Question

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

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

Hint:

The displacement of a point on a rolling disc combines both the rotational and translational motion. This results in a cycloidal path, and the displacement after half a rotation can be calculated using the Pythagorean theorem.

Answer 1

The Correct Option is A

Step 1: Understand the motion of the point A. When the disc rolls without slipping, every point on the disc follows a cycloidal path. Point A starts at the topmost point of the disc and traces an arc during the motion. 
Step 2: Displacement after half a rotation. When the disc completes half of its rotation, the point A will have traveled a horizontal distance equal to the arc length of half the disc. This arc length is equal to the circumference of the disc divided by two, which is \( \pi R \). The displacement of point A is the distance it has moved in both horizontal and vertical directions, and these form the two sides of a right triangle. 
Step 3: Calculating the displacement. The horizontal displacement is \( \pi R \), and the vertical displacement is also \( R \). Using the Pythagorean theorem, the displacement \( d \) is: \[ d = \sqrt{(\pi R)^2 + R^2} = R \sqrt{\pi^2 + 1} \] However, we also have an additional displacement because of the rolling motion, which contributes \( 2R \) to the total displacement. Hence, the total displacement is: \[ d = R \sqrt{\pi^2 + 4} \] Thus, the displacement of point A from its initial position is \( R \sqrt{\pi^2 + 4} \).