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![[Post New]](/templates/default/images/icon_minipost_new.gif) 29 Aug 2008 19:52:06 IST
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 30 Aug 2008 06:50:07 IST
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any replies
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 30 Aug 2008 11:37:41 IST
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wfinal = (4*L*w+24*u)/7L
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 30 Aug 2008 11:39:03 IST
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v final=u/2
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 30 Aug 2008 11:41:58 IST
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(7*u*L)/2(4*L*w+24u) from cm away from end b ie the length above the cm
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 30 Aug 2008 11:42:24 IST
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tell me whether i am correct
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 2 Sep 2008 17:17:38 IST
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I HAVE SOLVED IT AND ITS NOT THAT A TOUGH A SUM IST METHOD THE SIMPLEST ONE CONSERVING ANGULAR MOMENTUM .LET US CONSERVE IT ABT THE SURFACE F THE ROD mV1R=(m+m)V2R1 +IW HEREI=(ML^21/2+ML^2/4=ML^2/3) BUT SINCE V IS PARALEL TO R m+mV2R=0 SO MVIR=IW AND IF WE APPLY THE CONSERVATION OF LINEAR MOMENTUM MVV1=2MV2 WE GET V2 WE KNOW THAT THE VELCOITY OF ANY POINT OF AROTATIONG BODY CAN BE FOUND OUT FROM THE COM SO V=RW R=V/W PLS RATE ME IF U FNED ME USEFUL !!!!!!!!!!CHEERS!!!!!!!!!!
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From shadows a light shall spring
Renewed shall be blade that's broken
The crown less again shall be king.
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 3 Sep 2008 22:13:54 IST
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(4*L*w+24*u)/7L is not the correct answer paris.
The real answer is (2*L*w+6*u)/5L .
One mistake you coulda done is that you havnt taken the I of the system as I of plank + I of ball.
Second mistake is that you changed the point about which yiu took angular momentum conversation initially and after collision. Angular momentum once fixed about a point should not be altered even if it is conserved in all other points. Thats the only way someone will get your answer.
Peace
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 25 Sep 2008 21:35:41 IST
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Yes you are absolutly right it requires a good understanding of conservation of momentum and Mechanical energy . The solution can be found out by applying conservation of angular monentum about the COM and Linear momentum ( why about COM ? ) the explanation is beyond scope of Highschool physics but all that can be said COM is very special Point and you can conserve angular monetum even when its moving (Like in case of rolling).
and two points
First the while apply conserbation of angular monetum about any Point "O" You have to find MOI about that point "O"
second very important Point in problem like this is .. if the object sticks to the rod, It changes the COM and the MOI as well
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Bhupesh.M |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 28 Sep 2008 17:27:04 IST
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actually my answr is completely wrong i checked it we hv to calculate angular momentum abt the com of the systm that is at a distance of m/2m
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From shadows a light shall spring
Renewed shall be blade that's broken
The crown less again shall be king.
MY BLOG
http://my.opera.com/vu2rje/blog/ |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 28 Sep 2008 21:53:21 IST
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Yeah, after collision, the COM of the system changes. Otherwise, varun's concept is right. The reason why we conserve angular momentum about COM is because the only force that acts on the system here is the gravitational and the Normal force, and since both pass through COM, their net torque is zero.
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Will nip in at times to solve problems :)
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