Answer
$v=-r\omega \sin\theta$
Work Step by Step
We can determine the required velocity as follows:
$x=r+rcos\theta$
Differentiating with respect to time, we obtain:
$\frac{dx}{dt}=-rsin\theta\frac{d\theta}{dt}~~~$[eq(1)]
We know that
$\omega=\frac{d\theta}{dt}$ and $v=\frac{dx}{dt}$
Thus eq(1) becomes
$v=-r\omega \sin \theta$
