The Yo-Yo — Question 1

Question 1

An unwinding yo-yo falls more slowly than a dropped particle

A yo-yo of mass $0.20\,\mathrm{kg}$ has axle radius $r=0.010\,\mathrm m$ and inertia $I=4.0\times10^{-5}\,\mathrm{kg\,m^2}$. It descends from rest on a vertical fixed light cord without slipping. Use $g=9.8$.

1. Find acceleration and cord tension.

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

Question 1 - Solution

An unwinding yo-yo falls more slowly than a dropped particle

(a) Derivation and calculation

Choose downward translation and unwinding spin positive. Then $$Mg-T=Ma,\qquad Tr=I\alpha,\qquad a=r\alpha.$$ Eliminating $T,\alpha$ gives $$\boxed{a=\frac{Mg}{M+I/r^2}=\frac{.20(9.8)}{.20+.40}=\frac{49}{15}=3.267\ \mathrm{m/s^2}},$$ $$\boxed{T=M(g-a)=\frac{98}{75}=1.307\ \mathrm N}.$$

(b) Result and interpretation

The angular acceleration is $$\boxed{\alpha=a/r=3.267/.010=326.7\ \mathrm{rad/s^2}}.$$ For fixed $M,r$, the zero-inertia limit is $$I\to0:\qquad a\to g,\qquad T\to0.$$ The tension is below the weight because the center accelerates downward; rotational dynamics do not require the tension to balance gravity.

Physical check

The effective rotational mass $I/r^2=.40$ kg is twice the actual mass, making $a=g/3$.