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March 28, 2017

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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.

  • Physics- PLEASE HELP - ,

    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.

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