Question

A disc is rolling without slipping on a surface. The radius of the disc is R. At t = 0, the top most point on the disc is A as shown in figure. When the disc completes half of its rotation, the displacement of point A from its initial position is

• A. \(2R\sqrt{1+4π^2}\)
• B. 2R
• C. \(R\sqrt{π^2+1}\)
• D. \(R\sqrt{π^2+4}\)

Hint:

For rolling motion without slipping, calculate displacement considering both horizontal and vertical components. The horizontal motion is determined by the arc length, and the vertical motion depends on the rolling geometry.

Answer 1

The Correct Option is D

To find the displacement of point A after the disc completes half of its rotation, we need to consider both the rotational and translational movements of the disc.

1. The disc is rolling without slipping, meaning the distance traveled by the center of the disc is equal to the arc length covered by the rolling.

   For half a rotation, the center of the disc moves a distance of half the circumference of the disc, calculated as: \(\pi R\).

2. Initially, the top point A is at a height of R above the horizontal line passing through the center. After half a rotation, it moves to the bottom position, which is R below the horizontal line through the center.

   Therefore, the vertical displacement of point A is 2R.

3. The horizontal displacement of point A is the same as the distance traveled by the center, \(\pi R\).

4. The overall displacement is given by the Pythagorean theorem:

   Displacement = \sqrt{(\pi R)^2 + (2R)^2} = R\sqrt{\pi^2 + 4}

5. The correct answer is R\sqrt{\pi^2 + 4}, which matches the given option.

This involves converting rotational movement into horizontal and vertical components to find the net displacement.