Angular momentum of a multi-component system

Let us assume that the internal forces acting within the system are

(438) |

The equation of motion of the th particle can be written

(439) |

Now, we have already seen that

(441) |

(442) |

Consider the first expression on the right-hand side of Eq. (443). A general term,
, in this sum can always be paired with a
matching term,
, in which the indices have been swapped.
Making use of Eq. (437), the sum of a general matched pair can be written

(444) |

(445) |

where

(447) |

Suppose that the system is *isolated*, such that it is subject to *zero net external torque*.
It follows from Eq. (446) that, in this case, *the total angular momentum of the
system is a conserved quantity*. To be more exact, the components of the
total angular momentum taken about any three independent axes are individually conserved quantities.
Conservation of angular momentum is an extremely useful concept which greatly simplifies the
analysis of a wide range of rotating systems. Let us consider some
examples.

Suppose that two identical weights of mass are attached to a light rigid rod which rotates without friction about a perpendicular axis passing through its mid-point. Imagine that the two weights are equipped with small motors which allow them to travel along the rod: the motors are synchronized in such a manner that the distance of the two weights from the axis of rotation is always the same. Let us call this common distance , and let be the angular velocity of the rod. See Fig. 88. How does the angular velocity change as the distance is varied?

Note that there are no external torques acting on the system. It follows that the
system's angular momentum must remain constant as the weights move along the rod.
Neglecting the contribution of the rod, the moment of inertia of the system is
written

(448) |

(449) |

(450) |

Suppose that a bullet of mass and velocity strikes, and becomes embedded in, a stationary rod of mass and length which pivots about a frictionless perpendicular axle passing through its mid-point. Let the bullet strike the rod normally a distance from its axis of rotation. See Fig. 89. What is the instantaneous angular velocity of the rod (and bullet) immediately after the collision?

Taking the bullet and the rod as a whole, this is again a system upon which no
external torque acts. Thus, we expect the system's net angular momentum to
be the same before and after the collision. Before the collision, only the
bullet possesses angular momentum, since the rod is at rest. As is easily demonstrated, the
bullet's angular momentum about the pivot point is

(451) |

(452) |

(453) |

(454) |

(455) |