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![[Post New]](/templates/default/images/icon_minipost_new.gif) 27 Mar 2008 20:34:01 IST
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A rod of length L is hinged from one end. It is brought to a horizontal position and released.Find the angular velocity of rod inthe vertical position.
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27 Mar 2008 21:54:37 IST
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conserving energy
MgL/2 = 1/2(ML^2/3)w^2
=> w = root(3g/L)
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"Imagination is more important than knowledge."
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27 Mar 2008 22:22:03 IST
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another method
angular accn. at some angle theta = 3g cos (theta)/2l
write angular accn. = w dw/dt
integrate
u get w = sqrt. (3g sin (theta)/ l)
put theta = 90
u get v = sqrt. (3g/l)
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all the best ... |
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27 Mar 2008 22:45:17 IST
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Perfect answer by both of you. Well done
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Krishna Gopal Singh
B.Tech Chemical Engg
IIT Delhi 2002
Currently doing PhD from IIT Delhi |
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27 Mar 2008 22:46:21 IST
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Perfect answer by both of you. Well done
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Krishna Gopal Singh
B.Tech Chemical Engg
IIT Delhi 2002
Currently doing PhD from IIT Delhi |
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27 Mar 2008 23:10:01 IST
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sir why shouldnt we add 1/2mv^2 in rhs
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29 Mar 2008 07:31:15 IST
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That is because he is conserving angular momentum about the hinged point my dear friend.
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Will nip in at times to solve problems :)
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