Q: Equations of motion of a cylinder on a horizontal plane How would I go about deriving the equations of motion for the motion of the centre of mass of the cylinder in this system:
The cylinder has mass $M$ and radius $R$ and the small mass $m$ is being pulled down by gravity. I know the kinetic energy of the cylinder consists out of a translational and a rotational part: $T=\frac{1}{2}Mv^2+\frac{1}{2}I\omega^2$, where $I=\frac{1}{2}MR^2$ is the moment of inertia and $\omega=\frac{v}{R}$. I don't know how to go from here, could someone explain it to me?

A: Writing Newton's second law for $m$:
$$mg - T = ma$$
Where $T$ is the tension in the string. Newton's second law for (the center of mass of) $M$ is:
$$T-f=Ma$$
Where $f$ is the (static) friction felt by the contact point of the cylinder with the ground (which is essential for rolling motion) .Note that the magnitude of the acceleration of $m$ and the center of mass acceleration of $M$ is the same because the length of the string is assumed to be constant. Adding up these two equations we get:
$$mg-f=(m+M)a$$
Now we need to figure out what $f$ is. You can easily find another equation by using the relation $ {\tau} = I \alpha$ for the magnitude of the torque acting on the cylinder ($\tau$), angular acceleration ($\alpha$) and moment of inertia ($I$). Observe that whatever angle the cylinder rotates through, the displacement of its center of mass will be its radius ($R$) times that angle. This means that $a_{cm} = a = R \alpha$, or equivalently $\alpha = \frac {a}{R}$. Also, the only contribution to the torque on the cylinder comes from friction, which is $\tau = Rf$. Substituting these two equations in $\tau = I \alpha$ gives:
$$Rf = I \frac {a}{R}$$
In other words:
$$f=I \frac a{R^2}$$
For a cylinder rotating around its axis, $I= \frac 12 MR^2$. Thus:
$$f=\frac 12 Ma$$
Substituting this into our third equation, we get:
$$mg - \frac 12 Ma = (m+M)a$$
Finally, solving for the acceleration gives:
$$a = \frac {m}{m+\frac 32 M}g$$
So the center of mass of the cylinder has constant acceleration, with the value above. 
I believe you can take it from here using standard equations for uniformly accelerating motion.
