A Disk of Radius 1 m Rolls... | JEE Advanced 2022

JEE Advanced 2022

Paper-1 2022

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 $\alpha=\frac{2}{3} \mathrm{rads}^{-2}$. A small stone is stuck to the disk. At $\mathrm{t}=0$, 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 $\frac{1}{2}+\frac{\mathrm{x}}{10}$. The value of x is $\_\_\_\_$ .
[Take $\mathrm{g}=10 \mathrm{~ms}^{-2}$.]

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Solution

The angle rotated by disc in $\mathrm{s} \mathrm{t}=\sqrt{\pi} \mathrm{s}$ is
$ \begin{aligned} & \theta=\omega_0 t+\frac{1}{2} \alpha t^2 \\ & \Rightarrow \theta=\frac{1}{2} \times \frac{2}{3}(\sqrt{\pi})^2 \\ & =\frac{\pi}{3} \mathrm{rad} \end{aligned} $
and the angular velocity of disc is
$ \begin{aligned} & \omega=\omega_0+\alpha \mathrm{t} \\ & =\frac{2 \sqrt{\pi}}{3} \mathrm{rad} / \mathrm{s} \end{aligned} $

$ \begin{aligned} & \mathrm{v}=\sqrt{(\omega \mathrm{R})^2+\mathrm{v}_{\mathrm{cm}}^2+2(\omega \mathrm{R}) \mathrm{v}_{\mathrm{cm}} \cos 120^{\circ}} \\ & =\mathrm{v}_{\mathrm{cm}}=\frac{2 \sqrt{\pi}}{3} \mathrm{~m} / \mathrm{s} \\ & \text { and } \tan \theta=\frac{\omega \mathrm{R} \sin 120^{\circ}}{\mathrm{v}_{\mathrm{cm}}+\omega \mathrm{R} \cos 120^{\circ}} \\ & \Rightarrow \tan \theta=\sqrt{3} \\ & \Rightarrow \theta=\frac{\pi}{3} \mathrm{rad} \\ & \text { So, } \mathrm{H}_{\max }=\frac{\mathrm{u}^2 \sin ^2 \theta}{2 \mathrm{~g}}=\frac{\left(\frac{2 \sqrt{\pi}}{3}\right)^2 \times \sin ^2 60^{\circ}}{2 \times 10} \\ & =\frac{4 \pi \times 3}{9 \times 2 \times 10 \times 4}=\frac{\pi}{60} \mathrm{~m} \end{aligned} $
So, height from ground will be
$ R\left(1-\cos 60^{\circ}\right)+\frac{\pi}{60}=\frac{1}{2}+\frac{x}{10} \Rightarrow x=\frac{\pi}{6}=0.52 $
$ \begin{aligned} & \mathrm{v}=\sqrt{(\omega \mathrm{R})^2+\mathrm{v}_{\mathrm{cm}}^2+2(\omega \mathrm{R}) \mathrm{v}_{\mathrm{cm}} \cos 120^{\circ}} \\ & =\mathrm{v}_{\mathrm{cm}}=\frac{2 \sqrt{\pi}}{3} \mathrm{~m} / \mathrm{s} \\ & \text { and } \tan \theta=\frac{\omega \mathrm{R} \sin 120^{\circ}}{\mathrm{v}_{\mathrm{cm}}+\omega \mathrm{R} \cos 120^{\circ}} \\ & \Rightarrow \tan \theta=\sqrt{3} \\ & \Rightarrow \theta=\frac{\pi}{3} \mathrm{rad} \\ & \text { So, } \mathrm{H}_{\max }=\frac{\mathrm{u}^2 \sin ^2 \theta}{2 \mathrm{~g}}=\frac{\left(\frac{2 \sqrt{\pi}}{3}\right)^2 \times \sin ^2 60^{\circ}}{2 \times 10} \\ & =\frac{4 \pi \times 3}{9 \times 2 \times 10 \times 4}=\frac{\pi}{60} \mathrm{~m} \end{aligned} $
So, height from ground will be
$ R\left(1-\cos 60^{\circ}\right)+\frac{\pi}{60}=\frac{1}{2}+\frac{x}{10} \Rightarrow x=\frac{\pi}{6}=0.52 $

Question Tags

JEE Advance

Physics

Easy

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