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(C) The angular velocity $\omega$ of a particle moving with velocity $v$ at a position vector $\vec{r}$ relative to a point $O$ is given by $v_{\perp}

Vedclass · Accepted Answer

$\frac{v b}{a^2+b^2}$

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(C) The angular velocity $\omega$ of a particle moving with velocity $v$ at a position vector $\vec{r}$ relative to a point $O$ is given by $v_{\perp} = r\omega$,where $v_{\perp}$ is the component of velocity perpendicular to the position vector $\vec{r}$.From the geometry,the position vector $\vec{r}$ has a magnitude $r = \sqrt{a^2+b^2}$.The angle $	heta$ between the position vector and the $x$-axis is such that $\sin 	heta = \frac{b}{r} = \frac{b}{\sqrt{a^2+b^2}}$.The component of velocity $v$ perpendicular to the position vector is $v_{\perp} = v \sin 	heta$.Substituting the values,we get $v \sin 	heta = r \omega$.$v \left( \frac{b}{\sqrt{a^2+b^2}} ight) = \sqrt{a^2+b^2} \omega$.Solving for $\omega$,we get $\omega = \frac{v b}{a^2+b^2}$.

$A$ particle is moving with constant speed $v$ in the $xy$ plane as shown in the figure. The magnitude of its angular velocity about point $O$ is .........

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