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Moment of Inertia and KE

  • Thread starter Jax24
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  • #1
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1. A 3.0 kg vertical rod has a length of 80 cm. A 4.0 kg rectangle attached to the top of the rod has negligible dimensions. The entire object rotates counterclockwise about the bottom of the rod. Determine the angular speed of the apparatus at the instant the rod is horizontal.


2. Using energy --> mgh = .5Iωω --> ω= √(2mgh/I)


3. I tried to use the parallel axis theorem to solve for I. I got 1.28. This would give me ω=√(2*7*9.8*.8/1.28) = 6.25 rad/s. I have used I= (1/12)(3)(.8*.8) + (3)(.4*.4) + (4)(.4*.4) plus several different variations.
 

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  • #2
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Where is the effective mass of the rod? How far does it fall?
 
  • #3
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I think I have I have it figured out - I needed to use the center of gravity for my height in the mgh part of the problem --> so h = .625m not .4m, which is what I was doing before. Thanks!
 
  • #4
193
28
Are you sure it's 0.625 m?
 

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