The Yo-Yo — Question 4

Question 4

A falling support can make the cord go slack

A yo-yo has $I=2Mr^2$. Its cord support accelerates downward at $A$. While the cord stays taut and does not slip, downward acceleration and unwinding angular acceleration obey $a-r\alpha=A$.

1. Derive $a$ and tension in terms of $A$ and $g$.

2. Assess the cases $A=g$ and $A>g$.

Question 4 - Solution

A falling support can make the cord go slack

(a) Derivation and calculation

Combine the stated moving-support constraint with the force and torque laws: $$Mg-T=Ma,\qquad Tr=I\alpha=2Mr(a-A).$$ Thus $T=2M(a-A)$, so $$g-a=2(a-A)\Rightarrow\boxed{a=(g+2A)/3},$$ $$\boxed{T=2M(g-A)/3}.$$

(b) Result and interpretation

Cord tension must be nonnegative: $$T\ge0\Longleftrightarrow A\le g.$$ At $A=g$, $\boxed{a=g,\ T=0,\ \alpha=0}$, the boundary of taut contact. For $A>g$, the formal solution demands $T<0$ and is inadmissible. The cord slackens; the yo-yo then has $a=g$ and torque-free spin until contact changes again.

Physical check

Setting $A=0$ recovers the fixed-support result $a=g/3$.