JEE Advanced 2022 Paper 1 · Q05 · Rolling Motion

Question

At time t = 0, a disk of radius 1 m starts to roll without slipping on a horizontal plane with an angular acceleration of α = (2/3) rad s⁻². A small stone is stuck to the disk. At t = 0, it is at the contact point of the disk and the plane. Later, at time t = √π s, the stone detaches itself and flies off tangentially from the disk. The maximum height (in m) reached by the stone measured from the plane is (1/2) + x/10. The value of x is ___. [Take g = 10 m s⁻².]

SolveKar AI · Accepted Answer

Angular speed at t = √π: ω = αt = (2/3)√π rad/s. Angle turned: θ = (1/2)αt² = (1/2)(2/3)(π) = π/3 rad. At detachment, the stone is at angular position π/3 from the initial contact point (measured along rolling direction). Position of stone: x = R(θ − sin θ), y = R(1 − cos θ) ⇒ y = 1 − cos(π/3) = 1 − 1/2 = 1/2 m. Velocity of stone = velocity of contact-frame point + ωR tangentially: speed of the stone in lab frame, since centre moves with v_c = ωR, has magnitude 2ωR sin(θ/2) (for a point on a rolling rim). Direction makes angle (π/2 − θ/2) with horizontal. Vertical component of velocity at detachment v_y = ωR(1 − cos θ) ... carefully: for a point on rim of a rolling disk at angular position θ from contact, v_x = ωR(1 − cos θ), v_y = ωR sin θ. So v_y = (2/3)√π · 1 · sin(π/3) = (2/3)√π·(√3/2) = (√3/3)√π = √(π/3). Additional height gained after detachment = v_y²/(2g) = (π/3)/20 = π/60 m ≈ 0.05236 m. Total max height H = 1/2 + π/60 ≈ 0.5 + 0.0524 = 0.5524 m. Given H = 1/2 + x/10 ⇒ x/10 = π/60 ⇒ x = π/6 ≈ 0.524. Rounded: x = 0.52.

JEE Advanced 2022 Paper 1 · Q05 · Rolling Motion

Solution