- #1
cambo86
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The apparatus above has an initial angular velocity of [itex]\omega_{1}[/itex] as the rods are released. I need to find the angular velocity [itex]\omega_{2}[/itex] of the apparatus at the bottom.
I've tried 3 methods.
First I tried a work-energy balance where I included a gravitational potential energy.
[itex]KE_{1} + PE = KE_{2}[/itex]
[itex]0.5 I_{1}\omega_{1}^{2} + mgh = 0.5 I_{2}\omega_{2}^{2}[/itex]
Then I figured that the potential energy is going to only contribute to an increase in velocity of the arms falling down so I dropped the PE term and got,
[itex]0.5 I_{1}\omega_{1}^{2} = 0.5 I_{2}\omega_{2}^{2}[/itex]
I'm just wondering which is correct?
Then I spoke to another student in the class and he said he used conservation of angular momentum. So, he got,
[itex]I_{1}\omega_{1} = I_{2}\omega_{2}[/itex]
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