At time $t = 0,$ a disk of radius 1 m starts to roll without slipping on a horizontal plane with an angular acceleration of $α = \frac{2}{3}$ 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 $t = √π s,$ 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 $\frac{1}{2} + x/10.$ The value of x is ______.

Step 1: Identify the parameters of the rolling disk.

For a disk of radius $R = 1\text{ m}$ rolling without slipping from rest, the linear acceleration of the center of mass is related to the angular acceleration $\alpha = \frac{2}{3}\text{ rad s}^{-2}$.

$$a = \alpha R = \frac{2}{3}\text{ m s}^{-2}$$

Step 2: Determine the angular position and velocity at detachment.

At the detachment time $t = \sqrt{\pi}\text{ s}$, the angular displacement $\theta$ and angular velocity $\omega$ are calculated using rotational kinematics.

$$\theta = \frac{1}{2}\alpha t^2 = \frac{1}{2} \cdot \frac{2}{3} \cdot \pi = \frac{\pi}{3}\text{ rad}$$ $$\omega = \alpha t = \frac{2}{3}\sqrt{\pi}\text{ rad s}^{-1}$$

Step 3: Find the height of the stone at the instant of detachment.

Since the stone starts at the contact point at $t = 0$, its height $y_0$ from the ground when the disk has rotated by $\theta = \frac{\pi}{3}$ is given by geometry.

$$y_0 = R(1 - \cos\theta)$$ $$y_0 = 1 \cdot \left(1 - \cos\frac{\pi}{3}\right) = \frac{1}{2}\text{ m}$$

Step 4: Determine the vertical velocity component at detachment.

The vertical component of the velocity $v_y$ in the ground frame is due to the rotation of the disk.

$$v_y = \omega R \sin\theta$$ $$v_y = \omega \cdot 1 \cdot \sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}\omega$$

Step 5: Calculate the maximum height reached by the stone.

After detaching, the stone acts as a projectile under gravity with $g = 10\text{ m s}^{-2}$, rising by an additional height $\Delta h$.

$$\Delta h = \frac{v_y^2}{2g}$$ $$\Delta h = \frac{\frac{3}{4}\omega^2}{20}$$ $$\Delta h = \frac{\frac{3}{4} \left(\frac{4\pi}{9}\right)}{20} = \frac{\pi}{60}\text{ m}$$

Step 6: Find the total maximum height and solve for $x$.

The total maximum height $H$ measured from the horizontal plane is the sum of the initial height and the vertical rise.

$$H = y_0 + \Delta h = \frac{1}{2} + \frac{\pi}{60}\text{ m}$$

Comparing this to the given expression $H = \frac{1}{2} + \frac{x}{10}$ yields the value of $x$.

$$\frac{1}{2} + \frac{\pi}{60} = \frac{1}{2} + \frac{x}{10}$$ $$x = \frac{\pi}{6} \approx 0.52$$

Answer: The value of $x$ is $\dfrac{\pi}{6}$ (or approximately $0.52$).