Asked by Sarah
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.
Answers
Answered by
Elena
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.
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.
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