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![[Post New]](/templates/default/images/icon_minipost_new.gif) 3 Jan 2008 20:58:23 IST
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A solid sphere of mass M and radius R initially rotating with angular velocity o in anti clockwise direction is gently placed between smooth inclined plane having angle @ and ground as shown in figure. If coefficient of friction between the sphere and the ground is k, then find the time t after which rotation of the sphere ceases.
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Will nip in at times to solve problems :)
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 3 Jan 2008 21:01:26 IST
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answer will be posted tomorrow if no one is able to solve this
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Will nip in at times to solve problems :)
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 3 Jan 2008 21:27:24 IST
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is the answer t=wI(sec  -cosec  )/kmgsec  r
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 3 Jan 2008 21:30:45 IST
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sphere is already rotaing with some  so the sphere will have a tendency to move in the rightward direcrtion so friction will act leftwards. f = k(mg) a torque is produced by friction, it causes an angular retadation  =  /I = 5kg/2R so, 0 =  - (5kg/2R)t so, t =  (2R)/5Kg the other smooth surface that the sphere is in contact with will only disable the sphere from moving leftwards. i dont know if this is correct., waiting for conformation from Karthik2007:)
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S.Raghudevan
Everything that's happening as it should be happening, because of the simple fact that it's supposed to be happening just as it is happening. |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 3 Jan 2008 21:56:03 IST
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sorry... none are correct
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 3 Jan 2008 22:10:27 IST
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3 salutes for the correct answer
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Will nip in at times to solve problems :)
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 3 Jan 2008 22:11:58 IST
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can i know where i have gone wrong
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S.Raghudevan
Everything that's happening as it should be happening, because of the simple fact that it's supposed to be happening just as it is happening. |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 3 Jan 2008 22:27:16 IST
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You are wrong in the statement where you've found the force of friction... you have got the Normal reactions wrong
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 3 Jan 2008 23:40:20 IST
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t=2 oR((1+k2)sinA)/5kg(k(1+cosA)+sinA) if both surfaces have fricton otherwise... t=2 oR(1-kcotA)/5kg
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Be Strong Be Different. Just Be
    
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 3 Jan 2008 23:43:45 IST
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Nopes :(
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 4 Jan 2008 00:06:22 IST
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I got R(2/5-cos@sin@)w/gk
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science-
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 4 Jan 2008 00:08:11 IST
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is it correct
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science-
the most fundamental
the most eternal
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 4 Jan 2008 00:31:14 IST
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hey karthik2007 reply man am i correct??
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Be Strong Be Different. Just Be
    
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 4 Jan 2008 00:49:20 IST
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Let N be the normal force of upper surface with sphere and R be that of lower surface. So, kR is frictional force on the sphere by the ground towards the left.
Balance forces on both axis, kR=Nsin NCos + Mg=R
Replacing N from 1 eq to the other, we get R =Mg/(1-kCot ) So, friction = kR Torque = Ia = kRr 2Mr2a/5 = kMg/(1-kCot ) a=5kg/2r(1-kCot ) = angular deceleration =>0 = 0 - at t = 2 0r(1-kCot )/5kg
So, rohith291991's answer is right.
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 4 Jan 2008 02:32:16 IST
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