Question

A hoop with mass m and radius R is rolling without slipping on a horizontal surface with its center moving at a speed v. The rotational inertia of the hoop about its center is mR2. What is its total kinetic energy

Answer 1

Answer:

K.E. (Total) =  1/2 m (v² + R² ω²)

Explanation:

K.E. (Total) =Translation K.E. + Rotational K.E.

K.E. (Total) = 1/2 mv² + 1/2 I ω²

K.E. (Total) =  1/2 mv² + 1/2 I ω²        ( I = m R² given)

K.E. (Total) =  1/2 mv² + 1/2 m R² ω²

K.E. (Total) =  1/2 m (v² + R² ω²)