Irodov 1.42: Constant Speed Turns Curvature into Acceleration

Problem Irodov 1.42

A particle travels along a fixed plane curve $y(x)$ , its speed $v$ held constant throughout the motion. At the point $x=0$ , find the particle’s acceleration and the trajectory’s radius of curvature, when the curve is

the parabola $y=ax^2$ , with $a$ a constant
the ellipse $(x/a)^2+(y/b)^2=1$ , with $a$ and $b$ constants

(a) The parabola. The dashed line marks the x=0 symmetry axis, with the true origin O at its foot; the marked point is the vertex, and v is the tangent velocity there, of constant magnitude. — (a) The parabola. The dashed line marks the $x=0$ symmetry axis, with the true origin $O$ at its foot; the marked point is the vertex, and $\vect v$ is the tangent velocity there, of constant magnitude.

(b) The ellipse's upper half. The dashed line marks the x=0 symmetry axis, with the true origin O (the ellipse's center) at its foot; the marked point is the top vertex, directly above O, and v is the tangent velocity there. — (b) The ellipse’s upper half. The dashed line marks the $x=0$ symmetry axis, with the true origin $O$ (the ellipse’s center) at its foot; the marked point is the top vertex, directly above $O$ , and $\vect v$ is the tangent velocity there.

Given

$|\vect v|=v$ , constant
(a) $y=ax^2$
(b) $(x/a)^2+(y/b)^2=1$

Find

w

and

R

x=0

, for each curve

Solution

Step 1What constant speed forces at a point of zero slope

Parametrizing the curve by $x(t)$ , the chain rule gives $\dot y = y'\dot x$ , so the constant squared speed reads

v^2 = \dot x^2\bigl(1+y'^2\bigr). \tag{1}

Differentiating this constant with respect to time,

0 = 2\dot x\ddot x\bigl(1+y'^2\bigr) + 2\dot x^3 y'y''. \tag{2}

At a point where $y'=0$ this collapses, since $\dot x\neq0$ , to

\ddot x = 0.

With $\ddot x=0$ and $y'=0$ , the general $y$ -acceleration $\ddot y=y''\dot x^2+y'\ddot x$ reduces to a single term, and the speed relation with $y'=0$ gives $\dot x^2=v^2$ .

\ddot y = y''\dot x^2 = y''v^2

Since $\ddot x=0$ too, this is the whole acceleration, giving $w$ directly below; and because the speed is constant there is no tangential part anywhere on the curve, so the standard relation $w=v^2/R$ fixes $R$ as well.

w = v^2\,|y''(0)|,\qquad R=\frac{1}{|y''(0)|}

(3)

The parabola panel with the acceleration this step deduces: x=0 and y'=0 leave w perpendicular to v, pointing up into the curve's concavity. — The parabola panel with the acceleration this step deduces: $\ddot x=0$ and $y'=0$ leave $\vect w$ perpendicular to $\vect v$ , pointing up into the curve’s concavity.

The ellipse panel with the same deduction at its top vertex: w is again perpendicular to v, here pointing down into the concavity. — The ellipse panel with the same deduction at its top vertex: $\vect w$ is again perpendicular to $\vect v$ , here pointing down into the concavity.

Step 2The parabola: curvature from a constant second derivative

For $y=ax^2$ , differentiating twice gives $y'=2ax$ (zero at $x=0$ , confirming the tangent is horizontal there) and a constant second derivative.

y''(x) = 2a \tag{4}

Substituting $y''(0)=2a$ into (3):

w = 2av^2

(5)

R = \dfrac{1}{2a}

(6)

Step 3The ellipse: curvature at the top vertex

Implicit differentiation of $(x/a)^2+(y/b)^2=1$ gives the slope, and a second implicit differentiation gives $y''$ ; evaluating both at $x=0$ , $y=b$ (where the slope is already zero) isolates $y''(0)$ .

\begin{align*}\frac{2x}{a^2}+\frac{2yy'}{b^2} &= 0 \\ \frac{2}{a^2}+\frac{2(y'^2+yy'')}{b^2} &= 0\end{align*}

At $x=0$ , $y=b$ , $y'=0$ , the first equation confirms the horizontal tangent and the second gives

\frac{2}{a^2}+\frac{2by''(0)}{b^2} = 0 \;\Longrightarrow\; y''(0) = -\frac{b}{a^2}.

Substituting $|y''(0)|=b/a^2$ into (3):

w = \dfrac{bv^2}{a^2}

(7)

R = \dfrac{a^2}{b}

(8)

Check

Setting $a=b$ in the ellipse answers recovers a circle of radius $a$ : $w=bv^2/a^2=av^2/a^2=v^2/a$ and $R=a^2/a=a$ , exactly the familiar centripetal result $w=v^2/R$ for uniform circular motion, as it must for a circular special case. Both results also carry the right units: $[v^2][y'']=(\text{m/s})^2(\text{m}^{-1})=\text{m/s}^2$ , an acceleration.

Remark

Both answers are instances of the same fact: for a curve traced at constant speed, the whole acceleration is centripetal, $w=v^2/R$ , with no tangential part anywhere, not just at $x=0$ . What changes between the two curves is only how tightly they bend at that particular point, $y''(0)$ , which is why the parabola (curvature $2a$ at its vertex $x=0$ , the one point this problem asks about) and the ellipse hand back such structurally different radii from such similarly shaped equations. The ellipse’s point $(0,b)$ is its flattest only when $a>b$ (a long, low ellipse); when instead $b>a$ the semi-major axis runs along $y$ , curvature is largest there, $R=a^2/b$ is the smallest radius on the curve, and the flattest point sits at $(\pm a,0)$ with radius $b^2/a$ instead.