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![[Post New]](/templates/default/images/icon_minipost_new.gif) 28 Mar 2008 15:43:13 IST
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A uniform rod of mass m is hinged at its upper end. It is released from horizontal position. When it reached the vertical position, what force does it exert on the hinge?
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28 Mar 2008 15:47:03 IST
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The hinge duz zero work...so using wrk energy thoeorem or conservation of energy take out the speed....then the force applied by the hinge-force due to gravity is the centripetal force...that is mass*speed of COM^2/R(COM)
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28 Mar 2008 15:54:40 IST
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by conservation of energy
1/2mv^2 = mgr
v^2 = 2gr
r is the length of the rod
now for the rod at vertical position
T-mg = mv^2/r
= m ( 2gr)/r+ mg
= 3mg
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28 Mar 2008 15:56:22 IST
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uve fgten gravity?
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28 Mar 2008 15:58:31 IST
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oh sorry
i'll correct it
a rate for u for correcting me
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29 Mar 2008 07:31:13 IST
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Will any body tell me the final answer?
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29 Mar 2008 07:33:48 IST
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is it 7mg?
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Will nip in at times to solve problems :)
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29 Mar 2008 07:43:18 IST
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I am just going mad!! What a different variety of answer I am getting. I don't know how come you people think that I am so silly that I will ask that easy question which would have a straight forward answer. I have tried this question many times before and I am not getting the right answer. So please post your method and the "correct answer" if you can find.
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29 Mar 2008 07:44:07 IST
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A word to Ganesha. Your answer is wrong again!!!!
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29 Mar 2008 07:46:28 IST
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You get the angular velocity at the bottom as w2 = 12g/L
Hence, from newton's laws, F = mg+mw2L/2
which gives F = 7mg (dunno about calculation mistakes).
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Will nip in at times to solve problems :)
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29 Mar 2008 07:48:31 IST
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Still wrong! Try again!
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29 Mar 2008 08:18:21 IST
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I missed out on tension. Include that and get your answer
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Will nip in at times to solve problems :)
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29 Mar 2008 08:20:52 IST
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Tell me the final answer!
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29 Mar 2008 09:29:31 IST
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Perhaps the answer is 5mg/2 ...
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29 Mar 2008 09:32:33 IST
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i'm going mad
is it 3/2mg
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