Irodov 1.91: The Crest Is Really the Worst Case – Amit & Deepali Sinha Foundation

Problem Irodov 1.91

A car travels at a constant speed along a horizontal road whose shape, seen from above, is the sine curve $y = a\sin(x/\alpha)$ , with $a$ and $\alpha$ fixed constants. The coefficient of friction between the tires and the road is $k$ . Find the greatest constant speed at which the car can travel along this road without skidding.

Given

road shape $y=a\sin(x/\alpha)$
friction coefficient $k$

Find

Find the greatest speed

v

at which the car can take the curve without skidding

Solution

Step 1Friction is the only horizontal force

The road is level, so the weight and the normal reaction are both vertical and balance each other, and neither of them has a horizontal component. Writing that vertical equation gives the normal reaction, and the dry friction law then fixes the largest sideways force the contact can supply.

\begin{aligned} N - mg &= 0, \\ f_{\max} = kN &= kmg. \end{aligned} \tag{1}

The speed is constant, so at a point where the road curves with radius of curvature $R$ the acceleration is purely centripetal, and friction is the only force available to supply it. Newton’s second law along the horizontal normal to the path there reads

\frac{mv^2}{R} = f.

The car holds the curve at that point only while the friction it demands stays inside the ceiling (1),

\frac{mv^2}{R} \le kmg,

and canceling the mass leaves the local bound on the speed.

v^2 \le k g R

(2)

Since this must hold everywhere along the road, the true bound on $v$ is set by the smallest radius of curvature the road reaches, $R_{\min}$ .

Step 2The curvature of the road

The radius of curvature of a plane curve is given by the standard result $R=(1+y'^2)^{3/2}/|y''|$ . Differentiating $y=a\sin(x/\alpha)$ twice gives the two derivatives it needs,

\begin{aligned} y'(x) &= \frac{a}{\alpha}\cos\frac{x}{\alpha}, \\ y''(x) &= -\frac{a}{\alpha^2}\sin\frac{x}{\alpha}, \end{aligned}

R(x)=\frac{\left(1+\dfrac{a^2}{\alpha^2}\cos^2\dfrac{x}{\alpha}\right)^{3/2}}{\dfrac{a}{\alpha^2}\left|\sin\dfrac{x}{\alpha}\right|}.

Step 3Locating the sharpest bend

Two shorthands turn the curvature into a function of a single variable. Let $w$ track how far along the bend the point sits, and let $b$ absorb the two constants:

w = \cos^2\frac{x}{\alpha}, \qquad 0\le w\le 1, \qquad b = \left(\frac{a}{\alpha}\right)^2.

Rewriting $y'^2$ and $y''^2$ in terms of $w$ and substituting into the curvature $\kappa=1/R$ then isolates it as one function of $w$ alone.

\begin{aligned} y'^2 &= bw, \\ y''^2 &= \left(\frac{a}{\alpha^2}\right)^2(1-w), \\ \kappa(x)^2 = \frac{y''^2}{(1+y'^2)^3} &= \left(\frac{a}{\alpha^2}\right)^2h(w), \qquad h(w)=\frac{1-w}{(1+bw)^3}. \end{aligned}

Differentiating $h$ by the quotient rule and simplifying,

\begin{aligned} h'(w) &= \frac{-(1+bw)^3-(1-w)\cdot 3b(1+bw)^2}{(1+bw)^6} \\ &= \frac{-(1+bw)-3b(1-w)}{(1+bw)^4} = -\,\frac{1+3b-2bw}{(1+bw)^4}. \end{aligned}

Step 4Minimum radius, then the speed limit

The same road with the crest P, where (x/α)=0, the gray osculating circle there, whose radius is the smallest the road reaches, and the friction f supplying the centripetal pull. — The same road with the crest $P$ , where $\cos(x/\alpha)=0$ , the gray osculating circle there, whose radius is the smallest the road reaches, and the friction $f$ supplying the centripetal pull.

At $w=0$ , $\sin(x/\alpha)=\pm1$ and $y'=0$ , so

R_{\min}=\frac{\alpha^2}{a}

(3)

Putting $R=R_{\min}$ into the bound of (2) and using (3),

v\le\sqrt{kgR_{\min}}=\sqrt{\frac{kg\alpha^2}{a}}.

v \le \alpha\sqrt{\dfrac{kg}{a}}

(4)

Check

Dimensionally $\alpha$ and $a$ are lengths and $k$ is dimensionless, so $\alpha\sqrt{kg/a}$ has units of length times the square root of (length/time $^2$ )/length, that is length/time, a speed, as it must. And as $a\to0$ the sine curve flattens toward a straight road, $R_{\min}=\alpha^2/a\to\infty$ , and the bound on $v$ correctly diverges: a straight road puts no ceiling on the safe speed.

Remark

That the sharpest bend sits exactly at the crest is not a generic fact about wavy curves, it is special to a single sinusoid: Step 3 showed $h'(w)<0$ for every $b>0$ , but a road built from more than one harmonic can easily bend hardest somewhere between a zero crossing and a crest, so the same ”just check the peak” shortcut would fail there and the full curvature would have to be searched.