Irodov 1.39: A Constant Tangential Twist in Disguise

Problem Irodov 1.39

A point travels along a circular arc of radius $R$ . Its speed depends on the arc length already covered, $s$ , through $v = a\sqrt{s}$ , where $a$ is a positive constant. Find the angle $\alpha$ between the point’s total acceleration vector and its velocity vector, as a function of $s$ .

The setup: a point P on the circular arc of radius R centered at O, traveling counterclockwise with its velocity v tangent to the path. — The setup: a point $P$ on the circular arc of radius $R$ centered at $O$ , traveling counterclockwise with its velocity $\vect v$ tangent to the path.

Given

$v=a\sqrt{s}$
$R$

Find

\alpha(s)

Solution

Step 1Tangential acceleration from the speed law

The tangential acceleration is the rate of change of speed with time, rewritten with the chain rule as speed times the rate of change of speed with distance, which is the natural variable here since $v$ is given as a function of $s$ .

\begin{align*}w_\tau &= \frac{dv}{dt} = v\frac{dv}{ds} \\ &= a\sqrt{s}\cdot\frac{d}{ds}\bigl(a\sqrt{s}\bigr) \\ &= a\sqrt{s}\cdot\frac{a}{2\sqrt{s}}\end{align*}

w_\tau = \frac{a^2}{2}

(1)

constant, independent of $s$ .

Step 2Normal acceleration from the speed and the radius

\begin{align*}w_n &= \frac{v^2}{R} \\ &= \frac{(a\sqrt{s})^2}{R}\end{align*}

w_n = \frac{a^2 s}{R}

(2)

Step 3The angle between

\vect w

and

\vect v

What this step reads off: the total acceleration w at P, built from the tangential part w_τ along v and the normal part w_n toward O, with the angle α between w and v marked at their common tail. — What this step reads off: the total acceleration $\vect w$ at $P$ , built from the tangential part $w_\tau$ along $\vect v$ and the normal part $w_n$ toward $O$ , with the angle $\alpha$ between $\vect w$ and $\vect v$ marked at their common tail.

Since $\vect v$ is purely tangential, $w_\tau$ is exactly the component of $\vect w$ along $\vect v$ and $w_n$ is exactly the component perpendicular to it, so their ratio is the tangent of the angle between $\vect w$ and $\vect v$ .

\tan\alpha = \frac{w_n}{w_\tau} = \frac{a^2 s/R}{a^2/2}

\tan\alpha = \frac{2s}{R}

(3)

Check

As $s\to0^+$ the centripetal term $w_n=a^2s/R$ vanishes while $w_\tau=a^2/2$ stays fixed, so $\tan\alpha\to0$ : the acceleration becomes purely tangential just after the point sets off, which is the right limit even though $v=0$ exactly at $s=0$ makes “the direction of $\vect v$ ” itself undefined there. Both $\tan\alpha$ and $2s/R$ are dimensionless, as required.

Remark

Because $w_\tau$ came out constant, this speed law is secretly uniform acceleration along the arc: integrating $v=a\sqrt s$ gives $s=a^2t^2/4$ and hence $v=(a^2/2)t$ , linear in time with slope $w_\tau$ . The angle $\alpha$ simply tracks how far the point has traveled: it grows from $0$ at the start toward $90^\circ$ as $s$ increases, since the ever growing $w_n=a^2s/R$ eventually swamps the fixed $w_\tau$ .