Cylinder of mass 6.0 kg rolls without slipping on a horizontal surface. At a certain instant its center of mass has a speed of 15.0 m/s. (a) Determine the translational kinetic energy of its center of mass. J (b) Determine the rotational kinetic energy about its center of mass. J (c) Determine its total energy.

a). 675J

b). 337.5J

c). 1012.5J

Explanation:

M = 6.0kg

V = 15.0m/s

a). Translational energy

E = ½ *mv²

E = ½ * 6 * 15²

E = 675J

b). Rotational kinetic energy K.E(rot) = Iw²

But moment of inertia of a cylinder (I) = ½Mr²

I = ½mr²

V = wr, r = v / w

K.E(rot) = ¼ mv²

K.E(rot) = ¼* 6 * 15²

K.E(rot) = 337.5J

Total energy of the system = K.E(rot) + Translational energy = 337.5 + 675

T.E = 1012.5J

answer; i believe the correct answer is (40,000 j) good luck