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 w(omega). its centre of mass rises to a max. height of?

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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 w(omega). its centre of mass rises to a max. height of?

Physics
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options are \[1) 1l ^{2}\omega ^{2}\div3g\] \[2) 1l \omega \div6g\] \[3) 1l ^{2}\omega ^{2}\div2g\] \[4) 1l ^{2}\omega ^{2}\div6g\]
it comes in my xam and this qn move above from my head sso i didnt solve this :P and asking u guys how to solve
eve i have solution too but need sm1 to explain may i post the solution?

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Other answers:

i think this fig and think we use consevation of E
yes
|dw:1329508279870:dw|
i get C.M point for source point of E
this qn is so bad i hate such type qn really :/
i don't find solve from options
sory say that
np :)
loss in KE=gain in PE 1/2 Iw^2=mgh 1/2x Ml^3/3 x w^2=mgh h=l^2w^2/6g
i dont get the lat 2 steps
this formula is right but i can't understand why don't get result \[0.5mv^2+0.5I \omega^2=mgl 0.5(1-\cos \theta)\]
ohhhhhhh thats right i did a big mistake
rod had a net rotational energy so the 2nd statement in left hand's of my equation is zero
and in right hand (mgl/2)cos theta=h and I =(1/3)ml^2 hence moment of inertia by this axis equal to\[I=I _{C.M}+Ma ^{2}\] where "a" is perpendicular distance between that axis and C.M axis I(C.M)for rod is (1/12)ml^2
do you get it?
I is moment of inertia around arbitrary axis.
you'r made mistake in typing last realtion,i think this:\[(1/2)I \omega^2=mgh\] in gain point rod has only PE, hence in that point "omega"=0

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