Conservation of Rotational Energy Question

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SUMMARY

The discussion focuses on calculating the angular speed and linear speed of a hoop with a radius of 8.00 cm and mass of 0.180 kg, released from rest after descending 95.0 cm. The conservation of energy principle is applied, utilizing equations for gravitational potential energy (U = mgh), translational kinetic energy (K_trans = 1/2mv^2), and rotational kinetic energy (K_rot = 1/2Iω^2). The challenge arises from having two variables, linear velocity (v) and angular speed (ω), in the same equation, necessitating the use of the relationship between linear and angular motion for rolling objects.

PREREQUISITES
  • Understanding of gravitational potential energy (U = mgh)
  • Familiarity with kinetic energy equations for both translational and rotational motion
  • Knowledge of the moment of inertia for a hoop (I = m(r^2))
  • Concept of rolling without slipping and its implications on linear and angular velocity
NEXT STEPS
  • Study the relationship between linear velocity and angular velocity for rolling objects
  • Learn how to derive equations from conservation of energy principles in rotational dynamics
  • Explore the moment of inertia for different shapes and their impact on rotational motion
  • Practice solving problems involving energy conservation in systems with both translational and rotational motion
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Students studying physics, particularly those focusing on mechanics, as well as educators seeking to enhance their understanding of energy conservation in rotational systems.

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Homework Statement


A string is wrapped several times around the rim of a small hoop with radius 8.00 and mass 0.180 . The free end of the string is held in place and the hoop is released from rest. After the hoop has descended 95.0 , calculate the angular speed and speed of its center.


Homework Equations


U = mgh
K_trans = 1/2mv^2
K_rot = 1/2Iω^2


The Attempt at a Solution


I set the starting point was at a height of .95m.
K_i + U_i = K_f + U_f
0 + mgh = 1/2mv^2+1/2Iω^2 + 0
mgh = 1/2mv^2 + 1/2m(r^2)(ω^2)
(.18)(9.8)(.95) = (.5)(.18)(v^2) + (.5)(.18)(.08^2)(ω^2)

But then I have two variables in the same equation and I'm not sure where to go. I know that the speed at the center is going to be equal to v (the linear velocity) and angular speed is ω.
 
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It is as if the hoop rolled down on the string. There is a relation between the velocity of translation and angular speed of rotation for the case "rolling without slipping". You can figure it out if you answer the question: what distance does the hoop travel down on the string while it turns one?

ehild
 

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