Irodov 1.35: Curvature Radius from a Velocity Law – Amit & Deepali Sinha Foundation

Problem Irodov 1.35

A particle moves in the $xy$ plane with velocity components $v_x=a$ (constant) and $v_y=bx$ , where $a$ and $b$ are constants and $x$ is the particle’s own current $x$ coordinate. At the initial moment the particle is at the origin, $x=y=0$ . Find: (a) the trajectory $y(x)$ ; (b) the radius of curvature of the trajectory as a function of $x$ .

Given

$v_x = a$
$v_y=bx$
$x(0)=y(0)=0$

Find

(a)

y(x)

, (b)

R(x)

Solution

Step 1The trajectory

Since $v_x=a$ is constant, integrating from the origin gives $x$ linear in $t$ , which is then inverted for $t$ .

\begin{align*}x(t) &= \int_0^t a\,dt' = at \\ t &= \frac{x}{a}\end{align*}

With $v_y=bx$ , the chain rule turns $dy/dt$ into a separable equation in $x$ alone, since $dx/dt=a$ is already known.

\begin{align*}\frac{dy}{dt} &= \frac{dy}{dx}\frac{dx}{dt} = \frac{dy}{dx}\,a = bx \\ \frac{dy}{dx} &= \frac{bx}{a}\end{align*}

Integrating from the origin gives the trajectory directly.

\int_0^y dy' = \int_0^x \frac{bx'}{a}\,dx'

Evaluating both sides gives the equation of the path.

y(x) = \dfrac{b}{2a}x^2

(1)

The trajectory this step derives, drawn in the plane. Since v_x=a is constant, x grows linearly with time while v_y=bx grows with x, bending the path upward into a parabola; the velocity v at a marked point P is tangent to it. — The trajectory this step derives, drawn in the plane. Since $v_x=a$ is constant, $x$ grows linearly with time while $v_y=bx$ grows with $x$ , bending the path upward into a parabola; the velocity $\vect v$ at a marked point $P$ is tangent to it.

Step 2Acceleration, and its tangential part

Differentiating the given velocity components shows the acceleration is constant and purely vertical.

w_x = \frac{dv_x}{dt} = 0, \qquad w_y = \frac{dv_y}{dt} = b\frac{dx}{dt} = ab

The acceleration w, vertical and of constant magnitude ab, resolved into w_τ along the velocity and w_n across it, drawn at the point where bx=a, so the velocity makes φ=45° with w. — The acceleration $\vect w$ , vertical and of constant magnitude $ab$ , resolved into $w_\tau$ along the velocity and $w_n$ across it, drawn at the point where $bx=a$ , so the velocity makes $\varphi=45^\circ$ with $\vect w$ .

The tangential acceleration is the rate of change of the speed $v$ . Writing $v^2=v_x^2+v_y^2$ and differentiating in $t$ , using $dx/dt=a$ , isolates $dv/dt$ .

\begin{align*}v^2 &= a^2+b^2x^2 \\ 2v\frac{dv}{dt} &= 2v_y\frac{dv_y}{dt} \\ &= 2(bx)(ab)\end{align*}

Dividing through by $2v$ leaves the tangential acceleration.

w_\tau = \dfrac{dv}{dt} = \dfrac{ab^2x}{v}

(2)

Step 3The normal acceleration and the curvature radius

The normal acceleration is what Pythagoras leaves of $w=ab$ once the tangential part (2) is removed; the bracket then collapses because $v^2-b^2x^2=a^2$ .

\begin{align*}w_n^2 &= w^2-w_\tau^2 \\ &= a^2b^2-\frac{a^2b^4x^2}{v^2} \\ &= \frac{a^2b^2(v^2-b^2x^2)}{v^2} \\ &= \frac{a^2b^2\cdot a^2}{v^2}\end{align*}

Taking the positive square root,

w_n = \dfrac{a^2b}{v}

(3)

The normal acceleration and the local curvature radius of any smooth path are related by $w_n=v^2/R$ . Solving that for $R$ , putting (3) in for $w_n$ and then $v^2=a^2+b^2x^2$ for the speed gives the radius as a function of $x$ .

\begin{align*}R &= \frac{v^2}{w_n} = \frac{v^2}{a^2b/v} \\ &= \frac{v^3}{a^2b} \\ &= \frac{(a^2+b^2x^2)^{3/2}}{a^2b}\end{align*}

Pulling $a^2$ out of the bracket before the power is taken puts this in the book’s dimensionless form.

\begin{align*}(a^2+b^2x^2)^{3/2} &= \left[a^2\left(1+\frac{b^2x^2}{a^2}\right)\right]^{3/2} \\ &= a^3\left[1+\left(\frac{bx}{a}\right)^2\right]^{3/2}\end{align*}

Dividing that by $a^2b$ ,

R = \frac{a^3}{a^2b}\left[1+\left(\frac{bx}{a}\right)^2\right]^{3/2}

R(x) = \dfrac{a}{b}\left[1+\left(\dfrac{bx}{a}\right)^2\right]^{3/2}

(4)

Check

The boxed formula gives $R(0)=a/b$ , finite and positive, consistent with $v_x=a\neq0$ giving the trajectory a genuine, well-defined tangent direction to curve away from at the origin. As $x\to\infty$ , $R\propto x^3$ grows without bound: the parabola’s own curvature really does flatten out far from its vertex, as any parabola’s does.

Remark

$w=ab$ never changes, yet $w_\tau$ and $w_n$ trade off exactly as in a wind-bent balloon’s climb (1.34): the mechanism is the same one, a constant acceleration read against a turning velocity direction. There the constant acceleration $av_0$ came from the wind’s own coupling $v_x=ay$ against the steady climb $v_0$ ; here it is the constant $b$ coupling $v_y$ to $x$ against the steady $v_x=a$ that plays the equivalent role, and it is $y(x)$ that comes out a parabola rather than $x(y)$ .