Torque And Angular Acceleration

Question

A block of mass 2 kg hangs from a light string wound around a pulley of moment of inertia 0.1 kg m^2 and radius 0.2 m. Taking g as 10 m/s^2, what is the downward acceleration of the block?

RankGuru · Accepted Answer

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 $a = \frac{2 	imes 10}{2 + 2.5} = \frac{20}{4.5} \approx 4.44	ext{ m/s}^2$. 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 $I 	o 0$ would recover free fall, whereas a heavy pulley would slow the descent further, neatly capturing how rotational inertia opposes the motion.

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