JEE Main · 2023 · 13 Apr 2023 · Shift 1 · Rotational Mechanics

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) $R (π^{2} + 4)$ ✓ Correct
(B) $R (π^{2} + 1)$
(C)2 R
(D) $2 R (1 + 4 π^{2})$

Answer: (A)

Solution

For a disc rolling without slipping, when it completes half a rotation:

The center of the disc moves horizontally by a distance equal to half the circumference: $π R$
Point A, initially at the top, moves to the bottom of the disc (vertical displacement = $2 R$ downward)

The net displacement of point A is the vector sum:

Horizontal displacement:

π R

Vertical displacement:

2 R

Magnitude:

(π R)^{2} + (2 R)^{2} = R π^{2} + 4

JEE Main · 2023 · 13 Apr 2023 · Shift 1 · Single correct option

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JEE Main · 2023 · 13 Apr 2023 · Shift 1 · Rotational Mechanics

Solution

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