Answer:
Approximately
.
(Assuming that
and that the thickness of the loop is negligible.)
Step-by-step explanation:
Let
denote the mass of the hoop, and let
denote its radius.
Under the assumptions, the moment of inertia of this hoop would be:
.
Let
denote the linear velocity of the hoop at the bottom of the hill. The linear kinetic energy of the hoop would be:
.
Since the hoop is rolling without slipping, its angular velocity would be
. The rotational kinetic energy of the hoop would be:
.
The total kinetic energy of the hoop (linear and rotational) would be:
.
Assuming that total mechanical energy is conserved. Change in the Kinetic energy that the loop has gained would be the opposite of the change in the gravitational potential energy (GPE):
,
Where:
by assumption, and
is the change in the height of the hoop.
By the conservation of energy:
.
.
Solve for
:
.
.
In other words, the velocity of the loop would be approximately
at the bottom of the hill.