Newton's Second Law for Rotation — Question 2

Question 2

A falling mass must accelerate both itself and the pulley

A $2\,\mathrm{kg}$ mass hangs from a light cord wound around a fixed pulley of radius $0.20\,\mathrm m$ and inertia $0.040\,\mathrm{kg\,m^2}$. The cord does not slip. Neglect axle friction; use $g=9.8\,\mathrm{m/s^2}$.

1. Find the downward acceleration and cord tension.

2. Find angular acceleration and check the zero-inertia limit.

Question 2 - Solution

A falling mass must accelerate both itself and the pulley

(a) Derivation and calculation

Choose downward translation and the unwinding rotation as positive: $$mg-T=ma,\qquad TR=I\alpha,\qquad a=R\alpha.$$ Eliminate $T$ and $\alpha$: $$\boxed{a=\frac{mg}{m+I/R^2} =\frac{2(9.8)}{2+0.040/(0.20)^2}=6.533\ \mathrm{m/s^2}}.$$ $$\boxed{T=\frac{I}{R^2}a=6.533\ \mathrm N}.$$

(b) Result and interpretation

The pulley acceleration is $$\boxed{\alpha=\frac aR=\frac{6.533}{0.20}=32.67\ \mathrm{rad/s^2}}.$$ For the stated ideal unloaded pulley, $$I\to0\Rightarrow a\to g,\quad T\to0.$$ Using $T=mg$ would incorrectly assume zero acceleration of the hanging mass.

Physical check

The finite inertia reduces the mass acceleration below $g$ and requires a positive tension.