🚀 go-pugleaf

David,  Here is a brief answer to your question.
                                                              _
Tap on a long thin stick as shown at right with a force F    | |
for a (short) time t.  Assume the stick is initially         | |
motionless.  The change in the momentum equals the impulse   | |
                                                             | |
    ma t = F t                                            ___|.|
    mv = Ft       (where a and v are those of the CM)      | | |
                                                           x | |
The change in the intrinsic angular momentum (i.e.,        | | |
that about the CM) is equal to the torque about the      F-->| |
CM times t.                                                  |_|

 (rot. inertia about CM) (ang. acc. about CM) t = (Torque about CM) t
 (ml^2/12) w = Fxt

(where w is the angular velocity of the resulting rotation about
the CM.  Note that if the torque is calculated about points other
than the CM you must also deal with the change in the "orbital"
angular momentum--i.e., that due to the motion of the CM.)

Eliminate Ft from the above.

   w/v = 12x/l^2

So the larger x is, the larger is the ratio of w to v.  Makes
sense.

John Mallinckrodt
Cal Poly Pomona
ajmallinckro@csupomona.edu