The Yo-Yo — Question 10

Question 10

An opposing axle torque can halt unwinding before free fall

A descending yo-yo has $M=0.20\,\mathrm{kg}$, $r=0.010\,\mathrm m$, $I=4\times10^{-5}\,\mathrm{kg\,m^2}$ and constant opposing axle torque $b=0.006\,\mathrm{N\,m}$. Use $g=10$ and a fixed no-slip cord.

1. Find downward acceleration and tension while unwinding.

2. Find the torque magnitude that would maintain constant descending speed.

Question 10 - Solution

An opposing axle torque can halt unwinding before free fall

(a) Derivation and calculation

During unwinding, the opposing axle torque reduces the net moment: $$Mg-T=Ma,\qquad Tr-b=I\alpha=Ia/r.$$ Eliminating $T$, $$\boxed{a=\frac{Mg-b/r}{M+I/r^2}=\frac{2-.600}{.20+.40}=\frac73\ \mathrm{m/s^2}},$$ $$\boxed{T=M(g-a)=.20(10-7/3)=\frac{23}{15}=1.533\ \mathrm N}.$$

(b) Result and interpretation

At constant descending speed, $a=\alpha=0$: $$T=Mg,\qquad Tr-b=0.$$ Therefore $$\boxed{b_{\mathrm{constant}}=Mgr=.20(10)(.010)=.0200\ \mathrm{N\,m}}.$$ A larger opposing torque decelerates existing descent; the behavior once rest is reached requires a static-torque model.

Physical check

At constant speed the resisting power $b\omega=Mgr(v/r)=Mgv$ removes the released gravitational power.