A pendulum is made of a rod whose moment of inertia about its center of mass is 1/12ML^2 where its mass is 4.9 kg and and its length is 3.3 m, and a thin cylindrical disk, whose moment of inertia about its center of mass is 1/2MR^2, where its mass is 1.7 kg and its radius is 1.6 m. What is the moment of inertia of the pen- dulum about the pivot point? Answer in units of kg*m^2. I tried adding the moment of Inertias --> ([1/12 (4.9) (3.3)^2 ]+[1/2 (1.7)(1.6)^2] = 6.62275. This is obviously wrong. I have know clue how to approach this. I think there's a theorem (parallel theorem?) that will help but I don't know how to apply it. Help is much appreciated.