The Yo-Yo — Question 5

Question 5

A moving hand does work through the cord contact

A yo-yo has mass $M$, inertia $I$, downward center speed $v$, unwinding spin $\omega$ and tension $T$. Its hand support moves downward at speed $U$, with no slip: $v-r\omega=U$.

1. Find the instantaneous power of tension on the whole yo-yo.

2. Evaluate it for $T=2\,\mathrm N$ and a hand moving upward at $0.30\,\mathrm{m/s}$, and compare with a fixed hand.

Question 5 - Solution

A moving hand does work through the cord contact

(a) Derivation and calculation

Tension does negative translational work but positive unwinding rotational work: $$\mathcal P_{T,\mathrm{trans}}=-Tv,\qquad\mathcal P_{T,\mathrm{rot}}=(Tr)\omega.$$ Using $v-r\omega=U$, $$\boxed{\mathcal P_T=-Tv+Tr\omega=-T(v-r\omega)=-TU}.$$

(b) Result and interpretation

For upward hand motion, $U=-.30$ m/s. Therefore $$\boxed{\mathcal P_T=-(2)(-.30)=+.600\ \mathrm W}.$$ For a fixed hand, $$U=0\Rightarrow\boxed{\mathcal P_T=0}.$$ A fixed cord transfers energy between translation and rotation without changing their total through tension; a moving support can add or remove energy.

Physical check

The sign of support power reverses when hand velocity reverses while the cord remains taut.