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Ask iit jee aieee pet cbse icse state board community Community Discussion Question: Challenge - rotational mechanics
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karthik2007 (3349)

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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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karthik2007 (3349)

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answer will be posted tomorrow if no one is able to solve this

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srk (10)

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is the answer t=wI(sec-cosec)/kmgsecr
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Raghudevan (135)

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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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karthik2007 (3349)

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sorry... none are correct

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karthik2007 (3349)

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3 salutes for the correct answer

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Raghudevan (135)

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can i know where i have gone wrong

S.Raghudevan

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karthik2007 (3349)

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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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rohith291991 (511)

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t=2oR((1+k2)sinA)/5kg(k(1+cosA)+sinA) if both surfaces have fricton otherwise...
t=
2oR(1-kcotA)/5kg

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karthik2007 (3349)

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Nopes :(

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madman (239)

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I got R(2/5-cos@sin@)w/gk

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madman (239)

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is it correct

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rohith291991 (511)

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hey karthik2007 reply man am i correct??

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rooney (889)

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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 = 20r(1-kCot)/5kg

So, rohith291991's answer is right.


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karthik2007 (3349)