QJEEAdv22Phc8a2: JEE Advanced 2022 Physics - Class 11 JEEAdv22Phc8a2 Solution: At time , a disk of radius starts to roll without slipping on a …

Question

Verified Solution • 2025–26 Session moderate

At time

t=0

, a disk of radius

1 \mathrm{~m}

starts to roll without slipping on a horizontal plane with an angular acceleration of

\alpha=\frac{2}{3} \mathrm{rad} \,\mathrm{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 \mathrm{~m} \mathrm{~s}^{-2}

.]

Accepted Answer

The angle rotated by disc in

t=\sqrt{\pi} \mathrm{s}

is

\begin{aligned}&amp; \theta=\omega_0 t+\frac{1}{2} \alpha t^2 \\\\&amp; \Rightarrow \theta=\frac{1}{2} \times \frac{2}{3}(\sqrt{\pi})^2 \\\\&amp;=\frac{\pi}{3} \mathrm{rad}\end{aligned}

and the angular velocity of disc is

\begin{aligned}\omega &amp; =\omega_0+\alpha t \\\\= &amp; \frac{2 \sqrt{\pi}}{3} \mathrm{rad} / \mathrm{s} \\\\\text { and } v_{\mathrm{cm}} &amp; =\omega R=\frac{2 \sqrt{\pi}}{3} \times 1 \\\\&amp; =\frac{2 \sqrt{\pi}}{3} \mathrm{~m} / \mathrm{s}\end{aligned}

So, at the moment it detaches the situation is

JEE Advanced 2022 Paper 1 Online Physics - Rotational Motion Question 21 English Explanation

v=\sqrt{(\omega R)^2+v_{\mathrm{cm}}^2+2(\omega R) v_{\mathrm{cm}} \cos 120^{\circ}}

=v_{\mathrm{cm}}=\frac{2 \sqrt{\pi}}{3} \mathrm{~m} / \mathrm{s}

and

\tan \theta=\frac{\omega R \sin 120^{\circ}}{v_{\mathrm{cm}}+\omega R \cos 120^{\circ}}

\Rightarrow \tan \theta=\sqrt{3}

\Rightarrow \theta=\frac{\pi}{3} \mathrm{rad}

So,

H_{\max }=\frac{u^2 \sin ^2 \theta}{2 g}

=\frac{\left(\frac{2 \sqrt{\pi}}{3}\right)^2 \times \sin ^2 60^{\circ}}{2 \times 10}

\begin{aligned}&amp; =\frac{4 \pi \times 3}{9 \times 2 \times 10 \times 4} \\\\&amp; =\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

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