Newton's Second Law for Rotation — Question 3

Question 3

A massive Atwood pulley requires unequal tensions

Masses $3$ and $1\,\mathrm{kg}$ are connected over a fixed pulley with radius $0.20\,\mathrm m$ and inertia $0.080\,\mathrm{kg\,m^2}$. The light cord does not slip; axle friction is negligible. Use $g=9.8\,\mathrm{m/s^2}$.

1. Find the acceleration and the two tensions.

2. Explain why setting both tensions equal would contradict pulley dynamics.

Question 3 - Solution

A massive Atwood pulley requires unequal tensions

(a) Derivation and calculation

Take the heavier mass downward as positive and use $a=R\alpha$: $$m_hg-T_h=m_ha,\qquad T_l-m_lg=m_la,\qquad (T_h-T_l)R=I\frac aR.$$ $$\boxed{a=\frac{(m_h-m_l)g}{m_h+m_l+I/R^2} =\frac{2(9.8)}{3+1+2}=3.267\ \mathrm{m/s^2}}.$$ $$\boxed{T_h=3(9.8-3.267)=19.60\ \mathrm N},\qquad \boxed{T_l=1(9.8+3.267)=13.07\ \mathrm N}.$$

(b) Result and interpretation

The tension difference supplies the required pulley moment: $$(T_h-T_l)R=(19.60-13.0667)(0.20)=1.3067\ \mathrm{N\,m} =I\frac aR.$$ If $T_h=T_l$, then $I\alpha=0$, contradicting the nonzero no-slip acceleration for $I>0$.

Physical check

The pulley adds an effective translational inertia $I/R^2=2$ kg to the denominator.