At time $t=0$, a disk of radius 1 m starts to roll without slipping on a horizontal plane with an angular acceleration of $\alpha=\frac{2}{3} rad s^{-2}$. 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 $t=\sqrt{\pi} s$, 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 $\frac{1}{2}+\frac{x}{10}$. The value of $x$ is ____. [Take $g=10 m s^{-2}$.]

JEE Advanced · 2022 · 28 Aug 2022 · Paper 1 · Rotational Mechanics

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 = \frac{2}{3} r a d 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

\frac{1}{2} + \frac{x}{10}

. The value of

x

is ____. [Take

g = 10 m s^{- 2}

Answer: 0.52

t = π

Angular velocity: $ω = α t = \frac{2}{3} π$ rad/s
Linear velocity: $v = ω r = \frac{2}{3} π$ m/s
Angular displacement: $θ = \frac{1}{2} α t^{2} = \frac{1}{2} \times \frac{2}{3} \times π = \frac{π}{3} = 60°$

The stone is at angle 60° from the bottom when it detaches.

Vertical component of velocity:

v_{y} = v sin 60° = \frac{2}{3} π \times \frac{3}{2}

Height gained from detachment point:

h = \frac{v _{y}^{2}}{2 g} = \frac{3 v ^{2} /4}{2 g} = \frac{3 \times 4 π /9}{4 \times 20} = \frac{π}{60}

Maximum height from plane:

H = \frac{R}{2} + h = \frac{1}{2} + \frac{π}{60} = \frac{1}{2} + \frac{0.52}{10}

Therefore

x = 0.52

JEE Advanced · 2022 · 28 Aug 2022 · Paper 1 · Numerical answer