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![[Post New]](/templates/default/images/icon_minipost_new.gif) 20 Dec 2007 21:52:06 IST
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Q. A wheel rotates about an axle , due to friction it experiences a retardation which is proportional to its angular velocity (w).It makes 'n' revolutions before its angular velocity becomes half... then how many MORE revolutions before coming to rest.... (a) 2n (b) n (c) n/2 (d) n/3 Answer plzzzzzz, rates will b awarded.......
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SAARE JAHAAN SE ACHCHA;
GOIIT HAMARA!!!!!!!!!!!!!!!!!!!!!!!!! |
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20 Dec 2007 22:02:26 IST
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edit - wrong method :)
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Will nip in at times to solve problems :)
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20 Dec 2007 22:10:58 IST
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good one karthik.
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20 Dec 2007 22:18:41 IST
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Siddhart... whats the answer?
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20 Dec 2007 22:37:22 IST
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ans ( b ) n ; soln  = - k  from which the result follows .
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20 Dec 2007 22:40:33 IST
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@ karthik. it can't be n/3
We have variable force as we have variable retardation.
We need to integrate.
I think feynmann is right .
I'm still trying.
Sry to say Karthik your answer is wrong.
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<TABLE CELLSPACING="1" CELLPADDING="1" BORDER="0">
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<DIV ALIGN="right">Glitter Graphics</DIV></TD></TR></TABLE>
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20 Dec 2007 22:41:20 IST
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Karthik has made a mistake in calculating the work done by friction . See , the frictional force is not const . so you have to integrate to find out the total work done by friction .
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20 Dec 2007 22:43:00 IST
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yeah, I had a hunch about that, but I was initially wondering if it is a = -kw or @ = -kw.. though it doesn't make much of a difference to the final answer edit - we get dw = -kd  , which is valid only for values of w which change infinitesimaly, I doubt if it would work for a change such as w -> w/2, but again, I think this is the only way it can be done |
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20 Dec 2007 22:44:49 IST
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Anyway, very well done, take a salute
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25 Dec 2007 00:33:03 IST
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isnt it 2n?
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25 Dec 2007 16:26:14 IST
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the answe is 2n
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25 Dec 2007 19:47:05 IST
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Yea thnx
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