What is the Angular Speed of a Rotating Hoop with Given Radius and Mass?

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SUMMARY

The discussion focuses on calculating the angular speed of a rotating hoop with a radius of 8.00 cm and mass of 0.180 kg, released from rest and descending 95.0 cm. The key equations referenced include the relationship between linear velocity and angular velocity, specifically vcm = R(ω), and the derived formula vcm = √((4/3)gh) for the center of mass speed. The challenge lies in determining the angular speed (ω) without having initial values for linear speed or angular speed, highlighting the importance of understanding moments of inertia in rotational dynamics.

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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 0.180kg. The free end of the string is held in place and the hoop is released from rest (the figure View Figure ). After the hoop has descended 95.0cm, calculate

the angular speed of the rotating hoop and the speed of its center.


Homework Equations





The Attempt at a Solution



First attempt was like this

I thought that KE is 0 at initial ,and thought that I could get angular speed this way, but that would just give me that angular speed is 0. :O

Then I tried a formula given by the book for a yoyo, which is vcm = sqrt (4/3gh)
that did not result, and then is another formula that vcm = R(omega)

I don't have omega, and don't have vcm. So I can't use it at this instance. :(
 
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The hoop is decending under the force of gravity by virtue of its weight, mg. At the same time, it is unraveling the string. It's vertical velocity is the same as the tangential velocity (speed) at it's radius. And then there is the rotational inertia to consider.

For moments of inertia and rotational motion equations, see -

http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html
 
i don't get it..so the angular speed of the rotating hoop w = ?

and

the angular speed of the speed of its center v = Square_Root_Of_((4/3)gh) ... ??
 

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