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.

i need the answer to this too but idk it ; -;