Angular
Momentum for System of Particles
Definition: L
= r ´ p .
For the angular momentum of a single
particle, call it particle k, about a point, Q:
Lof k about Q = (rk
- rQ) ´ mk(r'k
- r'Q) where r' = dr/dt = v.
dLof k about
Q /dt = (r'k
- r'Q) ´ mk(r'k
- r'Q) + (rk - rQ)
´ mk(r''k
- r''Q) .
The first part of the above expression is
zero, since any vector cross itself will give zero. The quantity mkr''k is equal to the forces on
particle k (
dLof k about
Q /dt = (rk
- rQ) ´ (Fkext
+ Fkint) - (rk - rQ)
´ mkr''Q
.
The total angular momentum about a point Q is
just the sum of the individual angular momenta:
dLtotal
about Q/dt = Sk {dLof k about Q
/dt} =
Sk {(rk - rQ)
´ Fkext}
+ Sk
{rk
´ Fkint} - rQ ´ Sk {Fkint}
- Sk {(mkrk)} ´ r''Q + rQ ´ Sk {mk}r''Q .
The first term, Sk{(rk - rQ)
´ Fkext} is just the external torque (torque caused by the external
forces about the point Q), tQext .
The second term, Sk {rk ´ Fkint}, contains terms that usually cancel in pairs, for
example: r1 ´ F1by 2
+ r2 ´ F2by 1
, but by
The third term, -rQ
´ Sk{Fkint} is zero since Sk{Fkint}
= 0 by
The fourth term, -Sk {(mkrk)} ´ r''Q can be
rewritten as -MR ´ r''Q .
Thus the fourth and fifth terms can be
combined to give: M(rQ-R) ´ r''Q . Normally, we choose the point Q so that it is at the center of mass.
In this case, rQ = R, and this term is zero. Even if we don't choose
the center of mass as the reference point, as long as the reference point, rQ is not being accelerated, this term
gives zero.
Putting this altogether gives the result:
dLtotal about Q/dt = tQext + M(rQ-R) ´ r''Q.
IF
tQext = 0 and either rQ=R or r''Q
= 0, THEN dLtotal
about Q/dt = 0, which means that we have conservation
of angular momentum about Q:
THEN Ltotal about Q = constant .