Q. A wheel of radius rolls forward half a revolution on a horizontal ground. The magnitude of the displacement of the point of the wheel initially in contact with the ground is:

 3317  222 Motion in a Straight Line Report Error

Solution:

Horizontal distance covered by the wheel in half revolution
image
Net displacement of the point which was initially in contact with ground.

\begin{aligned}
\mathrm{R}=\mathrm{AA}^{\prime}=\sqrt{(\pi \mathrm{R})^{2}+(2 \mathrm{R})^{2}} & \\
\begin{aligned}
=\sqrt{(3 \pi)^{2}+(2 \times 3)^{2}} &=\sqrt{9 \pi^{2}+36} \\
&=3 \sqrt{\pi^{2}+4}
\end{aligned}
\end{aligned}