Q. A wheel of radius 1m rolls forward half a revolution on a horizontal ground. The magnitude to the displacement of the point of the wheel initially in contact with the ground is
A) $2\pi$
B) $\sqrt{2\pi}$
C) $\sqrt{\pi^{2} + 4}$
D) $\pi$
Explanation
(c) When wheel rolls half a revolution, the point (P ) of the wheel which is in contact with the ground initially, moves at the top of the wheel as shown.
[image:4_image94.emf]
Horizontal displacement of point P,
$y = 2r$
$Net\ displacement = \sqrt{x^{2} + y^{2}}$
$= \sqrt{(\pi r)^{2} + (2r)^{2}}$
$= r\sqrt{\pi^{2} + 4}$
$= \sqrt{\pi^{2} + 4}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.