$2.3.17.$ A cart of mass $m$ rolls with velocity $v$ along rails forming a horizontal circular path of radius $R$. A worker runs after it and begins to stop it by pulling a cable tied to the cart with a force $F$ at an angle of $\pi - \alpha$ to the direction of the cart's velocity. How many revolutions around the circle will the cart make before stopping? Neglect friction.
Solution
The cart will stop when all its initial kinetic energy is spent on the work done by the stopping force $F$. The projection of the force acting directly against the velocity of the cart is $F \cos\alpha$ (since the angle with the velocity vector is $\pi - \alpha$).
Equating the initial kinetic energy to the magnitude of the work done: $$\frac{mv^2}{2} = \int F \cos\alpha \, ds$$ Since the force is constant, we can take it out of the integral and integrate over the entire path distance $S = 2\pi n R$, where $n$ is the number of full revolutions: $$\frac{mv^2}{2} = F \cos\alpha \cdot 2\pi n R$$
From this, we express the number of revolutions $n$ around the circle before stopping: $$n = \frac{mv^2}{4\pi F R \cos\alpha}$$ where $n$ is the required number of full revolutions of the cart until it stops.
