The Yo-Yo — Question 7

Section 11-4 · Physics pages 1624–1625

Question 7 A sleeping yo-yo uses a different contact model A yo-yo spins with I = 8.0 × 10−5 kg m2 and angular speed 200 rad/s while its center hangs at rest from a frictionless axle bearing on a loop of string. Its mass is 0.10 kg and g = 9.8. Mg S ω bearing A loop supports the axle bearing while the rotor spins freely; the center remains stationary. (a) Find the total upward support force and spin energy. (b) Explain why the fixed-cord unwinding constraint cannot be imposed here.
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Question 7 - Solution A sleeping yo-yo uses a different contact model Mg S ω bearing (a) Derivation and calculation The center is in translational equilibrium: S −Mg = 0 ⇒S = .10(9.8) = .980 N . The spin energy is K = 1 2Iω2 = 1 2(8.0 × 10−5)(200)2 = 1.60 J . (b) Result and interpretation The frictionless bearing transmits no rotor torque: τC = 0 ⇒˙ω = 0, ˙K = 0. The rotor can rotate relative to the support, so the unwinding condition v = rω does not apply. Imposing it with v = 0 would incorrectly force ω = 0 and contradict the bearing model. Physical check An upward force can support the rotor weight without supplying a moment about its spin axis.

Question 7

A sleeping yo-yo uses a different contact model

A yo-yo spins with I=8.0×10−5kgm2I=8.0\times10^{-5}\,\mathrm{kg\,m^2} and angular speed 200rad/s200\,\mathrm{rad/s} while its center hangs at rest from a frictionless axle bearing on a loop of string. Its mass is 0.10kg0.10\,\mathrm{kg} and g=9.8g=9.8.

Diagram shown in the original SVG worksheet.

  1. Find the total upward support force and spin energy.

  2. Explain why the fixed-cord unwinding constraint cannot be imposed here.

Question 7 - Solution

A sleeping yo-yo uses a different contact model

Diagram shown in the original SVG worksheet.

(a) Derivation and calculation

The center is in translational equilibrium: S−Mg=0⇒S=.10(9.8)=.980N.S-Mg=0\Rightarrow\boxed{S=.10(9.8)=.980\ \mathrm N}. The spin energy is K=12Iω2=12(8.0×10−5)(200)2=1.60J.\boxed{K=\tfrac12I\omega^2=\tfrac12(8.0\times10^{-5})(200)^2=1.60\ \mathrm J}.

(b) Result and interpretation

The frictionless bearing transmits no rotor torque: τC=0⇒ω̇=0,K̇=0.\tau_C=0\Rightarrow\dot\omega=0,\qquad\dot K=0. The rotor can rotate relative to the support, so the unwinding condition v=rωv=r\omega does not apply. Imposing it with v=0v=0 would incorrectly force ω=0\omega=0 and contradict the bearing model.

Physical check

An upward force can support the rotor weight without supplying a moment about its spin axis.