A wheel of radius $R = 2 ext{ m}$ rolls strictly forward by exactly half a revolution without slip

Question

A wheel of radius $R = 2 	ext{ m}$ rolls strictly forward by exactly half a revolution without slipping on horizontal ground. What is the magnitude of the net displacement of the specific point on the wheel that was initially in contact with the ground?

Professor PC Thomas Classes · Answer

(c) : When the wheel completes exactly half a revolution, the center of mass moves horizontally by a distance equal to half the circumference, $x = \pi R$. Simultaneously, the point that was originally at the bottom (in contact with the ground) rotates to become the highest point on the wheel, so its vertical displacement is $y = 2R$. The total net displacement $s$ is the vector sum: $s = \sqrt{x^2 + y^2} = \sqrt{(\pi R)^2 + (2R)^2} = R\sqrt{\pi^2 + 4}$. Substituting $R = 2 	ext{ m}$, the displacement is $2\sqrt{\pi^2 + 4} 	ext{ m}$.