Irodov 1.38: Equal Twins: Slowing Down and Turning – Amit & Deepali Sinha Foundation

Problem Irodov 1.38

A point travels along a circle of radius $R$ , decelerating so that at every instant its tangential and normal accelerations have equal magnitude. At $t=0$ its speed is $v_0$ . Find: (a) the speed $v$ as a function of time $t$ , and separately as a function of the distance covered $s$ ; (b) the total acceleration $w$ as a function of the speed $v$ , and separately as a function of $s$ .

The setup: the point A on a circle of radius R about O, with v tangent to the path, the tangential acceleration w_τ against v and the centripetal w_n toward O, the two arrows equal in length. — The setup: the point $A$ on a circle of radius $R$ about $O$ , with $\vect v$ tangent to the path, the tangential acceleration $w_\tau$ against $\vect v$ and the centripetal $w_n$ toward $O$ , the two arrows equal in length.

Given

$R$
$|w_\tau| = w_n$ at every instant
$v(0) = v_0$

Find

(a)

v(t)

and

v(s)

, (b)

w(v)

and

w(s)

Solution

Step 1Speed as a function of time

Since the motion is a deceleration, $dv/dt<0$ ; setting its magnitude equal to $v^2/R$ gives

\frac{dv}{dt} = -\frac{v^2}{R}.

Separating variables and integrating from $v_0$ at $t=0$ ,

\int_{v_0}^{v}\frac{dv'}{v'^2} = -\int_0^t\frac{dt'}{R}.

Evaluating both integrals and clearing the overall minus sign,

\frac1v-\frac1{v_0} = \frac{t}{R}.

Collecting the two terms on the left over a common denominator gives the reciprocal speed in closed form.

\begin{align*}\frac1v &= \frac1{v_0}+\frac{t}{R} \\ &= \frac{R+v_0t}{v_0R}\end{align*}

Inverting that,

v(t) = \dfrac{v_0}{1+v_0t/R}

(1)

Step 2Speed as a function of distance covered

Writing $dv/dt=v\,dv/ds$ turns the same equation into one relating $v$ to $s$ directly.

\begin{align*}v\frac{dv}{ds} &= -\frac{v^2}{R} \\ \frac{dv}{v} &= -\frac{ds}{R}\end{align*}

Integrating from $v_0$ at $s=0$ ,

\int_{v_0}^{v}\frac{dv'}{v'} = -\int_0^s\frac{ds'}{R}.

Evaluating both integrals,

\ln\frac{v}{v_0} = -\frac{s}{R}.

Exponentiating both sides,

v(s) = v_0e^{-s/R}

(2)

Step 3Total acceleration

What this step assembles: the same instant with the vector sum w of w_τ and w_n drawn in, the total acceleration, at angle α from v. — What this step assembles: the same instant with the vector sum $\vect w$ of $w_\tau$ and $w_n$ drawn in, the total acceleration, at angle $\alpha$ from $\vect v$ .

The normal acceleration is $w_n=v^2/R$ regardless of the tangential law, and the problem’s own condition makes $w_\tau$ equal to it in magnitude; since the two are perpendicular, Pythagoras combines them.

w = \sqrt{w_\tau^2+w_n^2} = \sqrt{2}\,w_n

w(v) = \sqrt2\,\dfrac{v^2}{R}

(3)

Substituting (2) into this same relation gives the total acceleration as a function of distance covered.

w(s) = \sqrt2\,\frac{v(s)^2}{R} = \sqrt2\,\frac{\bigl(v_0e^{-s/R}\bigr)^2}{R}

w(s) = \sqrt2\,\dfrac{v_0^2}{R}e^{-2s/R}

(4)

Check

Both $v(t)$ and $v(s)$ decrease monotonically from $v_0$ toward $0$ , consistent with a deceleration that never reverses; $w(s)$ inherits that same decay, since $w\propto v^2$ , so the total acceleration also falls toward zero rather than diverging, as a decelerating point’s acceleration should once its speed is vanishing.

Remark

Because $w_n=v^2/R$ holds for any tangential law on a circle, the equal-magnitude condition here is really a statement about $w_\tau$ alone, $w_\tau=-v^2/R$ , the same differential equation that governs speed decay under quadratic (Newton) drag. The radius $R$ is not incidental, it sets the decay’s own scale from the start; what the circle’s geometry never demands is that the angular position be tracked at all, since $v(t)$ , $v(s)$ and $w$ all close without ever naming where on the circle the point currently sits.