Two points A and B on a rigid body are moving | vedclass

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(B) For a rigid body,the relative velocity of any two points $A$ and $B$ is given by $\vec{v}_B - \vec{v}_A = \vec{\omega} 	imes \vec{r}_{AB}$.Let th

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$\frac{v}{R}$

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(B) For a rigid body,the relative velocity of any two points $A$ and $B$ is given by $\vec{v}_B - \vec{v}_A = \vec{\omega} 	imes \vec{r}_{AB}$.Let the center of rotation be at a distance $x$ from point $A$ along the radius. The velocity of point $A$ is $v_A = \omega x$ and the velocity of point $B$ is $v_B = \omega (x+R)$.Given $v_A = v$ and $v_B = v$,we have:$v = \omega x$  --- $(1)$$v = \omega (x+R)$ --- $(2)$However,looking at the directions,the velocities are perpendicular. The instantaneous center of rotation $(ICR)$ is located at the intersection of the perpendiculars to the velocity vectors at $A$ and $B$.Since $v_A$ is vertical and $v_B$ is horizontal,the $ICR$ is at a distance $R$ from $A$ horizontally and $R$ from $B$ vertically.The distance of the $ICR$ from point $A$ is $r_A = R$ and from point $B$ is $r_B = R$.Thus,$v = \omega R$,which gives $\omega = \frac{v}{R}$.

Two points $A$ and $B$ on a rigid body are moving as shown in the figure. The angular velocity of the body is:

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