Re: [Vo]:Successful Mechanical OU

```On 5/06/2018 12:37 AM, Vibrator ! wrote:
Consider a 1 kg weight, connected by a pulley cord to another mass that slides
horizontally without friction.  You may verify that the rate of change of net
system momentum is a constant, invariant of the ratio of gravitating to
non-gravitating mass - taking gravity as 9.81 N, it is precisely thus 9.81
kg-m/s per kg of gravitating mass.```
```
So, the amount of non-gravitating mass could be anything from zero to infinity,
but regardless of whether the gravitating mass is rising or falling, the rate
of change of net system momentum is always 9.81 p/s/kg (where p=mV).

This is not, as one might suspect, a consequence of Galileo's principle - that
gravity defies F=mA - but rather a direct manifestation of it.  Same-same, no
matter what force we apply.

Now switch out that linear-sliding mass for an angular inertia instead.  If we
measure its angular inertia in terms of kg/m^2, and given that moment of
inertia (MoI) is equal to mass times radius squared, we can select a mass of 1
kg at 1 meter radius for an MoI of '1'.

If we measure its angular velocity in terms of radians per second, then we have
numerical parity with its linear equivalent for an equal distribution of
absolute momentum - that is, if we applied a linear to angular force between
them of 1 Newton for 1 second, we obtain 1 kg-m/s of linear momentum, and also

Likewise, if we employ a 1 kg drop-weight to torque up that MoI, the system
gains 9.81 p of net momentum per second.

Since they're equal absolute magnitudes of inertia, albeit in their respective
dimensions, the net system velocity remains equally-distributed between them.

Hence with 9.81 p of net system momentum, we have 4.905 p on each inertia - 1
kg dropping at 4.905 m/s, and an MoI of '1' rotating at 4.905 rad/s.

However, since the objective distance 1 radian corresponds to is dependent upon
the dimensions of the circle in question (it's a relative, not absolute,
quantity), this same point applies to the 'magnitudes' of angular momentum
we're measuring for any given angular velocity; so for instance if we double
All looks OK (even if rather strange language) to here.
but also halve the relative (angular?) velocity compared to the linear value
wherein inertia is a fixed function of rest mass.  Hence, repeating the 1
second, 1 kg drop, we'd again obtain 4.905 p on the weight, but '9.81' p on the
MoI - for a 'net' total of '14.715' p
This is numerically correct but dimensionally incorrect (which is maybe why you
use the quotes).  Angular momentum does not have the same dimensions as linear
momentum and so they really cannot be added in this fashion (just as you can't
add 4.905 meters to 9.81 square-meters and obtain a reasonable result as 14.715
somethings).
... i'm using scare-quotes there to highlight my point; the objective value of
the absolute magnitudes of momentum and their distribution remains 9.81 p/s for
the net system, regardless of how the angular component is represented.

.  .  .

```