Angular speed of wrapper string

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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 after it descends 60.0 cm. The key equations identified include the conservation of energy, where the linear kinetic energy (KE) is expressed as (1/2)*m*v^2 and gravitational potential energy (PE) as mgh. The relationship between angular speed (ω) and linear speed (v) is also established, emphasizing the need to integrate these concepts for a complete solution.

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Homework Statement




A string is wrapped several times around the rim of a small hoop with radius 8.00 cm and mass 0.180 kg . The free end of the string is held in place and the hoop is released from rest (the figure ). After the hoop has descended 60.0 cm, calculate angular speed and speed at the center

Homework Equations


w= omega

w = w+ at
K = 1/2 I w^2 ?



The Attempt at a Solution



I'm really unsure how to approach this problem and don't exactly know where to start
 
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Use energy conservation. Add linear KE=(1/2)*m*v^2 and gravitational potential energy PE=mgh to your list of equations.
 

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