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.
For problem $2.3.17$
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.
