Moment of Inertia - Need Guidance

[hoseA]

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.

 

[whozum]

The disc is at the END of the rod. Whats the distance of the disk from the pivot point then?

 

[FredGarvin]

Parallel axis theorem to be precise.

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

You have two masses that you will need to consider.