Kinetic Energy and Moment of Inertia

[RuthlessTB]

Homework Statement

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).

Homework Equations

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.

 

[milesyoung]

Have a look at this part again:

K = 1/2 m v^2
K = 1/2 (10) (8.5) = 42.5 J

Also, Kt = K + Kr = 1/2*m*v² + 1/2*m*r²*(v/r)² = ?