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JEE Advanced 2022 · Paper 1, Physics question 5

At time t=0t = 0, a disk of radius 1 m starts to roll without slipping on a horizontal plane with an angular acceleration of α=23 rad s−2\alpha = \dfrac23\ \mathrm{rad\,s^{-2}}. A small stone is stuck to the disk. At t=0t = 0, it is at the contact point of the disk and the plane. Later, at time t=πt = \sqrt\pi 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 12+x10\dfrac12 + \dfrac{x}{10}. The value of xx is _____. [Take g=10 m s−2g = 10\ \mathrm{m\,s^{-2}}.]

Answer and worked solution

Official answer: 0.48

At t=πt=\sqrt\pi s.

  • Angular speed: ω=αt=23π\omega=\alpha t=\frac23\sqrt\pi.
  • Angle turned: θ=12αt2=π3\theta=\frac12\alpha t^2=\frac\pi3.

Position of the stone. It started at the contact point. After turning through θ\theta it is at height R(1−cos⁡θ)=12R(1-\cos\theta)=\frac12 m.

Velocity. For rolling, v⃗=ωR(1−cos⁡θ, sin⁡θ)\vec v=\omega R(1-\cos\theta,\ \sin\theta). The vertical component is
vy=ωRsin⁡π3=23π⋅32=π3.v_y=\omega R\sin\frac\pi3=\frac23\sqrt\pi\cdot\frac{\sqrt3}2=\sqrt{\frac\pi3}.

Maximum height. The stone rises a further vy22g=π/320=π60\frac{v_y^2}{2g}=\frac{\pi/3}{20}=\frac{\pi}{60}:
H=12+π60=12+x10  ⟹  x=π6≈0.52.H=\frac12+\frac{\pi}{60}=\frac12+\frac{x}{10}\implies x=\frac\pi6\approx0.52.

Topics:Rolling motionProjectile motion
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