First compute the centre of mass of the object.
As the object is symmetrical we know the CM lies on this line of symmetry at distance
from O and so using polar coordinates we have that
where
is the mass per unit length of the semicircular ring. Consequently we have
Consider conservation of energy
If
is the angle by which the ring is initially displaced (about O) then subsequently we have
and the total K.E. is given by the addition of the rotational energy (about the C.M) with the translational energy of the C.M. Now, it is easy to show that the angular velocity about the CM is equal to that around O and that the translational velocity
is experienced by all parts of the semicircular ring.
So here's the tricky part: the translational energy of the CM is not simply due to a linear movement at speed
but movement on a curve given by the sum of its linear translation (due to rolling) and its rotational motion about O.
The rotational motion of the C.M (if O were fixed) would be given by:
and so the net component of the velocity of CM parallel to the ground,
, is
and so the net translational K.E. of the CM is given by
.
Note that I have defined
to be in the opposite sense to
.
Now using the parallel axis theorem we have that
so equating PE with KE we have
.
After tidying up and substituting for
we have
.
Deriving equation for
Differentiating wrt
and recalling
gives us:
so
which for small
and
gives us:
which is of the form
so then
.