Re: MOMENT OF INTERTIA

[James Mclean <jmclean@CHEM.UCSD.EDU>]

Bob Sciamanda wrote:
> ...
> all have the same total energy, Mgh.  As they proceed to roll down the
> plane without slipping, this gravitational potential energy is converted
> into both translational kinetic energy , (1/2)*M*V^2, and rotational
> kinetic energy, (1/2)*I*w^2=(1/2)*I*(V/R)^2.  The object with the

Picking up from here to treat the problem as initially set, in all the
hoop-to-disk cases we have
   Mgh = 0.5 Mv^2 +0.5 I(v/R)^2
or, dividing by M/2,
   2gh = v^2 + (I/M)(v/R)^2
For all the hoop-to-disk shapes, the left side is constant.  As you add
mass, I is increasing more slowly than M.  Therefore, (I/M) is decreasing,
and v must therefore increase ==> higher acceleration.

Intuitively speaking, although adding mass doesn't increase the
frictionless gravitational acceleration, it *does* enhance gravity's
ability to overcome impeding factors.  If two different masses were sliding
down an inclined plane with the same frictional force, the heavier one
would win.  (Note that this is *not* the same as with the same friction
*coefficient*.)

In this case, you are increasing mass more rapidly that you are increasing
the impeding factor I.  Therefore, gravity is more and more able to
overcome that impeding factor, and acceleration approaches the frictionless
value.

--
--James McLean
jmclean@chem.ucsd.edu
post doc
UC San Diego, Chemistry