Question

A uniform circular disc of radius r rolls without slipping with a constant velocity on a stationary horizontal surface. Find the distance covered by a point P on its circumference in one full rotation.

• A. 3R
• B. 7R
• C. 8R
• D. 2R

Answer 1

The correct option is C 8R

Net velocity$\Rightarrow |v|=\sqrt{v^2+v^2+2v\times \cos (180-0)}$

$|v|=\sqrt{2v^2-2v^2\cos \theta }=\sqrt{v^2(1-\cos \theta )}|v|=2v\sin \frac{\theta }{2}$

Distance transversed by point$\Rightarrow s={\int }_0^t|v|dt$

Thus, $s{\int }_0^{2\pi }2v\sin \frac{\theta }{2}dt\because$ we know$\omega =\frac{d\theta }{dt}\therefore \frac{d\theta }{\omega }=dt$

Hence, we get

$s={\int }_0^{2\pi }2v\sin \frac{\theta }{2}\frac{d\theta }{\omega }s=\frac{2v}{\omega }{\int }_0^{2\pi }\sin \frac{\theta }{2}d\theta =\frac{4\pi }{\omega }[\cos \theta -\cos \pi ]=\frac{4\pi }{\omega }[1+1]s=\frac{4\pi }{\omega }\times 2=\frac{8v}{\omega }$

Now,$v=R\omega \therefore \frac{v}{\omega }=R\Rightarrow s=8R$