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

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.

if you answer my question i will answer yours

the answer is f1 and d2 because if you put equal force on those two points the beam would balance trust me on this