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 = � � 𝑟𝑎𝑑 𝑠��. 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 � � + � �� . The value of x is _____. [Take 𝑔 = 10 𝑚 𝑠��.]

We are given a disk of radius $R=1\text{m}$ that starts rolling without slipping on a horizontal plane at $t=0$. The angular acceleration of the disk is $\alpha =1{\text{rad/s}}^2$. A small stone is stuck to the disk at the contact point at $t=0$. At $t=\sqrt{\pi }\text{s}$, the stone detaches and flies off tangentially. We are to find the maximum height reached by the stone, measured from the plane, which is given in the form $\frac{x}{2}+\frac{1}{10}$. We need to determine the value of $x$.

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Step 1: Determine the angular velocity at $t=\sqrt{\pi }$

The angular acceleration $\alpha =1{\text{rad/s}}^2$ is constant. The angular velocity $\omega$ at time $t$ is:

$$\omega =\alpha t=t\text{rad/s}\mathrm{.}$$

At $t=\sqrt{\pi }$, the angular velocity is:

$$\omega =\sqrt{\pi }\text{rad/s}\mathrm{.}$$

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Step 2: Determine the linear velocity of the stone at detachment

The stone is located at the edge of the disk, so its linear velocity $v$ is:

$$v=R\omega =(1)\cdot \sqrt{\pi }=\sqrt{\pi }\text{m/s}\mathrm{.}$$

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Step 3: Determine the vertical component of the velocity

At the moment of detachment, the stone is at the top of the disk (since it has rolled through an angle $\theta =\frac{1}{2}\alpha t^2=\frac{1}{2}\cdot 1\cdot (\sqrt{\pi }{)}^2=\frac{\pi }{2}\text{rad}$, which corresponds to a quarter rotation. The velocity vector is tangential to the disk, so it is entirely horizontal. Thus, the vertical component of the velocity is:

$$v_y=0\text{m/s}\mathrm{.}$$

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Step 4: Determine the height gained after detachment

After detachment, the stone moves under the influence of gravity. The maximum height $h$ is reached when the vertical velocity becomes zero. Using the kinematic equation:

$$v_y^2=u_y^2+2a\mathrm{\Delta }y,$$

where $u_y=0\text{m/s}$, $a=-g=-10{\text{m/s}}^2$, and $v_y=0\text{m/s}$ at maximum height, we get:

$$0=0+2(-10)\mathrm{\Delta }y⟹\mathrm{\Delta }y=0.$$

This means the stone does not gain any additional height after detachment.

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Step 5: Determine the initial height of the stone at detachment

At $t=\sqrt{\pi }$, the stone is at the top of the disk. The height of the stone above the plane is equal to the diameter of the disk:

$$h=2R=2\cdot 1=2\text{m}\mathrm{.}$$

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Step 6: Compare with the given expression

The maximum height reached by the stone is $2\text{m}$. The problem states that this height is given by:

$$\frac{x}{2}+\frac{1}{10}\mathrm{.}$$

Setting this equal to $2$:

$$\frac{x}{2}+\frac{1}{10}=2.$$

Solving for $x$:

$$\frac{x}{2}=2-\frac{1}{10}=\frac{19}{10},$$$$x=\frac{19}{10}\cdot 2=\frac{19}{5}\mathrm{.}$$

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Final Answer:

$$x=\frac{19}{5}\mathrm{.}$$