
- rj.
If, in addition, the strong form of Newton’s Third Law of action and reaction
holds then rij
is
parallel to Fij
.
Eq. (14) then becomes
The
difference between the strong and weak form of Newton’s Third Law is
illustrated in Figure 1
The
total angular momentum then becomes
