Combined Translation And Rotation

As the cylinder falls, the string tension both opposes gravity and exerts a torque that spins the cylinder, with the no-slip condition linking the two motions through $a=R\alpha$. Newton's second law along the vertical direction is $Mg-T=Ma$, while the torque equation about the centre is $TR=I\alpha =\frac{1}{2}M{R}^2\cdot \frac{a}{R}$, giving $T=\frac{1}{2}Ma$. Combining the two relations, $Mg-\frac{1}{2}Ma=Ma$, which leads to $g=\frac{3}{2}a$, hence $a=\frac{2g}{3}=\frac{2\times 9.8}{3}\approx 6.53{\text{ m/s}}^2$. The value 9.8 m/s${}^2$ ignores the tension and treats the cylinder as in free fall. The value 4.9 m/s${}^2$ halves $g$ without justification. The value 3.27 m/s${}^2$ corresponds to using the wrong inertia factor. This is the classic NCERT unwinding-cylinder problem. As a plausibility check, the acceleration must lie below $g$ because the tension retards the fall, and $6.53<9.8$ confirms it. Energetically, the descending cylinder converts gravitational potential energy into both translational and rotational kinetic energy, and since one-third of that energy is diverted into spin, only two-thirds drives the downward motion, fully consistent with the reduced acceleration of $\frac{2g}{3}$ obtained from the force analysis.