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![[Post New]](/templates/default/images/icon_minipost_new.gif) 26 Mar 2008 22:27:08 IST
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ques 1
two forces F1 and F2 are applied on a spool of mass M and moment of inertia I about an axis passing through its centre of mass . find the ratio F1/F2 so that the force of friction is zero. given I = 2Mr2
i am getting ans as INFINITY... but it is given as 1/2 .... pls solve
ques 2
a disc is placed on grpund . friction co-efficient is mue . what is the minimum force required to move the disc if it is applied at the topmost point.
ques...3
can mechanical energy b conserved if friction does work on a body.... give reasons in support of ur ans
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 26 Mar 2008 22:29:21 IST
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hey people ans .....
rates assured
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all the best ... |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 26 Mar 2008 23:05:55 IST
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pls answer buddies....
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 26 Mar 2008 23:16:33 IST
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hey buddies, why don't u all solve it
is it too hard or too easy
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 26 Mar 2008 23:25:18 IST
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arpan ans for 1st 1 is coming 1/3... wait il post my method
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Nitwit Blubber Odment Tweak
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 26 Mar 2008 23:27:09 IST
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are both forces applied only @ dist R from centre?? i mean lik this?
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 26 Mar 2008 23:28:44 IST
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F1 - F2 = Ma -----1
F1R + F2R = I alpha = Ia/r
F1 + F2 = Ia / r^2 = 2Ma --------2
adding 1 and 2
F1 = 3Ma / 2
F2 = Ma / 2
my answer -- 3:1
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Impossible To be Impossible is Impossible |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 26 Mar 2008 23:29:22 IST
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not given any solution dude
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 26 Mar 2008 23:31:12 IST
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lets say both forces are fwd... if 1 is back ans will come in negative but v care only abt mod value... so F1+F2 =ma now for frictional force to be 0 there shud be pure rolling so (F1-F2)R=I@ where @ is alpha (F1-F2)R=2mR^2*@ also a=R@ so v have F1-F2=2ma=2F1+2F2 so F1=-3F2 is backward but ratio is 1/3
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 26 Mar 2008 23:31:51 IST
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it is f1/f2 so tht shud be 3
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 26 Mar 2008 23:32:27 IST
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anchit i repeat: V BLEED THE SAME BLOOD :D
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 26 Mar 2008 23:34:01 IST
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Impossible To be Impossible is Impossible |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 26 Mar 2008 23:37:43 IST
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2) F 0 is enuf to make it move
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 26 Mar 2008 23:39:46 IST
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as for the smallest possible force applied itself..friction will have to act backward to prevent the linear motin...producing a torque making the body move
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Nitwit Blubber Odment Tweak
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 26 Mar 2008 23:45:01 IST
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3) i dont think so...due to dissipation of energy as heat... but in case of pure rolling pt of contact @ rel rest so in this case v can do mech energy cons... in this case W done by fric=0 but as the question states work is done...so it cant be used
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