Rolling mill rolls have radius R. Rotating, they retract the workpiece, if its thickness is small enough. The coefficient of friction between the rolls and the workpiece is \mu, and the gap between the rolls is d_0. Find the maximum thickness of the blank. The blank is not pushed.

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    <h3> Statement </h3>
    <p>
        $2.8.13.$ Rolling mill rolls have radius $R$. Rotating, they retract the workpiece, if its thickness is small enough. The coefficient of friction between the rolls and the workpiece is $\mu$, and the gap between the rolls is $d_0$. Find the maximum thickness of the blank. The blank is not pushed.

    <h3>Solution</h3>
    <p>

As the thickness of the workpiece increases, there will come a point where the mill rolls can no longer roll the workpiece.

It will happen until the vertical projection of friction force $\vec{F}_\text{fr}$ will exceed the vertical component of the support reaction force $\vec{N}$
$$F_\text{fr}\cos\alpha\geq N\sin\alpha$$
Considering the value of friction force $F_\text{fr} = \mu N$:
$$\mu = \tan\alpha\quad(1)$$
From the drawing
$$\frac{d-d_0}{2} = R(1-\cos\alpha) \Rightarrow \boxed{d = d_0 + 2R(1-\cos\alpha)}\quad(2)$$
From the expression $(1)$,
$$\cos\alpha = \frac{1}{\sqrt{1+\tan^2\alpha}}=\frac{1}{\sqrt{1+\mu^2}}\quad(3)$$
After substituting $(3)$ into $(2)$, we could obtain the maximum thickness of the blank
$$\boxed{d = d_0 + 2R\left(1-\frac{1}{\sqrt{1+\mu^2}}\right)}$$

    <h4>Answer</h4>
    <p>
        $$d_\text{max}=d_0+2R\left(1-1/\sqrt{1+\mu^2} \right)$$
    </p>

    <p style="text-align: right; font-style: italic; font-size: 14;">
      Lutfulloyev Shukurullo<br>
      Physics Department<br>
      National University of Uzbekistan<br>
    </p>



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