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Step-by-Step Detailed Solution
Step 1: Understand the scenario of rolling without slipping
When a wheel of radius 1 m completes half a rotation without slipping, the point of contact P (initially touching the ground) traces a path from the bottom to the top. We must find the net displacement between these two positions of point P.
Step 2: Determine the horizontal distance covered
During half a rotation, the center of the wheel moves horizontally by half the circumference of the wheel. Since the radius $r = 1\,\text{m}$, the circumference is $2\pi r = 2\pi \times 1 = 2\pi\,\text{m}$. Hence, for half a rotation, the center (and consequently the horizontal projection of point P) moves:
$$
\frac{1}{2} \times 2\pi\,\text{m} = \pi\,\text{m}.
$$
Therefore, the horizontal component of the displacement of point P is $\pi\,\text{m}$.
Step 3: Determine the vertical distance covered
In half a rotation, point P moves from the bottom of the wheel (in contact with the ground) to the top of the wheel. The center of the wheel remains at the same height (+1 m from the ground), but the point P effectively travels from the bottommost point to the topmost point of the circle. This change in vertical position corresponds to the diameter of the wheel. Since $r = 1\,\text{m}$, the diameter is $2\,\text{m}$. Thus, the vertical component of the displacement is $2\,\text{m}$.
Step 4: Calculate the resultant displacement
We treat the horizontal and vertical displacement components as perpendicular segments of a right triangle. Hence, the magnitude of the net displacement $d$ is found using the Pythagorean theorem:
$$
d = \sqrt{(\text{horizontal distance})^2 + (\text{vertical distance})^2}
= \sqrt{\pi^2 + 2^2}
= \sqrt{\pi^2 + 4}.
$$
Thus, the displacement of point P after half a rotation is:
$$
\sqrt{\pi^2 + 4}\,\text{m}.
$$
Final Answer: $ \sqrt{\pi^2 + 4}\,\text{m} $
In half rotation, point P has moved horizontally $\pi\,\text{m}$ and vertically $2\,\text{m}$. Therefore, the resultant displacement is $ \sqrt{\pi^2 + 4} \,\text{m}$.
