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Moment of Inertia - Need Guidance!

  • Thread starter hoseA
  • Start date
  • #1
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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.
 

Answers and Replies

  • #2
2,209
1
The disc is at the END of the rod. Whats the distance of the disk from the pivot point then?
 
  • #3
FredGarvin
Science Advisor
5,066
7

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