A 50 mm flat aluminium plate is reduced in thickness to 25 mm in a single pass cold-rolling operation with the rolling diameter of 1250 mm. For this case, the minimum required coefficient of friction between the plate and the roll (rounded off to two decimal places) is _______.
Show Hint
Use the rolling bite condition \(\mu_{min} = \sqrt{\Delta h / R}\), with \(\Delta h\) the draft (thickness reduction) and \(R\) the roll radius.
Step 1: Picture the roll-bite triangle directly.
Draw the roll of radius $R$ meeting the plate at the entry point. The plate surface drops by half the draft, $\Delta h/2$, over the arc of contact, because the reduction is shared equally above and below the rolling centerline. This drop, the roll radius, and the bite angle $\alpha$ form a right triangle relation at the entry point: the horizontal leg is the chord along the roll surface and the vertical leg is the drop $\Delta h/2$, measured from the roll center.
Step 2: Recall the roll-bite result and check it makes physical sense.
Working through that triangle geometry with a small-angle approximation leads to the standard rolling-mechanics result: the rolls can just grip the plate, with no slipping, when
\[
\mu_{min}=\sqrt{\frac{\Delta h}{R}}
\]
where $\Delta h$ is the draft and $R$ is the roll radius. Before using it, check the limits: if $\Delta h \to R$ (an extremely aggressive, physically unrealistic single pass), $\mu_{min}\to1$, which sits right at the practical upper bound for dry metal-on-metal friction. If $\Delta h \to 0$ (almost no reduction), $\mu_{min}\to0$, which also makes sense since a very light pass needs almost no grip to pull the plate through. The formula behaves sensibly at both ends, so it is safe to use here.
Step 3: Get the draft and the roll radius from the data.
The plate thins from $50$ mm to $25$ mm, so the draft is
\[
\Delta h=50-25=25\text{ mm}
\]
The roll diameter is $1250$ mm, so the radius is half of that,
\[
R=\frac{1250}{2}=625\text{ mm}
\]
Both $\Delta h$ and $R$ are in millimeters, so their ratio is a pure number and no unit conversion is needed before taking the square root.
Step 4: Form the ratio.
\[
\frac{\Delta h}{R}=\frac{25}{625}=0.04
\]
Step 5: Take the square root.
\[
\mu_{min}=\sqrt{0.04}
\]
Since $0.2\times0.2=0.04$, the square root is exactly $0.2$.
Step 6: Conclude.
Rounded to two decimal places, $\mu_{min}=0.20$, which sits inside the accepted range of 0.19 to 0.21. A coefficient of friction of about $0.2$ is a realistic, achievable value for a dry aluminium-on-steel roll contact, which is a good practical sanity check on the answer.
\[
\boxed{\mu_{min}=0.20}
\]
