Yup, comparing the moments of inertia should give you the correct
ordering, with the given conditions that the masses and radii of the
round objects are the same.
The full solution, assuming the objects do not slip as they roll down the inclined plane, is:
acceleration = g*sin(theta)/(1+I/(M*R^2))
where g is the acceleration due to gravity, 9.81 m/s^2, theta is
the angle the incline makes with respect to the horizontal, I is the
moment of inertia of the rolling object, M is the mass of the rolling
object, and R is the radius of the rolling object.
For a uniform solid sphere, the moment of inertia I=(2/5)*M*R^2,
for a hollow tube, I=M*R^2, and for a uniform solid cylinder
The interesting feature is that the acceleration of the object as
it rolls down the incline depends on what kind of object it is and not
how big or heavy it is.
You can solve for that acceleration using Newton's second law, and
also that torque=I*angular acceleration (there's a friction force which
needs to be solved for).
Another way to look at exactly the same result is to think of what
happens to the energy that the thing picks up as it rolls down. Some of
it goes into the net average motion and some into the turning. The more
that goes into turning, the less that's left to speed the object down
the slope. That's why the bigger "I" gives lower acceleration. /Mike W.
(republished on 07/13/06)