Question: If a solid cylinder rolls without slipping, what fraction of its total kinetic energy is translational?
Options:
Correct Answer: 2/3
Solution:
For a solid cylinder, the total kinetic energy is KE_total = KE_translational + KE_rotational = (1/2)mv^2 + (1/2)(1/2)mR^2(Ο^2). Since Ο = v/R, the translational part is 2/3 of the total.