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![[Post New]](/templates/default/images/icon_minipost_new.gif) 15 Mar 2007 21:49:21 IST
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A long horizontal rod has a bead which can slide along its length and initially placed at a distance L from one end A of the rod .The rod is set in angular motion about A with constant angular acceleration.If the coeff. of friction b/w rod and bead is mu, and gravity is neglected ,then the time after which the bead starts slipping is?
[ans: infinitesimal]
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16 Mar 2007 18:11:41 IST
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could u gimme the soln???
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19 Mar 2007 17:54:45 IST
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Any experts...please post the solution......
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19 Mar 2007 18:08:32 IST
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~~~~~~~~~~~~~~~~~~hey simple~~~~~~~~~~~~~~~~~~~~~~~~~~
~~~~~~~~~~~~~~~STAND ON THE rod ~~~~~~~~~~~~~~~~~~
~~~~~~~~~~~~~~~~~ apply pseudo force ~~~~~~~~~~~~
~~~~~~~~~~~~~~~ normal reaction becomes equal to this pesudo ~~~~~~~~~~~~~~~~~~~~~~``force~~~~~~~~~~~~~~~~~~
now calculate value of friction.
ANSWER U HAVE GIVEN IS 100 % WRONG ~~~~~~~~
AND KILL IT FINALLY ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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everything's fair in love , war and IIT JEE
   
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19 Mar 2007 18:10:03 IST
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``````` ANSWER IS SOMETHING FINTE~~~~~~~~~~~~~~~~~~~
~~~~~~~~~ NOT FOUND~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
~~~~~~~~~BUT 1000000000000000000000000000% SURE~~~~~~~~~
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everything's fair in love , war and IIT JEE
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19 Mar 2007 18:16:19 IST
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~~~~~~~~~~~~~~HEY U CLEAR~~~~~~~~~~~~~~~~~~~~~~~?????
~~~~~~~~~~~~~~~ OR ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
~~~~~~~~~~~ ALL CLEAR ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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everything's fair in love , war and IIT JEE
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19 Mar 2007 18:46:54 IST
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er
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EINSTIEN |
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19 Mar 2007 19:07:15 IST
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neeraj_agarwal_1990, there is something missing in your question the mass of the rod and the value of the coefficient of friction and the mass of the bead, the angular vel/acc .Such quantites have to be known or atleast there should exist some relation between them. Eg, if the mass of the bead too large and the angular velocity of the rod is too small the bead may never slide because of the very low centrifugal force, check the question again .
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19 Mar 2007 19:41:16 IST
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if gravity is neglected and if the angular motion is in the horizontal plane.... the normal rxn bw the rod and the bead will be very less.... so the the bead will start moving instantaneously as the max value of friction is very low and the the two pseudoforces mw^2r and (alpha)r act on it!!!
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--
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19 Mar 2007 20:19:43 IST
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subs , that is what i'm saying, if the mass of the bead too much then the normal reaction would be very large and the maximum value of friction would be very large and hence the bead would not move. the question here is not completel.  |
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19 Mar 2007 20:47:01 IST
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yet...I'm waiting for experts
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19 Mar 2007 21:17:59 IST
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if gravity is neglected then its weight would go to zero and hence normal reaction will be zero due to which frictional force will be zero so time for slip will be infinitelysmall
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21 Mar 2007 14:52:27 IST
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Well it will depend on angular acceleration (alpha) but not mass of the beads(m). Let me show you how it can be done.
Let angular acceleration be alpha.
So Normal reaction at any time =mL*alpha
Maximum Frictional force = mu*N=mu*mL*alpha.
Bead will start moving when centripetal force becomes equall to thisforce.
At any time t centripetal force = mL(alpha*t)^2
equating both we get t sqrt(mu/alpha)
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Krishna Gopal Singh
B.Tech Chemical Engg
IIT Delhi 2002
Currently doing PhD from IIT Delhi |
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