JEE Main 2023 · 13 April · morning · Single correct · +4 / −1
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 + 1)
- (b) 2R
- (c) R√(π ² + 4)
- (d) 2R√(1 + 4π ²)
Answer and solution
Answer (c) R√(π ² + 4)
Solution
Physics sir · 3 steps
Split the motion: the centre slides forward π R while A swings from the top to the bottom.
- Centre moves forward. In rolling without slipping, the centre moves one arc length for every angle turned. Half a rotation is an angle of π, so the centre moves π R horizontally.
- A reaches the bottom. Relative to the centre, half a turn takes A from the top of the disc to the bottom. So A drops 2R vertically. Its horizontal position relative to the centre is the same as before, so its only horizontal shift is the centre's π R.
|s⃗_A| = √((π R)² + (2R)²) = R√(π² + 4)
- So. The displacement of A is R√(π² + 4), which is option (c).
The idea to keep. For any point on a rolling body, add the centre's translation to the point's rotation about the centre. Track the start and end positions, not the curved path.
Answer from the NTA final answer key. Solution by PrepWiser.