Rotational Kinetic Energy and Work

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SUMMARY

The discussion focuses on calculating the work done on a puck with a mass of 0.120 kg as it moves closer to the center of rotation on a frictionless table. The puck initially has a speed of 85.0 cm/s and is positioned 40.0 cm from the center. When the string is pulled down by 15.0 cm, the new radius becomes 25.0 cm (40.0 cm - 15.0 cm). The work done is determined by analyzing the change in kinetic energy of the puck as it moves from its original radius to the new radius.

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  • Understanding of rotational kinetic energy concepts
  • Familiarity with the work-energy principle
  • Basic knowledge of mass, velocity, and radius in physics
  • Ability to perform calculations involving kinetic energy
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p8-51.gif


The puck in Figure P8.51 has a mass of 0.120 kg. Its original distance from the center of rotation is 40.0 cm, and it moves with a speed of 85.0 cm/s. The string is pulled downward 15.0 cm through the hole in the frictionless table. Determine the work done on the puck. (Hint: Consider the change of kinetic energy of the puck.)

I tried doing it by finding the Kinetic Rotational Energy before and after but its not coming out right. For the KE after do I use the radius as 15, or is it 40-15?
 
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The puck ends up 15 cm closer to the center, so its new radius will be 40-15 cm.
 

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