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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.