Acceleration of cylinder both rotating and moving under force from a pulley The following states the problem 1.2 from the book Lagrangians And Hamiltonians by Patrick Hamill.

A cylinder of mass $M$ and radius $R$ is set on end on a table at a distance $L$ from the edge, as shown in Figure 1.11. As the string is wound tightly around the cylinder, the free end of the string passes over a friction-less pulley and hangs off the edge of the table. A weight of mass $m$ is attached to the free end of the string. Determine the time required for the spool to reach the edge of the table.

The given answer for acceleration of cylinder across table is $\frac{mg}{M+3m}$ (using which time is calculated). Can we explain intuitively why the term $M+3m$ appears in the solution without resorting to the dynamical equations of motion?

 A: If the cylinder starts from rest and the acceleration, a, is given, then L = (1/2)a$t^2$.
A: Denoting the magnitude of the acceleration due to gravity by $g$, the magnitude of the rightward acceleration of the cylinder $M$ with respect to (w.r.t) the inertial (ground) reference frame by $a$, the magnitude of the angular acceleration of the cylinder $M$ w.r.t. the inertial reference frame by $\alpha$ (where a positive angular acceleration vector has a direction vertically upward), the magnitude of the downward acceleration of the mass $m$ w.r.t. the inertial reference frame by $a_m$, the magnitude of the tension force in the string by $T$ and noting that the moment of inertia of the cylinder about the vertical axis passing through the center of mass is given as $\frac{1}{2}MR^2$, the kinematics and dynamics of the cylinder and mass imply that $$a_m=a+\alpha R,$$ $$T=Ma=m(g-a_m),$$ $$\alpha R=\frac{TR}{\frac{1}{2}MR^2} R=2 \frac{Ma}{M} = 2 \frac{m(g-a_m)}{M},$$ so that the relevant algebra yields $$3 a = a_m,$$ $$a=\frac{m}{M}(g-3a),$$ implying that $$a=\frac{mg}{M+3m}.$$ Since the acceleration $0<a$ (we know the sign of the quantity by using our physical intuition about the system) is a constant w.r.t. time (time-invariant), we can find the time $0<t$ required for the cylinder to cover the length $0<L$ via $L=\frac{1}{2}at^2$ so that $t=\sqrt{\frac{2L(M+3m)}{mg}}$.

An intuitive explanation of the solution is obtained from a work-energy theorem viewpoint of the phenomenon, without resorting to the dynamical equations of motion. Let us presume to know the kinematics $a_m=3a$. External work done on the system is purely due to the gravity and is equal to the work done the system at any instant of time as computed using the applied net forces on the two masses. and is given as $mg\cdot \frac{1}{2}a_m t^2=\frac{1}{2}Mat^2 + \frac{1}{2}ma_mt^2$ which is identical to the dynamical equation $a=\frac{m}{M}(g-3a)$ obtained above. This underscores the fact that the term $M+3m$ appears in the solution as a consequence of the kinematics.
