A rim unwinding from a string due to gravity.

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SUMMARY

The discussion focuses on a physics problem involving a hoop unwinding from a string due to gravity. The hoop has a radius of 8 cm and a mass of 0.18 kg, and the string is released after the hoop descends 75 cm. The conservation of energy principle is applied, leading to the equation mgh = (1/2)mv² + (1/2)Iω², where I represents the moment of inertia of the hoop. Participants highlight the need to correctly calculate I to find the angular speed and linear speed of the hoop's center.

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  • Understanding of conservation of energy principles in physics
  • Knowledge of moment of inertia calculations for a hoop
  • Familiarity with angular velocity and linear velocity relationships
  • Basic algebra and equation manipulation skills
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  • Learn how to derive angular speed from linear speed using the formula ω = v/r
  • Explore energy conservation problems involving rotational dynamics
  • Practice solving similar physics problems using conservation of energy
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Students studying physics, particularly those focusing on mechanics and rotational motion, as well as educators seeking to enhance their understanding of energy conservation in dynamic systems.

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Homework Statement




A string is wrapped around a small hoop several times. the radius is 8cm and mass 0.18 kg. the string is released from rest after the hoop has descended 75 cm what is the angular speed of the rotating hoop and the speed at its centre?

Homework Equations




mgh 1/2 Iw^2

The Attempt at a Solution



MGH energy is conserved so mgh=1/2Iw^2
i sub in and solve except this doesn't give the right answer.
 
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If I is the moment of inertia of hoop about its centre then,
intial energy = final energy
GPE = KE of translation + KE of rotation
 
mgh = (1/2)mv^{2} + (1/2)I\omega^{2}
 

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