Irodov 1.120: Differentiating Energy Yields Only Tangential Force

Problem Irodov 1.120

A particle moves along a circular path of radius $R$ . Its kinetic energy $T$ depends on the distance $s$ it has covered along the circle as $T = as^2$ , where $a$ is a given constant. Find the magnitude of the total force acting on the particle as a function of $s$ .

The particle at a point A on the circle, with the net force F resolved into its tangential part F_τ and its normal (centripetal) part F_n, drawn perpendicular to each other. — The particle at a point $A$ on the circle, with the net force $\vect F$ resolved into its tangential part $F_\tau$ and its normal (centripetal) part $F_n$ , drawn perpendicular to each other.

Given

$T = as^2$
$R$

Find

F(s)

Solution

Step 1Tangential force from the growth of kinetic energy

Only the tangential component of the force does work as the particle advances along the path, so the work-energy theorem over an element $ds$ reads $dT = F_\tau\,ds$ , which identifies $F_\tau$ as the derivative of the given $T(s)$ .

F_\tau = \frac{dT}{ds} = \frac{d}{ds}\bigl(as^2\bigr)

F_\tau = 2as

(1)

Step 2Normal force from the curvature

The centripetal force is $F_n = mv^2/R$ , and the given kinetic energy is $T=mv^2/2$ , so $mv^2$ is just $2T$ ; substituting removes the mass and the speed together in favor of the given $T(s)$ .

F_n = \frac{mv^2}{R} = \frac{2T}{R}

F_n = \dfrac{2as^2}{R}

(2)

Step 3The resultant magnitude

The tangential and normal directions are perpendicular at every point of the path, so the total force is the Pythagorean sum of (1) and (2).

\begin{align*}F &= \sqrt{F_\tau^2+F_n^2} \\ &= \sqrt{(2as)^2+\left(\frac{2as^2}{R}\right)^2}\end{align*}

F = 2as\sqrt{1+(s/R)^2}

(3)

Check

Dimensionally, $T=as^2$ forces $[a]=\mathrm{J/m^2}$ , so $[a][s]=\mathrm{J/m}=\mathrm N$ , and the boxed $F$ is exactly such a product times a dimensionless root, a force as required. As $R\to\infty$ the path straightens and $F_n\to0$ , leaving $F\to F_\tau=2as$ , the whole force doing nothing but speeding the particle up along a straight line, as it must once curvature stops demanding anything.

Remark

The two components come from independent physics and only combine at the end: $F_\tau$ is fixed by how $T$ was given to grow with $s$ , through the work-energy theorem, while $F_n$ is fixed by the path’s geometry once $T$ is known, through $mv^2/R$ . Nothing about circular motion enters $F_\tau$ , and nothing about the law $T=as^2$ enters $F_n$ beyond supplying $mv^2$ ; the Pythagorean sum is the only place the two facts meet.