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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)

best_answer

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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