A 13.7-kg solid cylinder rolls without slipping on a rough surface. At an instan

Question

A 13.7-kg solid cylinder rolls without slipping on a rough surface. At an instant when its center of gravity has a speed of 8.03 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. If you substitute omega = v/r, the r's from omega and from I will cancel. Translational kinetic energy. J( plusminus 2 J) Rotational kinetic energy. J( plusminus 2 J)Total kinetic energy. J( plusminus 2 J) Source: Serway and Faughn, College Physics, 6th edition, Problem 8.39

Explanation / Answer

a) Translational kinetic energy = mv2/2 = 13.7 * (8.03)2 / 2 = 442 J

b) Rotational kinetic energy = I2/2 = (mr2/2) * (v/r)2 / 2 = mv2/4 = 221 J

c) Total kinetic energy = 442 + 221 = 663 J

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