Torque And Angular Acceleration

This problem couples linear and rotational dynamics through the constraint that the string does not slip on the pulley, giving $a=R\alpha$. For the hanging block, Newton's second law reads $mg-T=ma$. For the pulley, the tension provides the only torque, so $TR=I\alpha =I\frac{a}{R}$, which rearranges to $T=\frac{Ia}{R^2}$. Substituting this tension into the block equation gives $mg=ma+\frac{Ia}{R^2}$, so $a=\frac{mg}{m+I/R^2}$. Here $\frac{I}{R^2}=\frac{0.1}{0.04}=2.5$, hence $a=\frac{2\times 10}{2+2.5}=\frac{20}{4.5}\approx 4.44{\text{ m/s}}^2$. The value 10 m/s${}^2$ ignores the pulley inertia, treating the block as in free fall. The value 5 m/s${}^2$ arbitrarily halves $g$. The value 8 m/s${}^2$ underestimates the effective inertia of the pulley. This follows NCERT's combined translation-rotation method. As a check, the pulley's inertia must reduce the acceleration below $g$, and $4.44<10$ confirms this. The term $I/R^2$ behaves like an effective extra mass added to the falling block, so a very light pulley with $I\to 0$ would recover free fall, whereas a heavy pulley would slow the descent further, neatly capturing how rotational inertia opposes the motion.