Question

Two hoops, starting from rest, roll down identical inclined planes. The work done by nonconservative forces, such as air resistence, is zero (Wnc = 0J). Both have the same mass M, but, one hoop has the twice the radius of the other. The moment of inertia for each hoop is I = Mr sequared, where r is the radius. Which hoop, if either has the greater total kinetic energy (transational and rotational) at the bottom of the incline.
Hoop 1 Radius = R and Mass = M
Hoop 2 Radius = 1/2 R and Mass = M

Answer 1

Answer: both hoops have the same kinetic energy at the bottom of the incline.

Explanation:

If we assume no work done by non conservative forces (like friction) , the total mechanical energy must be conserved.

K1 + U1 = K2 + U2

If both hoops start from rest, and we choose the bottom of the incline to be the the zero reference level for gravitational potential energy, then

K1 = 0 and U2 = 0

⇒ ΔK = ΔU = m g. h

If both inclines have the same height, and both hoops have the same mass m, the  change in kinetic energy, must be the same for both hoops.