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Ask iit jee aieee pet cbse icse state board community Discussion Response Post to: Rotation
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anigo (7)

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Olaaa!! Perrrfect answer. 1  [2 rates]

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total posts: 11    
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ITS PRETTY SIMPLE
 
 
first make the translation motion equations
 
mg=kN1 + N2
N2 = kN1
 
therefore
N1 = mg/k(sq)+1
and N2=kmg/k(sq) +1
 
now make the rotational motion equations
 
 alpha=   -2k(N1 + N2)/mr
 
using third eqn of motion
w(final)=4k(N1 + N2) *theta/mr
 
 
therefore no of revolutions= theta/2 pi = rw(sq)(k(sq) + 1)/8* pi* k* g(1+k)
 
 this reply: 2 points  (with Olaaa!! Perrrfect answer.   in 1 votes )   [?]
 
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