A thin rod with mass 140 grams and 60 cm long rotates at an angular speed of 25 radians/s about an axle that is 20cm from one end of the rod (see Figure in Representations). What is its kinetic energy?

There is translational kinetic energy because the center of mass is moving, and there is rotational energy because there is rotation about the center of mass. From what we just showed, we have this $(r_{CM} = 0.1 m):$

$K_{tot} = K_{trans} + K_{rot} = \frac{1}{2}(Mr^{2}_{CM} + I_{CM})\omega^{2}$

$K_{tot} = \frac{1}{2}(Mr^{2}_{CM}\;+\;\frac{1}{12}ML^{2})\omega^{2}$

$K_{tot} = \frac{1}{2}M(r^{2}_{CM}\;+\;\frac{1}{12}L^{2})\omega^{2}$