A 10.0 kg cylinder rolls without slipping on a rough surface. At an instant when its center of gravity has a speed of 16.0 m/s, determine (a) the translational kinetic energy of its center of gravity.(b) the rotational kinetic energy about its center of gravity, and (c) its total kinetic energy.

Answer 1

KE(tr) =m•v²/2,
KE(rot) = I•ω²/2 =m•R²•v²/2•2• R²=m•v²/4.
KE= KE(tr)+ KE(rot)=
=m•v²/2 + m•v²/4= 3• m•v²/4.