The Yo-Yo — Question 2

Question 2

Descent timing measures rotational inertia

A yo-yo of mass $0.30\,\mathrm{kg}$ and axle radius $0.012\,\mathrm m$ descends $0.80\,\mathrm m$ from rest in $0.80\,\mathrm s$ on a fixed cord. Assume constant inertia, no slip and no losses; use $g=10$.

1. Infer its inertia.

2. Find tension and test whether a negative inferred inertia could be physical.

Question 2 - Solution

Descent timing measures rotational inertia

(a) Derivation and calculation

The measured constant-acceleration motion gives $$h=\tfrac12at^2\Rightarrow\boxed{a=\frac{2(.80)}{(.80)^2}=2.50\ \mathrm{m/s^2}}.$$ Rearrange $a=g/(1+I/Mr^2)$: $$\boxed{I=Mr^2\left(\frac ga-1\right) =.30(.012)^2(10/2.5-1)=1.296\times10^{-4}\ \mathrm{kg\,m^2}}.$$

(b) Result and interpretation

The tension is $$\boxed{T=M(g-a)=.30(10-2.5)=2.25\ \mathrm N}.$$ Physical inertia requires $$I\ge0\Longleftrightarrow g/a-1\ge0\Longleftrightarrow0<a\le g.$$ A negative inferred inertia would contradict the passive fixed-cord model or its measured inputs.

Physical check

The inferred ratio $I/(Mr^2)=3$ reproduces $a=g/4$ and a positive tension.