This is a problem from my introductory physics textbook:
A cylinder is released from rest from the top of an incline of inclination $\theta$ and length $l$. If the cylinder rolls without slipping, what will be its speed when it reaches the bottom.
This is of course easy to solve using the principle of conservation of mechanical energy. The change in potential energy, $mgl\sin \theta$, must be equal to the kinetic energy at the bottom, i.e., $$\begin{align} mgl\sin \theta &=\frac{1}{2}I\omega ^2 +\frac{1}{2}mv^2 \\ &= \frac{1}{4}mv^2+\frac{1}{2}mv^2 = \frac{3}{4}mv^2 \end{align}$$ Solving for v, $$v=\sqrt{4/3gl\sin \theta}$$
My question is, how do we solve this problem without using energy considerations, i.e., while taking a force-based approach?