Irodov 1.90: The Growing Centripetal Share – Amit & Deepali Sinha Foundation

Problem Irodov 1.90

A car starts from rest and moves with a constant tangential acceleration $w_\tau = 0.62$ m/s $^2$ along a horizontal circular track of radius $R = 40$ m. The coefficient of sliding friction between the wheels and the track surface is $k = 0.20$ . Find the distance $s$ the car can travel, measured along the track from the start, before the wheels begin to slide.

The car at some instant on the circular track of radius R, with its velocity v tangent to the track and the constant tangential acceleration w_τ the statement gives, along the direction of travel. — The car at some instant on the circular track of radius $R$ , with its velocity $\vect v$ tangent to the track and the constant tangential acceleration $w_\tau$ the statement gives, along the direction of travel.

Given

$w_\tau = 0.62$ m/s $^2$
$R = 40$ m
$k = 0.20$

Find

the distance

s

ridden without sliding, starting from rest

Solution

Step 1Speed built up after a distance

s

With $w_\tau$ constant and the car starting from rest, the speed after covering arc length $s$ follows from integrating $v\,dv = w_\tau\,ds$ along the track.

\begin{align*}\int_0^v v'\,dv' &= \int_0^s w_\tau\,ds' \\ \frac{v^2}{2} &= w_\tau s \\ v^2(s) &= 2w_\tau s\end{align*}

Step 2The no-slip ceiling on the total acceleration

The same instant with the acceleration completed: the constant tangential part w_τ and the normal part w_n, which grows with the speed already built up, add to a total w that friction alone has to supply. — The same instant with the acceleration completed: the constant tangential part $w_\tau$ and the normal part $w_n$ , which grows with the speed already built up, add to a total $\vect w$ that friction alone has to supply.

The car’s acceleration has a fixed tangential part $w_\tau$ and a normal (centripetal) part $w_n(s) = v^2(s)/R$ that grows with distance traveled; friction supplies both at once, so the wheels hold as long as the magnitude of the total acceleration stays within the friction’s ceiling $kg$ .

\begin{align*}w_n(s) &= \frac{v^2(s)}{R} = \frac{2w_\tau s}{R} \\ \sqrt{w_\tau^2 + w_n^2(s)} &\le kg\end{align*}

Step 3Solving for the critical distance

Set the inequality to equality at the critical distance $s$ and isolate the normal term.

\begin{align*}w_\tau^2 + \left(\frac{2w_\tau s}{R}\right)^2 &= (kg)^2 \\ \left(\frac{2w_\tau s}{R}\right)^2 &= (kg)^2 - w_\tau^2 \\ \frac{2w_\tau s}{R} &= \sqrt{(kg)^2-w_\tau^2}\end{align*}

Step 4Numerical evaluation

Isolate $s$ and substitute the given $w_\tau = 0.62$ m/s $^2$ , $R = 40$ m, $k = 0.20$ and $g = 9.8$ m/s $^2$ .

\begin{align*}kg &= (0.20)(9.8\ \text{m/s}^2) = 1.96\ \text{m/s}^2 \\ (kg)^2 - w_\tau^2 &= (1.96)^2-(0.62)^2 = 3.457\ \text{m}^2/\text{s}^4\end{align*}

s = \dfrac{R}{2w_\tau}\sqrt{(kg)^2-w_\tau^2} = 60\ \text{m}

(1)

Check

Dimensionally $[R/w_\tau]\cdot[\sqrt{(kg)^2-w_\tau^2}]$ is (m)/(m/s $^2$ ) times m/s $^2$ , which is a length, as it must be. The formula also requires $kg>w_\tau$ for a real, positive $s$ : a friction ceiling below the tangential acceleration alone could never be met even at $s=0$ , and the car would slip from the first instant, consistent with $w_n$ having to start at zero and only grow.

Remark

At $s=0$ the entire friction budget $kg$ is unused except for the fixed share $w_\tau$ takes; as $s$ grows the purely kinematic term $w_n=2w_\tau s/R$ eats steadily into what is left, and sliding is therefore inevitable eventually no matter how large $k$ is, only the distance to reach it changes with $k$ . A car built with more grip does not ride forever, it just delays the same crossing.