Determine the distance travelled by any point$P$ on the periphery of a rolling body  of radius R in one revolution.

The speed of the point $P$ at any angular position $\alpha$ as shown in below Figure.

$$\begin{gathered} v_p=2v_{cm}\cos \frac{\theta }{2} \\ =2v_{cm}\cos \left(\frac{180°-\alpha }{2}\right) \\ =2v_{cm}\cos (90°-\alpha /2) \\ =2v_{cm}\sin \frac{\alpha }{2} \end{gathered}$$

The distance travelled by point $P$in small time$dt$

$$\begin{gathered} dl=v_{p}dt \\ =2v_{cm}\sin \frac{\alpha }{2}\left(dt\right) \end{gathered}$$

We have $\omega =\frac{d\alpha }{dt}\Rightarrow dt=\frac{da}{\omega }$

$\therefore dl=2v_{cm}\sin \frac{\alpha }{2}\frac{d\alpha }{\omega }$

The distance travelled in one complete revolution

$$\begin{gathered} l={\int }_0^{2\mathrm{\pi }}2v_{cm}\sin \frac{\alpha }{2}\frac{d\alpha }{\omega } \\ =\frac{2v_{cm}}{\omega }{\int }_0^{2\mathrm{\pi }}\sin \frac{\alpha }{2}d\alpha \\ =\frac{2\omega R}{\omega }{\left|\frac{-\cos \alpha /2}{1/2}\right|}_0^{2\mathrm{\pi }} \\ =4R[-\mathrm{cos\pi }-(-\cos 0\left)\right] \\ =8R \end{gathered}$$