Prove a spinning yo-yo has angular velocity

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SUMMARY

The discussion centers on demonstrating that a yo-yo spinning with angular velocity can be analyzed through the conservation of energy. The key equations involved are the potential energy (PE = mgh) and the rotational kinetic energy (Krot = 0.5 I ω²). The relationship between the height change and angular velocity is established as ω = √(gL/R²), confirming that the yo-yo's potential energy converts into rotational energy as it unwinds.

PREREQUISITES
  • Understanding of conservation of energy principles
  • Familiarity with rotational dynamics and angular velocity
  • Knowledge of potential energy and its calculation
  • Basic grasp of moment of inertia (I) in rotational motion
NEXT STEPS
  • Study the derivation of angular velocity formulas in rotational dynamics
  • Explore the concept of moment of inertia for various shapes
  • Learn about energy conservation in mechanical systems
  • Investigate the relationship between linear and angular motion
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Students in physics, particularly those studying mechanics, as well as educators looking to explain the principles of energy conservation and rotational motion.

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Homework Statement


Show, by conservation of energy, that when a length, L, of string has unwound that a
yo-yo is spinnning with angular velocity

Remember the yo-yo has both linear and rotational velocity

Homework Equations



[tex]\omega[/tex]= [tex]\sqrt{}[/tex]gL/R[tex]^{2}[/tex]


The Attempt at a Solution


I do not know where to start, any help would be appreciated.
 
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The yo-yo converts it's potential energy to rotational energy. So the change in potential potential energy is how much the center of mass has dropped, and by energy conservation that equals the kinetic energy of rotation.

so you need
PE=mgh
Krot=.5 I w^2
 
Mindscrape said:
The yo-yo converts it's potential energy to rotational energy. So the change in potential potential energy is how much the center of mass has dropped, and by energy conservation that equals the kinetic energy of rotation.

so you need
PE=mgh
Krot=.5 I w^2

Thank You for your reply.

So am i correct in thinking mgh = 0.5 I w^2
and where the change of height = [gl/r^2], which is the angular velocity, or am I way off the mark??
 

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