Answer to Question #121136 in Classical Mechanics for Lizwi

Question

if the round object is a ring, using I=mR^2, what is the fraction of rotational kinetic energy over the total kinetic energy?
1 : 1/4
2 : 1/3
3 : 1/2
4 : 2/3
5 : 1

Answer 1

The total kinetic energy of a disk moving with a speed $v$ is the sum of kinetic energy of linear (L) and rotational (R) motion:

$KE=KE_L+KE_R,\\ KE_L=\frac{mv^2}{2},\space KE_R=\frac{I\omega^2}{2}=\frac{[mR^2][v/R]^2}{2}=\frac{mv^2}{2}.\\ KE=\frac{mv^2}{2}+\frac{mv^2}{2}=mv^2.$

Therefore, the fraction of rotational kinetic energy over the total kinetic energy is 1/2.