Irodov 1.41: Curvature Shrinks as Speed Runs Away – Amit & Deepali Sinha Foundation

Problem Irodov 1.41

A point moves along some plane curve. Its tangential acceleration is the constant $w_\tau = a$ and its normal acceleration grows with time as $w_n = bt^4$ , with $a$ and $b$ positive constants; the point starts from rest at $t=0$ . Express the trajectory’s radius of curvature $R$ and the magnitude of the total acceleration $w$ as functions of the distance $s$ already covered.

A general point P on the plane trajectory, with the dashed osculating circle of radius R centered along the inward normal at P. The total acceleration w sums the tangential w_τ=a and the normal w_n=bt^4. — A general point $P$ on the plane trajectory, with the dashed osculating circle of radius $R$ centered along the inward normal at $P$ . The total acceleration $\vect w$ sums the tangential $w_\tau=a$ and the normal $w_n=bt^4$ .

Given

$w_\tau=a$
$w_n=bt^4$
$v(0)=0,\ s(0)=0$

Find

R(s)

w(s)

Solution

Step 1Speed and distance from the constant tangential acceleration

The tangential acceleration $w_\tau=dv/dt$ is the constant $a$ , with $v(0)=0$ . One integration gives the speed; a second, using $v=ds/dt$ and $s(0)=0$ , gives the distance covered.

\begin{aligned} v(t) &= \int_0^t a\,d\tau = at \\ s(t) &= \int_0^t a\tau\,d\tau = \frac{at^2}{2} \end{aligned} \tag{1}

Step 2Eliminate time in favor of

s

Solving the distance relation of (1) for the time expresses $t$ through $s$ . Steps 3 and 4 both need this relation, so it is boxed here.

t = \sqrt{\dfrac{2s}{a}}

(2)

Substituting (2) into the speed $v=at$ of (1) puts the speed in terms of $s$ alone.

\begin{align*}v &= a\sqrt{\frac{2s}{a}} \\ &= \sqrt{a^2\cdot\frac{2s}{a}}\end{align*}

v = \sqrt{2as}

(3)

Step 3The curvature radius from the normal acceleration

Writing the curvature relation for $R$ , substituting the given $w_n=bt^4$ , then $v^2=2as$ from (3) and $t^2=2s/a$ from (2),

\begin{align*}R &= \frac{v^2}{w_n} \\ &= \frac{v^2}{bt^4} \\ &= \frac{2as}{b\left(2s/a\right)^2} \\ &= \frac{2as\cdot a^2}{4bs^2}\end{align*}

R = \frac{a^3}{2bs}

(4)

Step 4The total acceleration as a function of

s

The two parts are perpendicular, so $w$ is their root sum of squares. Squaring (2) and raising it to the fourth power puts $w_n=bt^4$ in terms of $s$ first.

\begin{align*}w_n &= b\left(\frac{2s}{a}\right)^2 = \frac{4bs^2}{a^2} \\ w &= \sqrt{w_\tau^2+w_n^2} = \sqrt{a^2+\left(\frac{4bs^2}{a^2}\right)^2} \\ &= a\sqrt{1+\left(\frac{4bs^2}{a^3}\right)^2}\end{align*}

w = a\sqrt{1+\left(\dfrac{4bs^2}{a^3}\right)^2}

(5)

Check

At $s=0$ (the instant of release) the boxed formula gives $w=a$ , matching $w_\tau=a$ directly since $w_n=0$ there; and $R\to\infty$ as $s\to0$ , meaning the trajectory starts out momentarily straight, consistent with zero normal acceleration at the start. Dimensionally $w_n=bt^4$ forces $b$ to carry units of $\text{m/s}^6$ , and with that, $a^3/(bs)$ works out to a plain length, matching what $R$ must be.

Remark

The curvature radius falls off as $1/s$ while the total acceleration grows like $s^2$ for large $s$ : the constant tangential push keeps building speed, and because $w_n=v^2/R$ must supply an ever-growing normal acceleration to match the quartic time dependence, the geometry is forced to curve more and more sharply as the point travels farther, rather than settling into any fixed curve.