Fraction of energy dissipated when solid cylinder placed on table

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SUMMARY

The discussion focuses on the energy dissipation of a uniform solid cylinder when placed on a rough horizontal table after being set in motion. The key equations involved are the kinetic energy equations, 0.5mv² for translational motion and 0.5Iw² for rotational motion. The conclusion drawn is that two-thirds (2/3) of the initial kinetic energy is dissipated due to sliding friction before the cylinder transitions to smooth rolling. The principle of conservation of angular momentum is crucial in understanding the energy transformation during this process.

PREREQUISITES
  • Understanding of kinetic energy equations (0.5mv² and 0.5Iw²)
  • Knowledge of angular momentum conservation principles
  • Familiarity with the concepts of friction and rolling motion
  • Basic physics of rigid body dynamics
NEXT STEPS
  • Study the principles of conservation of angular momentum in detail
  • Explore the effects of friction on rolling motion in different materials
  • Learn about energy dissipation in mechanical systems
  • Investigate the transition from sliding to rolling motion in rigid bodies
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Physics students, mechanical engineers, and anyone interested in the dynamics of rigid bodies and energy dissipation in motion.

looli9
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A uniform solid cylinder is set spinning about its axis and is then gently placed, with the axis horizontal, on a rough horizontal table. What fraction of its initial kinetic energy is dissipated in sliding friction before the cylinder eventually rolls smoothly along the plane?

The only equations I think are needed are 0.5mv^2 and 0.5Iw^2, and I can write down equations for the total energy before and after, but cannot relate the two.

Apparently the answer is 2/3.

Thanks
 
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Think conservation of angular momentum with respect to the point of contact.
 

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