JEE Challenger
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PhyHardAdv 2022 • Paper 1

Maximum Height of Stone Detaching from Rolling Disk

At time t=0t = 0, a disk of radius 1 m1\text{ m} starts to roll without slipping on a horizontal plane with an angular acceleration of α=23 rad s−2\alpha = \frac{2}{3}\text{ 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=π st = \sqrt{\pi}\text{ s}, the stone detaches itself and flies off tangentially from the disk. The maximum height (in m\text{m}) reached by the stone measured from the plane is 12+x10\frac{1}{2} + \frac{x}{10}. The value of xx is ______ . [Take g=10 m s−2g = 10\text{ m s}^{-2}.]

Official Numerical Answer0.52

Topics & Concepts

System of Particles and Rotational MotionRolling MotionProjectile Motion

Step-by-Step Solution

To find the maximum height reached by the stone measured from the horizontal plane, we analyze the kinematics of pure rolling motion followed by projectile motion.

1. Kinematics of the Disk

Given:

  • Radius of the disk, R=1 mR = 1\text{ m}
  • Angular acceleration, α=23 rad s−2\alpha = \frac{2}{3}\text{ rad s}^{-2}
  • Time of detachment, t=π st = \sqrt{\pi}\text{ s}
  • Acceleration due to gravity, g=10 m s−2g = 10\text{ m s}^{-2}

The disk starts rolling without slipping from rest at t=0t = 0.

  • Angular displacement (θ\theta) of the disk at t=π st = \sqrt{\pi}\text{ s}: θ=12αt2=12(23)(π)2=π3 rad=60∘\theta = \frac{1}{2}\alpha t^2 = \frac{1}{2} \left(\frac{2}{3}\right) (\sqrt{\pi})^2 = \frac{\pi}{3}\text{ rad} = 60^\circ

  • Angular velocity (ω\omega) of the disk at t=π st = \sqrt{\pi}\text{ s}: ω=αt=23π rad s−1\omega = \alpha t = \frac{2}{3}\sqrt{\pi}\text{ rad s}^{-1}

  • Speed of the center of mass (vcmv_{\text{cm}}): vcm=ωR=23π m s−1v_{\text{cm}} = \omega R = \frac{2}{3}\sqrt{\pi}\text{ m s}^{-1}

2. Position and Velocity of the Stone at Detachment

At t=0t = 0, the stone was at the bottom-most contact point. When the disk rotates by an angle θ=π3\theta = \frac{\pi}{3} in the direction of motion:

  1. Height of the stone above the plane at detachment (ydetachy_{\text{detach}}): ydetach=R−Rcos⁡θ=1−1⋅cos⁡(π3)=1−12=12 my_{\text{detach}} = R - R\cos\theta = 1 - 1 \cdot \cos\left(\frac{\pi}{3}\right) = 1 - \frac{1}{2} = \frac{1}{2}\text{ m}

  2. Vertical component of the velocity of the stone (vyv_y): The velocity of the stone is the vector sum of the translational velocity of the center of mass and the rotational velocity relative to the center of mass. The vertical velocity component comes entirely from the rotational motion relative to the center of mass: vy=ωRsin⁡θ=(23π)(1)sin⁡(π3)=23π⋅32=π3 m s−1v_y = \omega R \sin\theta = \left(\frac{2}{3}\sqrt{\pi}\right) (1) \sin\left(\frac{\pi}{3}\right) = \frac{2}{3}\sqrt{\pi} \cdot \frac{\sqrt{3}}{2} = \sqrt{\frac{\pi}{3}}\text{ m s}^{-1}

3. Maximum Height of the Stone

After detaching, the stone moves under gravity. The additional height (Δh\Delta h) reached by the stone above its detachment point is given by: Δh=vy22g=(π3)22×10=π320=π60 m\Delta h = \frac{v_y^2}{2g} = \frac{\left(\sqrt{\frac{\pi}{3}}\right)^2}{2 \times 10} = \frac{\frac{\pi}{3}}{20} = \frac{\pi}{60}\text{ m}

Therefore, the maximum height HmaxH_{\text{max}} measured from the ground plane is: Hmax=ydetach+Δh=12+π60 mH_{\text{max}} = y_{\text{detach}} + \Delta h = \frac{1}{2} + \frac{\pi}{60}\text{ m}

4. Calculation of xx

We are given that: Hmax=12+x10H_{\text{max}} = \frac{1}{2} + \frac{x}{10}

Comparing the two expressions for HmaxH_{\text{max}}: x10=π60  ⟹  x=π6\frac{x}{10} = \frac{\pi}{60} \implies x = \frac{\pi}{6}

Substituting π≈3.1416\pi \approx 3.1416: x=3.14166≈0.5236x = \frac{3.1416}{6} \approx 0.5236

Rounding off to two decimal places, x=0.52x = 0.52 (or in exact terms, x=π6x = \frac{\pi}{6}).

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