A hoop with mass m= 10 kg, radius r=0.6 m, and moment of inertia I=mr^2 is rolling without slipping with a linear velocity of 8.5 m/s. What is the total kinetic energy (in J).
K=1/2 m v^2
Kr=1/2 I ω^2
The Attempt at a Solution
The kinetic energy of the object is
K = 1/2 m v^2
K = 1/2 (10) (8.5) = 42.5 J
The rotational kinetic energy
Kr = 1/2 (10 * 0.6^2) (8.5/0.6)^2 = 361.25 J
Total kinetic energy
42.5 + 361.25 = 403.75 J
I am not sure about my solution since the answer I end up with is not included in the choices of the question.