Angular speed at bottom of loop

AI Thread Summary
To determine the angular speed at the bottom of a frictionless vertical loop for a circular ring, conservation of energy principles and the relationship between the center of mass speed and angular velocity are essential. The moment of inertia for the ring is calculated as I = M R². The required angular speed for the ring to just barely make it around the top of the loop is found to be approximately 25.99 rad/s. The discussion emphasizes the importance of applying these physics concepts correctly to solve the problem. Understanding these principles is crucial for achieving the correct answer.
Paulie71199
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Homework Statement


A circular ring with a mass of 11 kg and radius of 0.66 meters is to roll without slipping about a vertical loop of radius 10 meters which is frictionless. If it is to just barely make it around the top of the loop, what must its angular speed be at the bottom of the loop in rad/s?


Homework Equations



The moment of inertia for the ring about it's center is I = M R2.

The Attempt at a Solution



The answer comes to be 25.99 rads/sec. Although I don't know how to get to that answer
 
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Welcome to PF!

Hi Paulie! Welcome to PF! :smile:

(have a theta: θ and try using the X2 icon just above the Reply box :wink:)

Use conservation of energy (together with the rolling equation relating speed of centre of mass to angular velocity, and putting hte normal reaction force equal to zero at the top). :smile:
 
Thank you for the welcome and the help!

Thanks for the tip lol.
 

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Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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