# Joint Entrance Examination/AIEEE 2009/Q087 ENERGY CONSERVATION

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| Original Question by 164.100.170.4 Here's the question... Q87. A thin uniform rod of length l and mass m is swinging freely about a horizontal axis passing through its end. Its maximum angular speed is ω. Its centre of mass rises to a maximum height of (1)[1/6]Iω/g (2)[1/2]I$^2$ω$^2$/g (3)[1/6]I$^2$ω$^2$/g (4)[1/3]I$^2$ω$^2$/g Posted by 164.100.170.4 04:22, 25 September 2009 (UTC) Post a reply! | What's Up? Latest discussions on the forum Mechanics-Help Joint Entrance Examination/2000 Screening/Q018 Spinning rod with a bead on it, time when it starts slipping Joint Entrance Examination/2000 Screening/Q027 Wind power turbine relation between wind speed and output power Joint Entrance Examination/2000 Screening/Q003 A tank with two holes, one square and another round Joint Entrance Examination/2000 Screening/Q005 Toppling vs Sliding of a cube Joint Entrance Examination/2000 Screening/Q015 Pendulum sliding on a plane Physics-Help User:Thewinster/Sandbox/Forum-Watch User:Thewinster/Sandbox Joint Entrance Examination/2000 Screening/Q021 Ionized gas subjected to electric and magnetic fields Joint Entrance Examination/2000 Screening/Q022 Radioactive materials and decay constant Joint Entrance Examination/2000 Screening/Q006 Bohr model of an atom applied to an atom with some hypothetical particle |
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Reply to the thread, posted by 164.100.170.4
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Additional notes
SOLUTION 3
Loss in kinetic energy = Gain in potential energy
[1/2]Iω$^2$ =mgh
⇒[1/2][ml$^2$/3]ω$^2$=mgh
⇒h=l$^2$w$^2$/6g
These comments, along with the notes on the left, were contributed by 164.100.170.4 04:26, 25 September 2009 (UTC) :