In the manufacture of a cam, a uniform solid cylinder of radius R is first machined. Then an off-center hole of radius R/2 is drilled, parallel to the axis of the cylinder, and centered at a point a distance R/2 from the center of the cylinder. Then cam, of mass M, is then slipped onto the circular shaft and welded into place. What is the kinetic energy of the cam when it is rotating with angular speed omega about the axis of the shaft.

Use the parallel axis theorem

http://hyperphysics.phy-astr.gsu.edu/hbase/parax.html

to compute the moment of inertia about the new axis. Call it I'. The parallel axis theorem tells you how to relate I' to the moment of inertia about the center of mass, which is
I = (1/2) M R^2
In your case, the parallex axis theorem says that
I' = I + M (R/2)^2 = (3/4)MR^2

Finally, the kinetic energy is
(1/2) I' w^2
where w is the angular velocity, "omega"

I see now, Thanks.

You're welcome! I'm glad I could help you understand. Do you have any other questions?