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30 Nov 2008 16:59:07 IST
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STATEMENT 1: in pure rolling motion net work done by friction is ZERO
STATEMENT 2: sum of translational work done by friction and rotaional work done by friction is zero.
answer soon whether A B C or D
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30 Nov 2008 17:17:03 IST
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any one reply yaar
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a big shot is a number of small shots which were kept hitting |
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30 Nov 2008 17:32:05 IST
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it is B..... pls rate.......... Mr. hottie.....
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ANTI-RAGGING HELPLINE NUMBER ---- 1800-180-5522 |
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30 Nov 2008 17:36:47 IST
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The answer is B.
That is St1 and St2 are correct and St1 is not a correct explanation of St2.
This Question was asked in the vidyamandir Test Series -3 , this year.
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30 Nov 2008 17:43:15 IST
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it is b....
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30 Nov 2008 19:03:09 IST
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no the ans given is A .think of it
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a big shot is a number of small shots which were kept hitting |
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30 Nov 2008 19:06:07 IST
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any one reply.explain why it is B?
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a big shot is a number of small shots which were kept hitting |
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30 Nov 2008 19:37:23 IST
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frinstion acts at lowest point which remain at rest in case of pure rolling and hence work done by it is zero
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Beat others otherwise they will beat u |
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3 Dec 2008 22:30:54 IST
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here ans. is (b)
at the lowest point , relative motion is zero!! so no friction acts!!
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4 Dec 2008 13:44:57 IST
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ya it should be a
consider a body rolling down an inclined plane
friction acts in upard direction at point of contct
consider only translational motion find work done by f (its negative , as it tries to stop the translational motion0
and work done in ROTATION (its + as friction tries to incr the rotational velocity of the body)
hence net work is 0
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30 Apr 2009 19:16:39 IST
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ans. is A..this was the doubt i had for a long time...actually friction does work for both translation motion...n for the rotational motion if u analyze the plane motion as combination of both....wen u add both the work u will get zero ans...as for one motion it will do + work n for other -ve...u can observe for any rolling body...as their rotational n translational motion r connected by v=wR..the both components of work r of same magnitude n of opposite direction..
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30 Apr 2009 19:28:51 IST
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in rolling motion work done by friction =0 ans-b there is no slipping of the body in rolling so force of friction acting at the contact point=0
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30 Apr 2009 19:33:16 IST
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how can the wok done by friction be 0 in case of inclined plane frictional force produces torque which gives rotational kinetic energy to the disc
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1 May 2009 23:45:43 IST
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CLEAR EXPLANATION : ans is ( B ) POINTS ARE : 1. net WORK DON BY FRCTION IS ZERO AS POINT OF CONTACT AT REST...KNOWN 2 ALL... 2. NOW ONLY TWO WORKS ARE POSSIBLE...ROTATIONAL AND TRANSLATIONAL...SO THEIR SUM WILL ALSO BE ZERO... 3. BUT THE REASON IS THAT POINT OF CONTACT IS AT REST...
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1 May 2009 23:46:19 IST
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RATE PLS...
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5 May 2009 16:10:38 IST
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POINT OF CONTACT AT REST MEANS F.D MEIN D=0;
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21 Feb 2011 14:41:52 IST
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STATEMENT 1: in pure rolling motion net work done by friction is ZERO
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21 Feb 2011 14:49:35 IST
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The answer is clearly a) as the net work done by friction is zero because the translational and rotational work done by friction cancel each other and the point of contact remains at rest only instantaneously..................so the clear answer is a)
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21 Feb 2011 16:59:16 IST
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friction acts tangentially so shouldn't it be F.S equal to zeropls let me know if its correct
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