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Hence $$x=R\cos \theta$$ $$\theta=\cos^{-1} (\frac{x}{R})$$ Now we can get, $$\frac{d\theta}{dt}=\frac{-1}{\sqrt{R^2-x^2}}\frac{dx}{dt}$$ Now using basic knowledge of circular motion, $$v_{\text{surface}}=\color{red}{-}R\frac{d\theta}{dt}=\frac{R}{\sqrt{R^2-x^2}}\frac{dx}{dt}$$ $$v_{\text{surface}}=\frac{R}{\sqrt{R^2-x^2}}v_{object}$$
> We have the $\color{red}{\text{minus}}$ because $\omega$ is anticlockwise, which would give us $v$ in opposite direction of what we want.