PrayogX › Physics › System of Particles and Rotational Motion
Two rollers on the rim of a fixed disk, rolling apart at omega and 2 omega: rolling without slipping, the centres' circle R + r and the time until they touch again
JEE Advanced 2026
Paper 1 · Question 1
Physics
Rolling without slipping on a fixed circular surface - angular speed of the centre and relative angular motion
Moderate
Two small disks touching on the rim of a clamped disk roll apart at omega and 2 omega. Magnifying lenses show each contact point at rest and v = omega r; the centres ride on the circle R + r, so they go round at omega/51 and 2 omega/51; a stopwatch times the long way round until the centres are again 2r apart, and all four options are audited.
Open the interactive simulationSub-topics covered
- Rolling without slipping: the contact point is instantaneously at rest, v = omega r
- Rolling on the outside of a fixed circle: the centre moves on radius R + r
- Coin-rotation paradox: a roller spins 1 + R/r times per lap
- Relative angular motion of two bodies moving round a circle in opposite directions
- Contact angle between two touching disks on a circle (small-angle approximation)
What it teaches
- The contact point of a disk rolling on a fixed surface is at rest, so its centre moves at v = omega r
- Rolling round the outside of a fixed disk, the centre is at distance R + r from O, so Omega = omega r/(R + r) = omega/51, not omega/50
- Moving in opposite directions, the angle between the centres changes at Omega1 + Omega2 = 3 omega/51
- Touching disks of finite size meet again when the far-side gap closes to the contact angle: the sweep is 2 pi - 2 dTheta, not 2 pi
- With sin dTheta = dTheta the chord 2r equals the arc (R + r) dTheta, so dTheta = 2/51
Simulation ID
- ADV-2026-P1-PHY-Q01 · revision 1
See this simulation in the library → · open the simulation file on its own