The kinetic energy of a body moving on a circle of radius R…

The kinetic energy of a body moving on a circle of radius $R$ depends on the distance travelled as $E_k = ks^2$ ($k$ constant). How does the force acting on the body depend on $s$?

Explanation of Method

Force from kinetic energy on a circle.

Step 1: Tangential force

By the work–energy theorem, the tangential force is the rate of change of kinetic energy with distance: $F_t = \frac{dE_k}{ds} = 2ks$.

Step 2: Normal force

\[ F_n = \frac{mv^2}{R} = \frac{2E_k}{R} = \frac{2ks^2}{R}. \]

Step 3: Total force

\[F = \sqrt{F_t^2 + F_n^2} = 2ks\sqrt{1 + \frac{s^2}{R^2}}\]

$F_t = 2ks$ $F_n = \tfrac{2ks^2}{R}$ $\vec F$ $s$ $F = 2ks\sqrt{1 + s^2/R^2}$

The tangential force changes the kinetic energy ($F_t = dE_k/ds$); the normal force keeps the body on the circle ($F_n = mv^2/R = 2E_k/R$).

Result

\[\boxed{\,F = \sqrt{F_t^2 + F_n^2} = 2ks\sqrt{1 + \frac{s^2}{R^2}}\,}\]

Remark

Check. The original result $\tfrac{2ks}{R}\sqrt{R^2 + s^2}$ is correct (the same expression). Note that the mass is not needed.