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A ring of radius R is rolling over rough horizontal surface with velocity v_0. Two points located at A and B on rim of ring. The angular velocity of point A w.r.t. point B will be


 

Option: 1

\frac{v_0}{2R}


Option: 2

\frac{v_0}{R}


Option: 3

\frac{v_0}{\sqrt{2}R}

 


Option: 4

\frac{\sqrt{2}v_0}{R}


Answers (1)

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Answer (2)

\\V_{AB}=2V_0\\ \omega _{AB}=\frac{V_{AB}}{2R}\\

            \\=\frac{2V_{0}}{2R}\\=\frac{V_{0}}{R}
 

 

 

Posted by

jitender.kumar

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