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$$^{-2}$$.]

The disk begins from rest and rolls without slipping. Hence the angle turned by the disk in time $$t$$ is obtained from the constant-angular-acceleration relation

$$\theta = \frac12 \alpha t^{2}$$

With $$\alpha = \frac23 \text{ rad s}^{-2}$$ and $$t = \sqrt{\pi}\ \text{s}$$,

$$\theta = \frac12 \left(\frac23\right)(\sqrt{\pi})^{2}= \frac{\pi}{3} \text{ rad}$$

For pure rolling, a point that was at the initial contact position follows the cycloid

$$x = R(\theta - \sin\theta), \qquad y = R(1-\cos\theta)$$

with $$R = 1\ \text{m}$$. At $$\theta = \pi/3$$,

$$y = 1-\cos\left(\frac{\pi}{3}\right)=1-\frac12 = 0.5\ \text{m}$$

Thus the stone is $$0.5\ \text{m}$$ above the plane when it detaches.

The instantaneous angular speed is

$$\omega = \alpha t = \frac23\sqrt{\pi}\ \text{rad s}^{-1}$$

Differentiate the cycloid to find the velocity components (ground frame):

$$v_x = (1-\cos\theta)\,\omega,\qquad v_y = (\sin\theta)\,\omega$$

At $$\theta = \pi/3$$, $$1-\cos\theta = 0.5$$ and $$\sin\theta = \frac{\sqrt3}{2}$$, so

$$v_y = \frac{\sqrt3}{2}\,\omega = \frac{\sqrt3}{2}\left(\frac23\sqrt{\pi}\right) = \frac{\sqrt{3\pi}}{3}\ \text{m s}^{-1}$$

The stone now behaves like a projectile. The extra height it can rise is

$$\Delta h = \frac{v_y^{2}}{2g} = \frac{\left(\frac{\sqrt{3\pi}}{3}\right)^{2}}{2\times10} = \frac{\pi/3}{20}= \frac{\pi}{60}\ \text{m}$$

Total maximum height measured from the plane:

$$H = 0.5 + \frac{\pi}{60}\ \text{m}$$

The question writes this as $$H = \frac12 + \frac{x}{10}$$, hence

$$\frac{x}{10} = \frac{\pi}{60}\;\Longrightarrow\; x = \frac{\pi}{6} \approx 0.5236 \; \text{(m)}$$

Rounded to two decimal places, $$x = 0.52$$.

Answer: 0.52