Therefore kinetic energy is k ½ mv 2 cm ½ i cm w 2

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Therefore, kinetic energy is: K = ½ Mv 2 cm + ½ I cm w 2 Where M, vcm, Icm and w are the mass, speed of the center of the mass, moment of inertia and angular speed of the rigid body and w = vcm/R if the rigid bofy rolls along a plane without slipping. In this case, no work is being done by kinetic friction if the rigid body rolls along a plane without slipping and the effect of rolling can be ignored provided that the body and the surface on which it rolls are perfectly rigid. As a result, the only force that works on the rigid body is gravity. If the height of the incline is h, then we have: 0 +Mgh = ½ Mv 2 cm + ½ I cm w2 = ½ Mv 2 cm + ½ cMR 2 (v cm /R) 2 = ½ (1+c)Mv 2 cm and the speed of the rigid body at the bottom of the incline is: vcm= 2gh/1+c Which indicates that the speed doesn’t depend on mass or radius, but instead depends on c. The smaller the value of c, the faster the body is moving on the way down. Small c-bodies always beat larger c-bodies because they have less kinetic energy tied up in rotation about its axis and have more energy available for translation. This is evidenced in the resulting Table 1 attached.

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