Irodov 1.89: Fading Friction Picks an Optimal Radius

Problem Irodov 1.89

A cyclist rides at constant speed around a circle of radius $r$ centered at the center $O$ of a horizontal circular plane of radius $R$ . The friction coefficient between the tires and the plane depends only on the distance from $O$ , as $k(r) = k_0(1 - r/R)$ for a constant $k_0$ . Find the radius $r$ of the circle on which the cyclist can maintain the largest possible constant speed, and find that maximum speed.

The cyclist A rides at constant speed on a circle of radius r about the center O of the plate of radius R, with the velocity v tangent to that circle. — The cyclist $A$ rides at constant speed on a circle of radius $r$ about the center $O$ of the plate of radius $R$ , with the velocity $\vect v$ tangent to that circle.

Given

$k(r) = k_0(1-r/R)$
$R$

Find

(a) the radius

r

of maximum-speed riding, (b) the maximum speed

v_{\max}

there

Solution

Step 1The speed limit at a given radius

The same scene with the one horizontal force acting on the bicycle: friction f, pointing along the inward radius, because on a circle ridden at constant speed the centripetal pull is all there is for it to supply. — The same scene with the one horizontal force acting on the bicycle: friction $\vect f$ , pointing along the inward radius, because on a circle ridden at constant speed the centripetal pull is all there is for it to supply.

At radius $r$ the only horizontal force on the bicycle is friction, which must supply the entire centripetal force for uniform circular motion; the fastest the cyclist can go there is the speed at which friction is working at its full available value $k(r)mg$ .

\frac{mv^2}{r} = k(r)mg \quad\Longrightarrow\quad v_{\max}^2(r) = k(r)\,g\,r.

Step 2The speed ceiling as a function of

r

Substituting the given $k(r) = k_0(1-r/R)$ turns the ceiling into an explicit function of $r$ alone, one this problem now asks to maximize.

v_{\max}^2(r) = k_0 g\left(r - \frac{r^2}{R}\right). \tag{1}

Step 3Maximizing over the riding radius

Differentiate $v_{\max}^2(r)$ with respect to $r$ and set the result to zero; since $v_{\max}^2\ge 0$ and vanishes at both $r=0$ and $r=R$ , the interior stationary point is the maximum.

\begin{align*}\frac{d}{dr}\left[k_0 g\left(r-\frac{r^2}{R}\right)\right] &= k_0 g\left(1-\frac{2r}{R}\right) \\ 0 &= 1-\frac{2r}{R}\end{align*}

r = \dfrac{R}{2}

(2)

Step 4The maximum speed itself

Substitute $r = R/2$ back into (1) and simplify.

\begin{align*}v_{\max}^2 &= k_0 g\left(\frac{R}{2}-\frac{(R/2)^2}{R}\right) \\ &= k_0 g\left(\frac{R}{2}-\frac{R}{4}\right) \\ &= \frac{k_0 g R}{4}\end{align*}

v_{\max} = \dfrac{1}{2}\sqrt{k_0 g R}

(3)

Check

Dimensionally $[k_0 g R] = (\text{m/s}^2)(\text{m}) = \text{m}^2/\text{s}^2$ , and its square root is a speed. At both ends of the plate the ceiling correctly collapses: $r\to 0$ gives $v_{\max}\to 0$ because there is no radius left to curve around, and $r\to R$ gives $v_{\max}\to 0$ because $k(R)=0$ , the friction itself has faded away there.

Remark

The interior maximum exists only because $k(r)$ fades to zero at the rim; a friction coefficient held constant at $k_0$ instead would make $v_{\max}^2(r) = k_0 g r$ increase all the way to the edge, pushing the optimal radius out to $r=R$ with no interior stationary point to find. Here the two competing effects, more room to curve in at larger $r$ against less friction to curve with, are what cross to produce the answer.