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Q.

A uniform circular disc of radius ‘R’ is rolling without slipping on a rough horizontal surface with a constant acceleration ‘a’. Then the radius of curvature of trajectory of point ‘A’ of the disc relative to the ground at the given instant as shown in the figure is found to be abR. Then a+ b =   

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answer is 4.

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Detailed Solution

Velocity of point ‘A’ VA=V2+ω2R2=v2 normal acceleration of point A, 
 

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 aA(n)=ω2Rcos450+αRcos450acos450,                  aA(n)=ω2R2=V22R                     
 Radius of curvature of trajectory of point ‘A’ relative to the ground is 
 r=(VA)2 aA(n)=(V2)2V22R=22R

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