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![[Post New]](/templates/default/images/icon_minipost_new.gif) 23 Apr 2008 13:06:38 IST
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A coin is placed on a turn table rotating with a angular velocity 'w' at a distance of 4cm from the edge . It takes time 't' to just fall . Now the angular velocity is doubled , so what should be the distance of the coin from the edge to take the same time to fall.
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23 Apr 2008 13:15:13 IST
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velocity doubled acceleration 4 time so distance 4 times so 16cm
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24 Apr 2008 13:05:58 IST
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the answer is 1 cm for sure
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24 Apr 2008 13:15:11 IST
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dude r u stpid its mw^2r nd r is constantly changing integrate this twice mw^2r*dr=dv*v ull find the time then chagne w and see
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24 Apr 2008 13:20:37 IST
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rad. acc = v^2/r
v = rw so it is = w^2.r
now it is doubled so
(2w)^2.r
= 4w^2.r
so radius has to be increased by factor of 4.
4x4 = 16 cm.
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25 Apr 2008 11:08:14 IST
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ur saying that distance is 16 cm from edge..
but the ans. given was 1cm
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25 Apr 2008 11:44:52 IST
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answer is 16 cm
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25 Apr 2008 13:56:58 IST
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the velocity is doubled the ans is 16 perfect
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1 May 2008 05:48:27 IST
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First the coin will fall after covering a displacement horizontal to its point . This movemement and the changing position of the coin v=wr since r is changing it has changing linear speed also . Actually You have bring Coriolis force in to action which is beyond JEE Syllabus . I would suggest do not take this problem seriously .
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Bhupesh.M |
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1 May 2008 12:24:19 IST
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SEE THE TANGENTIAL FORCE IS MV^2/R=
=MW^2R
NOW THE TANGENTIAL FORCE SHOULD BE SAME
MW^2R1=HERE W=2W
4MW^2R1=MW^2R
4R1=R
R1=R/4=4/4=1
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3 May 2008 00:58:01 IST
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IS MY ANSWER RIGHT
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4 May 2008 18:50:44 IST
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AM I RIGHT
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